AP-Level Calculus

A calculator that refuses to answer

Ask any calculator to evaluate f(x) = (x² - 9) / (x - 3) at x = 3 and it will refuse. The top becomes 9 - 9 = 0. The bottom becomes 3 - 3 = 0. And 0/0 is not a number, not because anyone forbade it but because it does not pin down an answer: 0 × 5 = 0 and 0 × 71 = 0 are equally true, so "the number you multiply 0 by to get 0" has no single value.

Now ask the same calculator for values near 3 instead of at 3.

x	f(x)
2.9	5.9
2.99	5.99
2.999	5.999
3	undefined
3.001	6.001
3.01	6.01
3.1	6.1

The hole in the middle of that table is real and it is permanent. No amount of squeezing will make f(3) exist. But every row on both sides is aiming at the same target, 6, and getting closer to it the closer x gets to 3. That target is what calculus calls the limit, and the whole subject is built on the observation that the target can be perfectly definite even when the value at the point is missing.

The algebra confirms it. Factor the top: x² - 9 = (x - 3)(x + 3). So for every x except 3,

f(x) = (x - 3)(x + 3) / (x - 3) = x + 3.

You are allowed to cancel that (x - 3) precisely because x is not 3. The limit process never lets x equal 3, so the cancellation is legal for every value the limit ever looks at. And x + 3 at 3 is 6.

Key idea: a limit describes where a function is headed as the input approaches a number, not what it does when the input arrives.

The notation, word by word

We write this as

lim(x → 3) f(x) = 6

and read it aloud as "the limit, as x approaches 3, of f of x, equals 6." Four separate claims are packed into that line, and it pays to separate them:

• x approaches 3 means x takes values nearer and nearer to 3 from both sides, and never equals 3.
• f(x) equals 6 in the limit means the outputs get and stay arbitrarily close to 6.
• The statement says nothing whatsoever about f(3). It may not exist. It may exist and be something else entirely.
• The equals sign is genuine: the limit is one specific number when it exists, not a range or an approximation.

Three functions, one limit

Here is the experiment that makes the point permanent. Compare three functions near x = 2.

Function	Value at 2	Limit as x approaches 2
f(x) = x + 1	3	3
g(x) = (x² - x - 2)/(x - 2)	undefined	3
h(x) = x + 1 for x not 2, and h(2) = 10	10	3

Three different situations at the point; the same limit at the point. For g, factor the top as (x - 2)(x + 1) and cancel. For h, somebody has simply reached in and moved one point of the graph up to height 10; the rest of the curve has no idea and keeps aiming at 3. Whatever the limit is measuring, it is not the value.

This is exactly the property we will need in the next module. The derivative is defined as a limit of quotients that are all of the form 0/0 at the point of interest, and if limits cared about the value at the point, the derivative could never have been defined at all.

Three ways a limit fails

One: the two sides disagree. Take f(x) = |x| / x. For any positive x this is x/x = 1. For any negative x it is -x/x = -1. Approach 0 from the right and the outputs sit at 1; approach from the left and they sit at −1. Nothing is broken, nothing blows up, and yet there is no single target. The limit does not exist.

Two: the outputs run away. Take f(x) = 1/x² near 0. At x = 0.1 the output is 100; at 0.01 it is 10,000; at 0.001 it is 1,000,000. The outputs do not approach a number, they exceed every number. We write lim(x → 0) 1/x² = ∞, but read that carefully: infinity is not a value the function reaches, and the sentence is shorthand for "the outputs grow without bound." The limit still does not exist as a number.

Three: the outputs never settle. Take f(x) = sin(1/x) near 0. As x shrinks, 1/x races off toward infinity, so the sine cycles between −1 and 1 faster and faster. In any interval around 0, no matter how small, the graph completes infinitely many full oscillations. There is no target. This one is worth remembering because it defeats the table method entirely, as we will see in a moment.

Berkeley's ghosts, and why anyone wanted a definition

Newton and Leibniz built calculus in the 1660s and 1670s using quantities that were somehow small enough to discard at the end of a calculation but not zero during it. In 1734 the philosopher George Berkeley published The Analyst, a genuinely funny attack on that reasoning, asking whether these vanishing increments were the ghosts of departed quantities. His complaint was not that the answers were wrong; everyone could see the answers were right. His complaint was that no one could say what the objects were.

The repair took roughly a century. Augustin-Louis Cauchy, lecturing at the Ecole Polytechnique and publishing Cours d'analyse in 1821, began stating limits in terms of quantities becoming smaller than any assignable amount. Karl Weierstrass and his students turned that into the definition below in the 1860s. What they produced does not mention anything infinitely small. It only mentions ordinary numbers and inequalities, which is why it settled the argument.

The definition, read as a promise

Here it is in words. To say that lim(x → a) f(x) = L means: for every tolerance you name, however small, I can name a closeness for x that keeps f(x) inside your tolerance.

In symbols, using the Greek letters everyone uses, epsilon for your tolerance and delta for my closeness:

for every ε > 0 there is a δ > 0 such that if 0 < |x - a| < δ then |f(x) - L| < ε

Take it apart:

• |f(x) - L| < ε is "the output is within epsilon of the target." Epsilon is the tolerance you demand.
• |x - a| < δ is "the input is within delta of a." Delta is how close I promise to hold the input.
• 0 < |x - a| is the small clause that carries the entire idea of this lesson: x is close to a but not equal to a. That is why the value at the point never enters.
• The order matters absolutely. You choose epsilon first; I must then produce a delta that works. If I had to choose delta first, the definition would say almost nothing.

Think about a shop cutting steel rods to 40 cm on a lathe whose dial sets the stock length. The customer says the finished rod must be within 0.5 mm of 40 cm. That is epsilon. The machinist has to answer: how accurately must I set the dial? That answer is delta. If for every tolerance the customer can name there is a dial accuracy that meets it, the process is in control. That is precisely what the limit statement asserts about a function.

Why this matters: the definition is written as a challenge and a response because that is the only honest way to say "as close as you like" without ever mentioning an infinitely small quantity.

Working one, slowly

Show that lim(x → 4) (2x - 1) = 7.

1. Write down what has to be made small: |f(x) - L| = |(2x - 1) - 7| = |2x - 8|.
2. Factor out the 2: |2x - 8| = 2|x - 4|. Notice that |x - 4| is exactly the quantity delta controls. This is the whole trick: get the expression into the form (something) times |x - a|.
3. We need 2|x - 4| < ε, which happens exactly when |x - 4| < ε/2.
4. So choose δ = ε/2. Now check it forward: if 0 < |x - 4| < ε/2, then |f(x) - 7| = 2|x - 4| < 2(ε/2) = ε. Done.

Try it with a number to feel it. If the customer demands ε = 0.01, the machinist sets δ = 0.005. Any x within 0.005 of 4 gives an output within 0.01 of 7. Check x = 4.004: f(4.004) = 7.008, and |7.008 - 7| = 0.008, comfortably under 0.01.

You will not be asked to produce these proofs for complicated functions in this course, and neither is an AP student. What matters is that you can read the definition and see it is not mystical. It is a guarantee about tolerances.

When a table lies to you

Numerical tables are how most people first meet limits, and they have a failure mode you should see once. Investigate lim(x → 0) sin(π/x) with the sensible-looking inputs 1, 0.5, 0.25, 0.1, 0.01:

x	π/x	sin(π/x)
1	π	0
0.5	2π	0
0.25	4π	0
0.1	10π	0
0.01	100π	0

Every row says 0. A student reports that the limit is 0. The student is wrong: at x = 2/5 = 0.4 the angle is 2.5π and the sine is 1, and at x = 2/7 the sine is −1. Those inputs exist arbitrarily close to 0 as well; the table simply did not happen to land on them, because every chosen x was a reciprocal of a whole number. The function oscillates forever and the limit does not exist.

The lesson is not that tables are useless. It is that a table is evidence, and algebra or a theorem is proof. Use the table to form a guess, then justify the guess some other way.

Common misconceptions

• "The limit is just the value, so plug in." Substitution works for a large and friendly class of functions, and Lesson 3 will name that class exactly: the continuous ones. It fails for every interesting case in calculus, all of which are 0/0.
• "0/0 equals 1, or equals 0." Neither. It is called an indeterminate form because the answer depends entirely on which functions produced the zeros. Near 3, (x² - 9)/(x - 3) tends to 6, while (x - 3)²/(x - 3) tends to 0 and (x - 3)/(x - 3)³ blows up. Same form, three different answers.
• "The function must get closer and closer without ever reaching L." Not required. The constant function f(x) = 5 sits exactly on its limit everywhere. Nothing in the definition forbids equality; it only requires closeness.
• "Writing the limit equals infinity means the limit exists and is infinity." No. It is a description of how the limit fails, and it is more informative than saying nothing, but infinity is not a number the function approaches.
• "Delta comes first." Epsilon is named first by the challenger; delta is your reply. Reversing them turns a strong statement into a nearly empty one.

Where this leaves us

A limit is the number a function's outputs approach as the input approaches a point, and it deliberately ignores what happens at the point, which is what makes it able to describe 0/0 situations that algebra cannot touch. Evaluate one by simplifying the expression when a common factor cancels, by reading a table or graph when it does not, and by remembering that a table is a hint rather than a proof. Limits fail in exactly three ways: the two sides disagree, the outputs grow without bound, or the outputs oscillate forever. The epsilon-delta definition exists because Berkeley was right that "infinitely small" explained nothing, and it works by turning the phrase "as close as you like" into a promise about tolerances that mentions only ordinary numbers. Every derivative and every integral in the rest of this course is a limit underneath.

Sources

1. OpenStax. (2016). 2.2 The limit of a function. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 2.5 The precise definition of a limit. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). The limit. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Limit of a function. en.wikipedia.org
5. Berkeley, G. (1734). The Analyst: A Discourse Addressed to an Infidel Mathematician. London: J. Tonson.

Four fractions that all read 0/0

Here are four limits. Every one of them produces 0/0 on direct substitution, so by the argument of the previous lesson every one of them is a genuine question rather than a computation. And every one of them needs a different tool.

• lim(x → 2) (x² - 4)/(x² - x - 2)
• lim(x → 0) (√(x + 9) - 3)/x
• lim(x → 3) (1/x - 1/3)/(x - 3)
• lim(x → 0) x² sin(1/x)

By the end of this lesson you will have all four, and the answers are 4/3, 1/6, −1/9 and 0. None of them is guessable from the form. That is the point: 0/0 is a question, and these are the standard ways of answering it.

The limit laws, and the string attached

Before the repairs, the routine machinery. If lim(x → a) f(x) = L and lim(x → a) g(x) = M, both existing as numbers, then limits pass straight through the arithmetic:

Law	Statement
Sum and difference	lim (f ± g) = L ± M
Constant multiple	lim (c f) = cL
Product	lim (f g) = L M
Quotient	lim (f/g) = L/M, provided M ≠ 0
Power and root	lim (f)² = L², and lim √f = √L when L ≥ 0

Two consequences do most of the everyday work. First, for any polynomial, substitution simply works: lim(x → 4) (3x² - 5x + 2) = 48 - 20 + 2 = 30, because the laws let you take the limit of each piece and the limit of x is 4. Second, for a rational function, substitution works whenever the denominator does not vanish. The quotient law carries the condition M ≠ 0, and every hard limit in calculus is hard exactly because that condition fails.

The point: the limit laws are not a technique. They are permission to compute normally, and the technique begins where the permission runs out.

Repair one: factor and cancel

Evaluate lim(x → 2) (x² - 4)/(x² - x - 2).

1. Substitute to confirm the form: top is 4 - 4 = 0, bottom is 4 - 2 - 2 = 0. It is 0/0, so the quotient law does not apply and we must rewrite.
2. The zeros are not an accident. If a polynomial vanishes at 2, then (x - 2) is a factor of it. Both do, so both have it.
3. Factor: top is (x - 2)(x + 2), bottom is (x - 2)(x + 1).
4. Cancel the common factor, legal because x never equals 2 during a limit: the expression is (x + 2)/(x + 1).
5. Now the denominator limit is 3 ≠ 0, so substitute: 4/3.

The structural fact worth carrying away: in a 0/0 rational limit at x = a, the factor (x - a) is guaranteed to be sitting in both the top and the bottom, waiting to be cancelled. You are never searching blindly.

Repair two: multiply by the conjugate

Evaluate lim(x → 0) (√(x + 9) - 3)/x. Substitution gives (3 - 3)/0 = 0/0. Nothing factors, because of the root. The trick is to multiply top and bottom by the conjugate of the top, which is the same expression with the sign flipped, √(x + 9) + 3. Multiplying by that over itself is multiplying by 1, so nothing changes except the appearance.

1. The top becomes a difference of squares: (√(x + 9) - 3)(√(x + 9) + 3) = (x + 9) - 9 = x. The root is gone, which was the entire objective.
2. The bottom becomes x(√(x + 9) + 3).
3. Cancel the x: 1/(√(x + 9) + 3).
4. Substitute now: 1/(3 + 3) = 1/6.

Check it numerically to be sure. At x = 0.01, √9.01 ≈ 3.0016662, so the quotient is about 0.0016662/0.01 = 0.16662, and 1/6 = 0.1666.... The algebra and the arithmetic agree.

Repair three: clear the compound fraction

Evaluate lim(x → 3) (1/x - 1/3)/(x - 3). Again 0/0. The obstacle is a fraction inside a fraction, so combine the top over a common denominator first.

1. Top: 1/x - 1/3 = (3 - x)/(3x).
2. The whole expression is now [(3 - x)/(3x)] ÷ (x - 3), which is (3 - x) / (3x(x - 3)).
3. Notice 3 - x = -(x - 3). Substitute that and cancel: -(x - 3) / (3x(x - 3)) = -1/(3x).
4. Substitute: -1/9.

That sign flip catches almost everyone once. When you see (3 - x) against (x - 3), factor out the −1 deliberately and write it down rather than doing it in your head. Keep this example in mind: in Module 2 it turns out to be the derivative of 1/x at x = 3, and the answer -1/9 will reappear as -1/x² evaluated at 3.

One side at a time

Some functions behave differently depending on which direction you come from, so we need notation for direction. lim(x → 2−) f(x) means the limit as x approaches 2 from the left, through values smaller than 2. lim(x → 2+) f(x) means from the right, through larger values. The connecting theorem is short and does all the work:

The two-sided limit exists and equals L exactly when both one-sided limits exist and both equal L.

Consider a delivery charge that is 6 dollars for any parcel up to and including 2 kilograms and 4 dollars per kilogram after that:

C(w) = 6 for 0 < w ≤ 2, and C(w) = 4w for w > 2.

Approaching 2 from below, the cost is 6 the whole way, so the left limit is 6. Approaching from above, 4w heads to 8, so the right limit is 8. The two disagree, so lim(w → 2) C(w) does not exist, and the graph has a jump of 2 dollars at that weight. A customer with a 2.001 kilogram parcel pays 8.004 while one with a 2.000 kilogram parcel pays 6. Jump discontinuities are not pathological curiosities; they are how tariffs, tax brackets and postage are written.

Solving for the constant that removes the jump

The standard exam version of this runs backward. Find k so that lim(x → 1) f(x) exists, where f(x) = x² + k for x < 1 and f(x) = 3x - 2 for x ≥ 1.

1. Left limit: as x approaches 1 from below we use the first rule, giving 1 + k.
2. Right limit: from above we use the second rule, giving 3 - 2 = 1.
3. For the limit to exist the two must be equal: 1 + k = 1, so k = 0.

Check: with k = 0 the left piece is x², which approaches 1, matching the right piece. The limit is 1. Any other k leaves a jump.

Out at infinity, where the graph flattens or runs

A different question: not what happens near a point, but what happens far from every point. lim(x → ∞) f(x) = L means the outputs settle toward L as x grows without bound, which is the definition of a horizontal asymptote at y = L.

The reliable method for a rational function is to divide every term by the highest power of x in the denominator. Evaluate lim(x → ∞) (3x² - 5x + 1)/(7x² + 2):

1. Highest power below is x². Divide every term by it: (3 - 5/x + 1/x²) / (7 + 2/x²).
2. Each term of the form c/x² or c/x tends to 0 as x grows, because a fixed numerator over a runaway denominator shrinks.
3. What is left is 3/7.

The same method applied to three shapes gives a rule you can quote. Compare the degree of the top with the degree of the bottom:

Situation	Example	Limit at infinity
Bottom degree wins	(2x + 1)/(x² - 3)	0
Degrees tie	(3x² - 5x + 1)/(7x² + 2)	ratio of leading coefficients, 3/7
Top degree wins	x³/(x² + 1)	grows without bound, no horizontal asymptote

One trap deserves its own line, because it is the most commonly missed limit in this topic. Evaluate lim(x → -∞) √(4x² + 3)/x. Pulling x² out of the root gives √(x²) √(4 + 3/x²), and √(x²) is |x|, not x. When x is negative, |x| = -x. So the expression is -x √(4 + 3/x²) / x = -√(4 + 3/x²), which tends to −2. Heading to positive infinity the same function tends to +2. One curve, two horizontal asymptotes.

The squeeze theorem, and the fourth fraction

The last of the four opening limits, lim(x → 0) x² sin(1/x), resists every algebraic repair. The sin(1/x) factor oscillates forever, exactly as in the previous lesson, so no cancelling and no conjugate will help. What saves it is that the oscillation is trapped.

The squeeze theorem says: if g(x) ≤ f(x) ≤ h(x) near a, and g and h both have limit L at a, then f is forced to have limit L too. There is nowhere else for it to go.

1. Sine of anything lies between −1 and 1: -1 ≤ sin(1/x) ≤ 1 for every x except 0.
2. Multiply through by x², which is positive, so the inequalities keep their direction: -x² ≤ x² sin(1/x) ≤ x².
3. Both -x² and x² tend to 0 as x tends to 0.
4. The function is squeezed between two things heading to 0, so it heads to 0.

The same theorem proves the one limit you will need most in this whole course, the trigonometric limit lim(θ → 0) (sin θ)/θ = 1, where the angle is in radians. The numbers make it believable: at θ = 0.1 the quotient is 0.998334, at 0.01 it is 0.999983, at 0.001 it is 0.9999998. Module 3 uses this limit to produce the derivative of sine, and everything trigonometric in calculus rests on it.

Common misconceptions

• "Cancelling changes the function, so it is cheating." It does change the function at exactly one point, and that is the one point a limit never looks at. The cancelled form and the original agree everywhere the limit is evaluated.
• "If substitution gives 0/0, the limit does not exist." It means the question is still open. Three of this lesson's four opening limits are 0/0 and all three exist.
• "If substitution gives a nonzero number over 0, the limit is infinity." It is unbounded, but check the sides. For 1/(x - 2) at 2, the left side runs to negative infinity and the right side to positive infinity, so there is not even a consistent infinite behaviour.
• "The square root of x squared is x." It is |x|. Ignoring that is what turns the −2 asymptote above into a wrong answer of +2.
• "A function cannot cross its horizontal asymptote." It can, as often as it likes. (sin x)/x crosses y = 0 infinitely many times on the way out and still has that horizontal asymptote.

Putting it together

Start every limit by substituting, because if the denominator does not vanish the limit laws finish the job in one line. When substitution gives 0/0, choose the repair that matches the obstacle: factor and cancel for polynomials, multiply by the conjugate to kill a root, combine over a common denominator to clear a fraction inside a fraction. Direction matters at a break, so compute one-sided limits separately and remember that the two-sided limit exists only when they agree, which is also how you solve for a constant that makes a piecewise function fit together. For behaviour far out, divide by the highest power in the denominator and compare degrees, watching for the absolute value that appears when a square root meets a negative x. When the function oscillates but is trapped between two curves that agree in the limit, the squeeze theorem finishes it, and it is the squeeze theorem that will hand you (sin θ)/θ → 1 when you need it.

Sources

1. OpenStax. (2016). 2.3 The limit laws. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 4.6 Limits at infinity and asymptotes. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). One-sided limits. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. MIT OpenCourseWare. (2010). 18.01SC Single Variable Calculus. Massachusetts Institute of Technology. ocw.mit.edu

The year someone bothered to prove the obvious

In 1817, in Prague, Bernard Bolzano published a pamphlet with a title longer than some of its proofs: a purely analytic proof of the theorem that between any two values giving results of opposite sign there lies at least one real root of the equation. He was proving something every draughtsman already believed. If a curve is below the axis here and above it there, and you drew it without lifting the pen, it crossed.

Bolzano's point was that "without lifting the pen" is a statement about hands, not about functions, and that the theorem is false for functions that do lift the pen. This lesson supplies the missing definition, which is continuity, and then collects the payoff, which is that continuous functions can be trusted to do several things nothing else can.

Continuity at a point has three conditions, not one

A function f is continuous at a when all three of these hold:

1. f(a) exists. The point is on the graph.
2. lim(x → a) f(x) exists. The neighbours agree on a target.
3. lim(x → a) f(x) = f(a). The target is the point.

Three conditions means three ways to fail, and the classification of discontinuities is nothing more than which condition broke first.

Type	Example at the marked point	Condition that fails
Removable	(x² - 9)/(x - 3) at x = 3	Condition 1: the limit is 6 but there is no value
Jump	the parcel charge from the last lesson at w = 2	Condition 2: the one-sided limits are 6 and 8
Infinite	1/(x - 2) at x = 2	Condition 2: the outputs are unbounded
Removable, second flavour	h(x) = x + 1 except h(2) = 10	Condition 3: the value and the limit disagree

Removable is the interesting word. It means the damage is confined to a single point and could be repaired by redefining the function there: setting f(3) = 6 in the first row produces a function continuous everywhere. Nothing you can do to one point will repair a jump, because the trouble is on both sides of it.

In short: continuity at a point says the limit and the value exist and match. Everything else in this lesson is a consequence.

The list that makes substitution legal

Here is the theorem that quietly justified every "just substitute" in the previous lesson. Each of these is continuous everywhere on its natural domain:

• Polynomials, on all real numbers.
• Rational functions, everywhere the denominator is not zero.
• Root functions, on their domain (so √x is continuous for x ≥ 0).
• Sine and cosine, on all real numbers; tangent, everywhere except the odd multiples of π/2.
• Exponential functions such as 2^x and e^x, on all real numbers; logarithms, for positive inputs.

And continuity survives construction: sums, differences, products, quotients (where the denominator is nonzero) and compositions of continuous functions are continuous. So f(x) = √(x² + 1) / (x - 4) is continuous everywhere except x = 4, and you know that without computing a single limit. This is why the practical algorithm for a limit is: substitute first, and only reach for a repair when substitution fails.

Worked example: find and classify every break

Let f(x) = (x² - 1)/(x² - 3x + 2). Where is it discontinuous, and of what kind?

1. Discontinuities of a rational function can only occur where the denominator is zero. Factor it: x² - 3x + 2 = (x - 1)(x - 2). So the candidates are 1 and 2.
2. Factor the top too: x² - 1 = (x - 1)(x + 1).
3. The factor (x - 1) appears above and below, so it cancels: for x other than 1 and 2, f(x) = (x + 1)/(x - 2).
4. At x = 1: the limit is (1 + 1)/(1 - 2) = -2, a perfectly good number, but f(1) is undefined. Removable. Define f(1) = -2 and the hole is filled.
5. At x = 2: nothing cancels, the top is 3 and the bottom is 0, so the outputs are unbounded. Approaching from the right the denominator is a small positive, so f runs to positive infinity; from the left it runs to negative infinity. Infinite discontinuity, and there is a vertical asymptote at x = 2.

The general principle: a factor that cancels leaves a hole, a factor that survives in the denominator leaves an asymptote.

Fitting the pieces together

Piecewise functions are where continuity gets solved for rather than checked. Find a and b so that this function is continuous everywhere:

f(x) = x² for x ≤ 2, f(x) = ax + b for 2 < x < 5, f(x) = 13 for x ≥ 5.

1. Each piece is continuous on its own interval, so only the seams at 2 and 5 can fail.
2. At the seam x = 2: the left value is 2² = 4, the right limit is 2a + b. Continuity demands 2a + b = 4.
3. At the seam x = 5: the left limit is 5a + b, the right value is 13. Continuity demands 5a + b = 13.
4. Subtract the first equation from the second: 3a = 9, so a = 3. Then b = 4 - 6 = -2.

Check the answer, always. With a = 3, b = -2 the middle piece is 3x - 2, which gives 4 at x = 2 and 13 at x = 5. Both seams close.

The intermediate value theorem, hypotheses first

If f is continuous on the closed interval [a, b], and N is any number between f(a) and f(b), then there is at least one c in [a, b] with f(c) = N.

Read the hypotheses as seriously as the conclusion, because they are doing real work. The interval must be closed, so the endpoints belong. The function must be continuous on all of it, not merely at most of it. And the conclusion is deliberately weak in two ways: it says at least one c, not exactly one, and it gives no method for finding c. It is a pure existence theorem, which is exactly what Bolzano set out to establish.

Break continuity and the theorem dies immediately. Take f(x) = 1/x on the interval from −1 to 1. It is −1 at the left end and +1 at the right end, so 0 lies between them, and yet f(x) is never 0 anywhere. There is no contradiction, because f is not continuous on that interval: it is not even defined at 0. The hypothesis was not decoration.

Using it, then chasing the root down

Show that x³ - x - 1 = 0 has a solution between 1 and 2, then find it to one decimal place.

1. Let g(x) = x³ - x - 1. It is a polynomial, so it is continuous on every interval, including [1, 2]. Hypothesis satisfied.
2. g(1) = 1 - 1 - 1 = -1, and g(2) = 8 - 2 - 1 = 5. Zero lies between −1 and 5.
3. By the intermediate value theorem there is some c in [1, 2] with g(c) = 0. A root exists.

Existence proved, now hunt. The bisection method is nothing but the theorem applied over and over to half of the previous interval.

Midpoint	g(midpoint)	Sign	New interval
1.5	3.375 - 1.5 - 1 = 0.875	positive	[1, 1.5]
1.25	1.953125 - 1.25 - 1 = -0.296875	negative	[1.25, 1.5]
1.375	2.599609 - 1.375 - 1 = 0.224609	positive	[1.25, 1.375]
1.3125	2.261353 - 1.3125 - 1 = -0.051147	negative	[1.3125, 1.375]

Four halvings and the root is pinned between 1.3125 and 1.375, so to one decimal place it is 1.3. The true value is about 1.32472, a number called the plastic ratio, and each further bisection buys one more binary digit of it. Notice what the machine is doing: it never solves the cubic. It only ever checks a sign and applies Bolzano's theorem again.

The other theorem that needs a closed interval

One companion result, because Module 5 will lean on it hard. The extreme value theorem says a function continuous on a closed interval [a, b] attains an absolute maximum and an absolute minimum somewhere on that interval. Both hypotheses matter again: f(x) = x on the open interval from 0 to 1 has no maximum, since for any candidate you can name a larger value still below 1, and f(x) = 1/x on the half-open interval from 0 to 1 has no maximum for a different reason. When you are told to optimise something on a closed interval, this theorem is the reason a best answer is guaranteed to be there to find.

Common misconceptions

• "Continuous means you can draw it without lifting your pen." A serviceable picture, and it is where Bolzano started, but it is not a definition. The three conditions are, and they are what let you decide the question for a function you cannot draw.
• "A function with a vertical asymptote is discontinuous there, so it is not continuous." Careful with the phrasing. 1/x is continuous at every point of its domain. It is not defined at 0, so 0 is not a point where continuity could hold or fail in the ordinary sense; what we say is that it has an infinite discontinuity at 0 and is continuous on each side.
• "The intermediate value theorem finds the root." It proves one exists. Bisection, or Newton's method later, does the finding.
• "If f(a) and f(b) have the same sign, there is no root between them." False. x² - 4 is positive at −3 and positive at 3, with two roots in between. The theorem gives a sufficient condition, not a necessary one.
• "Removable means unimportant." A removable discontinuity is precisely where a formula has been algebraically simplified past a hole. Every derivative you compute will start life as a function with exactly this kind of hole at the point of interest.

The short version

Continuity at a point means the value exists, the limit exists, and they are the same number, and the three ways a function can break correspond exactly to the three ways that can fail. Polynomials, rational functions off their zeros, roots, sines, cosines, exponentials and logarithms are continuous on their domains, and so is anything built from them by adding, multiplying, dividing or composing, which is the theorem that makes substitution a legal first move on any limit. In a rational function, a cancelling factor leaves a removable hole and a surviving one leaves a vertical asymptote. Continuity is what the intermediate value theorem requires and what it is really about: on a closed interval a continuous function hits every value between its endpoint values, which proves roots exist and, applied repeatedly, lets bisection squeeze one down to any accuracy you like. The extreme value theorem, resting on the same hypotheses, guarantees the maximum and minimum you will go looking for in Module 5.

Sources

1. OpenStax. (2016). 2.4 Continuity. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Continuity. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Wikipedia. (n.d.). Intermediate value theorem. en.wikipedia.org
4. Wikipedia. (n.d.). Continuous function. en.wikipedia.org
5. Bolzano, B. (1817). Rein analytischer Beweis des Lehrsatzes, dass zwischen je zwey Werthen, die ein entgegengesetztes Resultat gewaehren, wenigstens eine reelle Wurzel der Gleichung liege. Prague: Gottlieb Haase.

How fast is it falling at exactly two seconds?

Drop a stone from a high ledge. Ignoring air resistance, it falls s(t) = 4.9t² metres in t seconds, which is Galileo's result written with the modern value of gravity. After 2 seconds it has fallen 19.6 m; after 3 seconds, 44.1 m. So over that second its average speed was (44.1 - 19.6)/1 = 24.5 metres per second.

But that is the average over a whole second, and the stone is speeding up the entire time. What we want is the speedometer reading at the instant t = 2. Shrink the interval and watch.

Interval	Distance fallen	Average speed (m/s)
2 to 3 s	44.1 - 19.6 = 24.5 m	24.5
2 to 2.1 s	21.609 - 19.6 = 2.009 m	20.09
2 to 2.01 s	19.79649 - 19.6 = 0.19649 m	19.649
2 to 2.001 s	19.6196049 - 19.6 = 0.0196049 m	19.6049

The numbers are converging on 19.6 m/s. But look at what happens if you try to reach the instant directly by setting the interval length to zero: the distance fallen is 0 and the time elapsed is 0, and the average speed is 0/0. The instantaneous question is a 0/0 question. That is why calculus needed limits before it could have derivatives, and why we spent three lessons on limits first.

Key idea: instantaneous rate of change is the limit of average rates of change over intervals shrinking to nothing.

The difference quotient

Write the shrinking interval as [a, a + h], where h is the small step. The average rate of change of f across it is

(f(a + h) - f(a)) / h,

which is called the difference quotient. Every one of the numbers in that table is a difference quotient with a = 2 and h equal to 1, 0.1, 0.01, 0.001 in turn. The derivative of f at a is what those approach:

f'(a) = lim(h → 0) (f(a + h) - f(a)) / h

read aloud as "f prime of a". There is an equivalent form that is sometimes more convenient, obtained by writing x = a + h:

f'(a) = lim(x → a) (f(x) - f(a)) / (x - a)

Both say the same thing. If the limit exists, f is differentiable at a; if it does not, f has no derivative there.

Three derivatives from the definition, worked out

Nobody computes derivatives this way for long, but you have to do it a few times to believe the shortcuts of the next lesson.

One: f(x) = x².

1. Build the top: f(a + h) - f(a) = (a + h)² - a².
2. Expand: (a² + 2ah + h²) - a² = 2ah + h². The a² terms cancel, which is what always happens and is what makes this workable.
3. Divide by h: (2ah + h²)/h = 2a + h. The division is legal because the limit never lets h be 0.
4. Take the limit as h → 0: 2a.

So f'(a) = 2a. Test it against the falling stone: s(t) = 4.9t² has derivative 9.8t by the same computation with a factor of 4.9 riding along, and at t = 2 that is 19.6 m/s, exactly the number the table was aiming at.

Two: f(x) = 1/x.

1. Top: 1/(a + h) - 1/a. Combine over the common denominator a(a + h): (a - (a + h)) / (a(a + h)) = -h / (a(a + h)).
2. Divide by h, which means multiply by 1/h: -1 / (a(a + h)).
3. Let h → 0: -1/a².

You have already met this answer. In Lesson 2 we evaluated lim(x → 3) (1/x - 1/3)/(x - 3) and got −1/9. That limit was the derivative of 1/x at 3, and -1/3² = -1/9 agrees. The limit exercises of Module 1 were derivatives in disguise the whole time.

Three: f(x) = √x.

1. Top: √(a + h) - √a. There is a root, so use the conjugate: multiply top and bottom by √(a + h) + √a.
2. The top becomes (a + h) - a = h.
3. The whole quotient is now h / (h(√(a + h) + √a)), and the h cancels: 1/(√(a + h) + √a).
4. Let h → 0: 1/(2√a).

Sanity check the shape of that answer. At a = 0.01 it gives 1/(2 × 0.1) = 5, a steep slope, and at a = 100 it gives 1/20 = 0.05, nearly flat. That matches the square root graph, which rises sharply near the origin and flattens out. An answer whose behaviour matches the picture is probably right.

Two readings of one number

The difference quotient (f(a + h) - f(a))/h is rise over run between the points (a, f(a)) and (a + h, f(a + h)), so it is the slope of the secant line through them. As h shrinks, the second point slides along the curve toward the first, and the secant pivots. Its limiting position is the tangent line. So:

• Geometrically, f'(a) is the slope of the curve at that point, meaning the slope of the tangent line there.
• Physically, f'(a) is the instantaneous rate at which the output changes per unit of input.

These are not two facts. They are one number described twice, and switching between the descriptions is most of the skill in applying calculus.

Units, which are half the answer

A derivative always carries the units of output divided by the units of input, and reading them off tells you what the number means.

Function	Derivative	Meaning of the derivative
s(t), metres at time t seconds	metres per second	velocity
v(t), metres per second	metres per second per second	acceleration
C(q), dollars to make q units	dollars per unit	marginal cost, the cost of one more unit
P(t), bacteria at time t hours	bacteria per hour	growth rate

Suppose a workshop's cost function satisfies C(200) = 4800 dollars and C'(200) = 17 dollars per unit. That second statement is the useful one for a manager: the 201st unit will cost about 17 dollars to make. Not exactly 17, because the derivative is the rate at the instant rather than over the next whole unit, but close enough to price a decision on.

Four notations, because two people invented this

Newton and Leibniz developed calculus independently, and their notations both survived.

• f'(x) and y', the Lagrange notation, compact and good for stating rules.
• dy/dx, the Leibniz notation, which keeps the units visible and is the reason the chain rule looks obvious later.
• d/dx [f(x)], the same thing written as an instruction: differentiate what follows with respect to x.
• f'(a) or dy/dx evaluated at x = a, for the number at one specific point.

One caution about dy/dx: it is not a fraction. It is a single symbol for a limit. It happens to behave like a fraction in several important ways, which is a real theorem and not a coincidence, but treating it as literal division will eventually mislead you.

The tangent line, written down

Find the equation of the tangent to y = x² at the point where x = 3.

1. The point: y = 3² = 9, so the point is (3, 9).
2. The slope: f'(x) = 2x, so f'(3) = 6.
3. Point-slope form: y - 9 = 6(x - 3).
4. Simplify: y = 6x - 9.

Check that the line touches rather than crosses. At x = 3.1 the parabola is at 9.61 and the line is at 9.6: the gap is 0.01. At x = 3.01 the parabola is at 9.0601 and the line is at 9.06, a gap of 0.0001. The gap shrinks like the square of the step, which is what "tangent" means in practice, and which is exactly why linear approximation works.

Where the derivative does not exist

Differentiability is a stronger demand than continuity, and there are four standard ways it fails.

Failure	Example	What goes wrong
Corner	f(x) = |x| at 0	The difference quotient is −1 from the left and +1 from the right, so the limit does not exist
Cusp	f(x) = x to the power 2/3, at 0	Slopes run to negative infinity on one side and positive infinity on the other
Vertical tangent	f(x) = x to the power 1/3, at 0	Slopes run to positive infinity from both sides; the tangent line is vertical and has no slope
Discontinuity	any jump, such as the parcel charge at 2 kg	The numerator of the difference quotient does not go to 0, so the quotient blows up

That last row is a special case of a theorem worth memorising: if f is differentiable at a, then f is continuous at a. The converse is false, and |x| at 0 is the standard counterexample: continuous, with a perfectly good value, and no derivative. Differentiability is smoothness, which is more than unbrokenness.

Common misconceptions

• "The derivative is the slope of the graph, so I can just look." Looking gives you the sign and rough size, which is worth having. The number requires the limit or a rule derived from it.
• "You cannot divide by h, since h goes to zero." During the limit h is never zero, so the division is legal at every step. Only afterwards do we ask what the simplified expression approaches.
• "dy/dx is dy divided by dx." It is one symbol. Its fraction-like behaviour is a theorem, not a definition, and there is no separate number called dy in this course.
• "Continuous means differentiable." Backwards. Differentiable implies continuous; the absolute value function is continuous everywhere and not differentiable at 0.
• "The tangent line touches the curve at only one point." A tangent line at one point may cross the curve elsewhere, and for y = x³ at the origin it even crosses at the point of tangency. Tangency is a local statement about slope, not a global statement about intersections.

What to carry forward

The derivative is the limit of the difference quotient (f(a + h) - f(a))/h as h goes to 0, and it exists exactly when that 0/0 limit exists. Computed from the definition it gives 2a for x², -1/a² for 1/x and 1/(2√a) for √x, each by one of the algebraic repairs from Module 1. The same number is both the slope of the tangent line and the instantaneous rate of change, carrying units of output per unit of input, which is what makes it the right tool for velocity, marginal cost and growth rate alike. To write a tangent line you need only the point and that slope. And differentiability is stronger than continuity: corners, cusps, vertical tangents and jumps all kill the derivative while leaving the function otherwise well behaved.

Sources

1. OpenStax. (2016). 3.1 Defining the derivative. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 3.2 The derivative as a function. In Calculus Volume 1. openstax.org
3. OpenStax. (2016). 3.4 Derivatives as rates of change. In Calculus Volume 1. openstax.org
4. Dawkins, P. (n.d.). The definition of the derivative. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Wikipedia. (n.d.). Derivative. en.wikipedia.org

The rule everyone tries first, and why it fails

Take f(x) = x and g(x) = x². Their derivatives are 1 and 2x. If the derivative of a product were the product of the derivatives, then the derivative of f(x)g(x) would be 1 × 2x = 2x. But f(x)g(x) = x³, and we can check its derivative another way: it is 3x². At x = 2 the guess says 4 and the truth says 12. The guess is not slightly off; it is a different function.

Derivatives pass cleanly through addition and refuse to pass through multiplication, and this lesson is the set of rules that say exactly what happens instead. Each one can be proved from the limit definition, and each one exists so that you never have to use the limit definition again.

The power rule

d/dx [xⁿ] = n xⁿ⁻¹, or in plain words: bring the exponent down in front, then reduce the exponent by one.

Where does it come from? Expand (a + h)³ in the difference quotient: a³ + 3a²h + 3ah² + h³ - a³ = 3a²h + 3ah² + h³. Divide by h: 3a² + 3ah + h². Let h go to 0 and everything with an h in it vanishes, leaving 3a². The general case is the same argument with the binomial theorem: exactly one term of the expansion survives the limit, the one with a single h in it, and there are n of them.

The rule holds for every real exponent, not just whole numbers, which is why rewriting is such a useful habit.

Function	Rewritten as a power	Derivative
x⁵	already a power	5x⁴
1/x	x⁻¹	-x⁻² = -1/x²
1/x³	x⁻³	-3x⁻⁴ = -3/x⁴
√x	x to the power 1/2	(1/2)x to the power −1/2, that is 1/(2√x)
x	x¹	1
7	constant	0

Two of those rows should look familiar: in the previous lesson we ground out -1/a² and 1/(2√a) from the limit definition, with a conjugate and a compound fraction respectively. The power rule produces both in about four seconds. That is the trade this lesson is offering.

Why this matters: the derivative of a constant is zero because a constant function has no rise anywhere, and every horizontal line has slope 0. That single fact is why the +C will appear when we reverse the process in Module 6.

Sums, differences and constant multiples

These are the well-behaved ones. (f ± g)' = f' ± g', and (cf)' = c f' for any constant c. Together with the power rule they differentiate every polynomial in one pass.

Differentiate f(x) = 5x⁴ - 3x³ + 7x - 12.

1. 5x⁴ gives 5 × 4x³ = 20x³.
2. -3x³ gives -3 × 3x² = -9x².
3. 7x gives 7.
4. -12 gives 0.

So f'(x) = 20x³ - 9x² + 7. Nothing here needs thought once the pattern is in your hands, which is exactly the point.

Rewrite first, differentiate second

A large fraction of the errors in this topic come from reaching for a heavy rule when algebra would have made the problem trivial. Look for the simplification before you start.

• y = (2x³ + x)/x. Divide first: y = 2x² + 1, so y' = 4x. Using the quotient rule here is legal and slow and error-prone.
• y = √x · x². Combine first: y = x to the power 5/2, so y' = (5/2)x to the power 3/2.
• y = (x + 3)². Expand first: y = x² + 6x + 9, so y' = 2x + 6. (The chain rule will do this in one step next lesson, and will agree.)

The product rule

(fg)' = f'g + fg'. Said aloud: the derivative of the first times the second, plus the first times the derivative of the second.

The proof is a trick worth seeing once. In the difference quotient for f(x+h)g(x+h) - f(x)g(x), add and subtract f(x+h)g(x). That splits the expression into f(x+h)[g(x+h) - g(x)] + g(x)[f(x+h) - f(x)], and dividing by h turns each bracket into a difference quotient. Take the limit and you have f g' + g f'. Adding zero in a well-chosen form is one of the most productive moves in all of analysis.

Worked example. Differentiate y = (3x² + 1)(x³ - 4x).

1. Name the parts: f = 3x² + 1 so f' = 6x; g = x³ - 4x so g' = 3x² - 4.
2. Assemble: y' = (6x)(x³ - 4x) + (3x² + 1)(3x² - 4).
3. Expand the first product: 6x⁴ - 24x².
4. Expand the second: 9x⁴ - 12x² + 3x² - 4.
5. Add: 15x⁴ - 33x² - 4.

Verify it independently. Multiply out first: y = 3x⁵ - 12x³ + x³ - 4x = 3x⁵ - 11x³ - 4x, whose derivative by the power rule is 15x⁴ - 33x² - 4. The two methods agree, which is a check you can always run on a polynomial product and should run the first several times.

The quotient rule

(f/g)' = (f'g - fg') / g². The order in the numerator is not negotiable: subtraction is not commutative, and swapping the two terms flips the sign of every answer. A mnemonic that has survived generations of students, with "high" for the top and "low" for the bottom: low d-high minus high d-low, all over low squared.

Worked example one. Differentiate y = (x² + 1)/(x - 3).

1. f = x² + 1, f' = 2x; g = x - 3, g' = 1.
2. Numerator: (2x)(x - 3) - (x² + 1)(1) = 2x² - 6x - x² - 1 = x² - 6x - 1.
3. Denominator: (x - 3)².
4. y' = (x² - 6x - 1)/(x - 3)².

Worked example two. Differentiate y = (3x - 2)/(x² + 4).

1. f' = 3, g' = 2x.
2. Numerator: 3(x² + 4) - (3x - 2)(2x) = 3x² + 12 - 6x² + 4x = -3x² + 4x + 12.
3. y' = (-3x² + 4x + 12)/(x² + 4)².

Watch the sign in step 2 of that second example. The distributed minus turns -(-2)(2x) into +4x, and dropping that is the single most common slip in the entire topic. Write the subtraction with brackets and expand deliberately.

Derivatives of derivatives

The derivative of a function is a function, so it has a derivative of its own. That is the second derivative, written f''(x) or d²y/dx², and it measures how fast the rate is changing.

Let a particle's position be s(t) = t³ - 6t² + 9t metres, t in seconds.

1. Velocity: v(t) = s'(t) = 3t² - 12t + 9 m/s.
2. Acceleration: a(t) = v'(t) = s''(t) = 6t - 12 m/s per second.
3. At t = 1: v(1) = 3 - 12 + 9 = 0 and a(1) = -6. The particle is momentarily stopped and being pushed in the negative direction, so it is about to reverse.
4. Factor the velocity: 3(t² - 4t + 3) = 3(t - 1)(t - 3), so it is also at rest at t = 3, where a(3) = 6, positive, so it reverses again the other way.

Between 1 and 3 seconds the velocity is negative, so the particle moves backwards, and outside that window it moves forward. You have just read the entire motion off two derivatives without plotting anything, which is what Module 4 will generalise.

Common misconceptions

• "The derivative of a product is the product of the derivatives." The opening example kills this: 2x against 3x². There is no version of this that works.
• "The quotient rule numerator can go either way round." It cannot. f'g - fg' and fg' - f'g differ by a sign on every problem.
• "The power rule needs a whole number exponent." It works for negative and fractional exponents too, which is why rewriting 1/x³ as x⁻³ and √x as x to the power 1/2 is worth doing.
• "The derivative of 7x is 7x, or the derivative of 7 is 7." The derivative of 7x is 7 and the derivative of the constant 7 is 0. A constant function is a horizontal line and horizontal lines have no slope to speak of.
• "You should always use the quotient rule on a fraction." Only when the algebra will not simplify it away first. (2x³ + x)/x becomes 2x² + 1 in one line.

Pulling it together

The power rule d/dx xⁿ = n xⁿ⁻¹ holds for every real exponent, and combined with the constant multiple, sum and difference rules it differentiates any polynomial or rewritten power function immediately. Products and quotients need their own rules because differentiation does not distribute over multiplication: (fg)' = f'g + fg', and (f/g)' = (f'g - fg')/g² with the order in that numerator fixed. Simplify before differentiating whenever the algebra allows, since a rewritten expression is faster and safer than a heavy rule. Differentiating twice gives the second derivative, which for a position function is acceleration and which tells you, at a moment when the velocity is zero, whether the object is turning around and which way.

Sources

1. OpenStax. (2016). 3.3 Differentiation rules. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Differentiation formulas. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Dawkins, P. (n.d.). Product and quotient rule. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Product rule. en.wikipedia.org
5. Wikipedia. (n.d.). Quotient rule. en.wikipedia.org

A balloon, a lapse rate, and two rates multiplied

In the international standard atmosphere, air temperature falls about 6.5 degrees Celsius for every kilometre you climb. A weather balloon rises at 5 metres per second. How fast is the temperature at the balloon dropping, in degrees per second?

Nobody needs calculus for this one, which is what makes it a good way in. In one second the balloon climbs 5 m, which is 0.005 km, and each kilometre costs 6.5 degrees, so the loss is 6.5 × 0.005 = 0.0325 degrees per second. You multiplied two rates. Written in Leibniz notation:

dT/dt = (dT/dh) × (dh/dt) = (-6.5 °C per km) × (0.005 km per s) = -0.0325 °C per s

Notice that the kilometres cancel, top and bottom, exactly as they would in a fraction. That cancellation is not a coincidence and it is not a proof, but it is the reason the chain rule is easy to remember in this notation: dy/dx = (dy/du)(du/dx).

The point: when a quantity depends on something that depends on something else, the rates multiply.

The rule, in the notation you will actually use

If y = f(g(x)), then

y' = f'(g(x)) · g'(x)

which is worth saying in words until it is automatic: differentiate the outside function, leave the inside alone, then multiply by the derivative of the inside. That last multiplication is the part people forget, and forgetting it is the single most common error in first-year calculus.

Half the work is identifying which function is which. Ask: if I were evaluating this at a number, what would I compute first? That is the inside.

Function	Inside g	Outside f	Derivative
(3x + 1)⁵	3x + 1	fifth power	5(3x + 1)⁴ · 3
√(x² + 1)	x² + 1	square root	(1/2)(x² + 1) to the −1/2 · 2x
1/(2x - 7)³	2x - 7	−3 power	-3(2x - 7)⁻⁴ · 2

Two worked all the way down

One. Differentiate y = (3x² - 4x + 1)⁷.

1. Inside: 3x² - 4x + 1. Outside: seventh power.
2. Outside derivative, inside untouched: 7(3x² - 4x + 1)⁶.
3. Derivative of the inside: 6x - 4.
4. Multiply: y' = 7(3x² - 4x + 1)⁶(6x - 4).

Do not expand that. Nobody wants a fourteenth-degree polynomial, and the factored form is more useful for everything Module 4 will do with it.

Two. Differentiate y = √(x² + 1).

1. Rewrite as a power: y = (x² + 1) to the power 1/2. Rewriting first is what turns a root into an ordinary chain rule problem.
2. Outside derivative: (1/2)(x² + 1) to the power −1/2.
3. Times the inside derivative 2x: (1/2)(2x)(x² + 1) to the power −1/2.
4. Simplify: y' = x / √(x² + 1).

Sanity check the answer. At x = 0 it gives 0, and the graph of √(x² + 1) does have a horizontal tangent at its lowest point (0, 1). For very large x the expression tends to 1, and indeed the curve becomes nearly the line y = x far to the right. Both checks pass.

Confirming it against something you can expand

Trust the rule once by testing it where you already know the answer. Differentiate y = (x + 3)² two ways.

• Chain rule: 2(x + 3)¹ · 1 = 2x + 6.
• Expand first: y = x² + 6x + 9, so y' = 2x + 6.

Agreement. Now watch what happens if you forget the inside derivative on y = (2x + 3)²: you would write 2(2x + 3) = 4x + 6. Expanding gives y = 4x² + 12x + 9 with derivative 8x + 12, which is twice as big. The missing factor was the 2 from the inside. When the inside has a coefficient other than 1, the error is visible immediately, which makes such cases good practice.

Chains inside chains

Compositions nest, and so does the rule: keep peeling from the outside in, multiplying as you go. Differentiate y = ((x² + 1)³ + 5)².

1. Outermost: something squared. Derivative: 2((x² + 1)³ + 5).
2. Times the derivative of that something. Its outermost layer is a cube: 3(x² + 1)², and the 5 contributes nothing.
3. Times the derivative of x² + 1, which is 2x.
4. Multiply the three: y' = 2((x² + 1)³ + 5) · 3(x² + 1)² · 2x = 12x(x² + 1)²((x² + 1)³ + 5).

All three rules in one problem

Differentiate y = x²√(2x + 3). This is a product whose second factor needs the chain rule.

1. Product rule frame: y' = (2x)√(2x + 3) + x² · d/dx[√(2x + 3)].
2. The chain rule piece: d/dx[(2x + 3) to the power 1/2] = (1/2)(2x + 3) to the power −1/2 × 2 = 1/√(2x + 3).
3. Assemble: y' = 2x√(2x + 3) + x²/√(2x + 3).
4. Combine over a common denominator: y' = (2x(2x + 3) + x²) / √(2x + 3) = (5x² + 6x)/√(2x + 3).

That last simplification is optional for correctness and essential for what comes next: in the factored form x(5x + 6)/√(2x + 3) you can read off immediately that the derivative is zero at x = 0 and at x = -6/5, which is exactly the information Module 4 needs.

When y refuses to stand on its own

The equation x² + y² = 25 describes a circle of radius 5. It is not a function: the vertical line at x = 3 meets it at y = 4 and y = -4. So you cannot write y = f(x) and differentiate, at least not without splitting the circle into a top half and a bottom half and doing everything twice.

Implicit differentiation avoids the split. Treat y as an unnamed function of x, differentiate every term of the equation with respect to x, and use the chain rule wherever a y appears.

1. Differentiate both sides: d/dx[x²] + d/dx[y²] = d/dx[25].
2. The first term is 2x. The last is 0.
3. The middle term is the whole idea. Since y is a function of x, y² is a composition, so its derivative is 2y · dy/dx. That extra dy/dx is the chain rule's inside derivative, and it is what distinguishes implicit differentiation from ordinary differentiation.
4. So 2x + 2y (dy/dx) = 0.
5. Solve: dy/dx = -x/y.

Check it against the explicit route on the upper half, where y = √(25 - x²). The chain rule gives dy/dx = (1/2)(25 - x²) to the power −1/2 × (-2x) = -x/√(25 - x²), and since √(25 - x²) is y, that is -x/y. Same answer, and the implicit version covers the lower half at no extra cost.

At the point (3, 4) the slope is -3/4. At (3, -4) it is 3/4. Both are visibly right on a circle: the tangent is perpendicular to the radius, and the radius to (3, 4) has slope 4/3.

A curve you could not solve for y at all

The folium of Descartes is x³ + y³ = 6xy, a loop that Descartes and Fermat argued over in 1638. Solving it for y means solving a cubic, which is possible and horrible. Implicitly it takes four lines. Find the slope at (3, 3), a point you can check lies on the curve since 27 + 27 = 54 and 6 × 9 = 54.

1. Differentiate: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). The right side needed the product rule, because 6xy is x times y and both depend on x.
2. Gather the dy/dx terms on one side: 3y²(dy/dx) - 6x(dy/dx) = 6y - 3x².
3. Factor and divide: dy/dx = (6y - 3x²)/(3y² - 6x) = (2y - x²)/(y² - 2x).
4. At (3, 3): (6 - 9)/(9 - 6) = -3/3 = -1.

The tangent there is y - 3 = -(x - 3), or y = 6 - x. The loop of the folium is symmetric about the line y = x, and (3, 3) sits on that line of symmetry, so a tangent perpendicular to it is exactly what the picture demands.

Reading vertical tangents off the formula

Implicit derivatives usually have y in them, and that is a feature. For the circle, dy/dx = -x/y is undefined when y = 0, which happens at (5, 0) and (-5, 0). Those are precisely the two points where the circle has a vertical tangent. The formula told you where the geometry breaks without your having to look at a graph.

Common misconceptions

• "The derivative of (3x + 1)⁵ is 5(3x + 1)⁴." Missing the inside derivative 3. This is the error the chain rule exists to prevent, and it is worth checking every composition for it.
• "Implicit differentiation is a different kind of differentiation." It is the chain rule, applied to a function whose formula you never write down. Every step is a rule you already know.
• "An answer with y in it is unfinished." For an implicit curve it is normal and often more useful, since it exposes vertical tangents and works on both branches at once.
• "d/dx of y² is 2y." Only if y were the variable of differentiation. With respect to x it is 2y (dy/dx).
• "Differentiating 6xy gives 6(dy/dx)." It is a product of two x-dependent quantities, so the product rule applies: 6y + 6x(dy/dx).

What you now know

The chain rule says composed functions have multiplied rates: differentiate the outside, leave the inside alone, multiply by the derivative of the inside, and repeat for every nested layer. Rewriting roots and reciprocals as powers turns most compositions into routine work, and combining the chain rule with the product or quotient rule in one problem is normal rather than exceptional. Implicit differentiation is the same rule pointed at an equation instead of a formula: differentiate every term with respect to x, remember that any y carries a factor of dy/dx, then collect and solve. It handles curves like circles and the folium of Descartes that are not functions at all, and the y left in the answer is what lets you spot vertical tangents by asking where the denominator vanishes.

Sources

1. OpenStax. (2016). 3.6 The chain rule. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 3.8 Implicit differentiation. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). Chain rule. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Dawkins, P. (n.d.). Implicit differentiation. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Wikipedia. (n.d.). Chain rule. en.wikipedia.org

Two calculators, two different answers

Set a calculator to degree mode and compute sin(1°). You get 0.0174524. Divide by the angle, 1, and the difference quotient near zero is about 0.017. Now set it to radian mode and compute sin(0.001), which is 0.00099999983. Divide by 0.001 and you get 0.99999983, which is heading for 1.

Same function, same idea, two utterly different slopes at the origin: about 0.0175 in degrees, exactly 1 in radians. The derivative of sine depends on the unit of the angle, and 0.0174532925 is π/180, the conversion factor. In degrees, the derivative of sine is (π/180) cos x, and that stray constant would then infect every formula in physics that ever touches an oscillation.

Radians are chosen because they are the unit that makes the constant equal to 1. Everything in this lesson, and every trigonometric derivative in the rest of the course, assumes radians.

Remember: in calculus, angles are radians. There is no exception and no convention to negotiate.

The two limits everything rests on

Two limits do all the work. Both are 0/0 forms, and neither can be repaired by factoring.

Limit one: lim(θ → 0) (sin θ)/θ = 1.

θ (radians)	sin θ	(sin θ)/θ
0.1	0.0998334	0.998334
0.01	0.00999983	0.999983
0.001	0.000999999833	0.999999833

The proof is the squeeze theorem from Module 1, applied to a picture. On a unit circle, for a small positive angle, compare three areas: the triangle with area (1/2)sin θ, the circular sector with area (1/2)θ, and the larger triangle with area (1/2)tan θ. The middle one sits between the other two, which gives sin θ ≤ θ ≤ tan θ. Dividing through by sin θ and inverting turns that into cos θ ≤ (sin θ)/θ ≤ 1. Since cos θ tends to 1, the quantity in the middle is squeezed to 1.

Limit two: lim(θ → 0) (cos θ - 1)/θ = 0. This one follows from the first by the conjugate trick.

1. Multiply top and bottom by cos θ + 1: the top becomes cos²θ - 1.
2. The Pythagorean identity turns that into -sin²θ.
3. So the expression is -(sin θ/θ) × (sin θ/(cos θ + 1)).
4. The first factor tends to 1 and the second tends to 0/2 = 0, so the product tends to 0.

Numerically, at θ = 0.1 the quotient is (0.995004 - 1)/0.1 = -0.0500, and at 0.01 it is -0.0050. Shrinking toward zero, as promised.

Deriving the derivative of sine

Now the payoff. Start from the definition and use the angle addition identity sin(x + h) = sin x cos h + cos x sin h.

1. The difference quotient is (sin x cos h + cos x sin h - sin x)/h.
2. Group the terms containing sin x: [sin x (cos h - 1) + cos x sin h] / h.
3. Split into two fractions: sin x · (cos h - 1)/h + cos x · (sin h)/h. Notice that sin x and cos x are constants as far as the h limit is concerned.
4. Apply the two limits: the first bracket goes to 0, the second to 1.
5. What is left is sin x · 0 + cos x · 1 = cos x.

So d/dx [sin x] = cos x. The identical argument with cos(x + h) = cos x cos h - sin x sin h gives d/dx [cos x] = -sin x, and the minus sign there is not decorative: cosine starts at its maximum and falls, so its slope at 0 must be 0 and its slope just after must be negative.

The other four, by the rules you already have

Nothing else needs a new limit. Tangent is a quotient of two functions we can now differentiate.

1. tan x = sin x / cos x, so by the quotient rule the derivative is [(cos x)(cos x) - (sin x)(-sin x)] / cos²x.
2. The numerator is cos²x + sin²x = 1, by the Pythagorean identity.
3. So the derivative is 1/cos²x = sec²x.

Secant is a chain rule problem: sec x = (cos x)⁻¹, so its derivative is -(cos x)⁻² × (-sin x) = sin x / cos²x, which regroups as (1/cos x)(sin x/cos x) = sec x tan x. Cosecant and cotangent go the same way and pick up minus signs.

Function	Derivative
sin x	cos x
cos x	-sin x
tan x	sec²x
cot x	-csc²x
sec x	sec x tan x
csc x	-csc x cot x

There is a pattern that halves the memorising: every function whose name starts with "co" (cosine, cotangent, cosecant) has a minus sign in its derivative, and every one that does not, does not. Learn three rows and the sign rule.

Now combine them with everything else

These derivatives are ingredients, not answers. Almost every real problem wraps them in a product, a quotient or a composition.

• y = x² sin x. Product rule: y' = 2x sin x + x² cos x.
• y = sin(3x²). Chain rule, inside is 3x²: y' = cos(3x²) · 6x.
• y = sin³x, which means (sin x)³. Chain rule, inside is sin x: y' = 3 sin²x · cos x.
• y = tan(√x). Chain rule: y' = sec²(√x) · 1/(2√x).
• y = (sin x)/x. Quotient rule: y' = (x cos x - sin x)/x².

The third and fourth items are where notation causes real casualties. sin²x means square the sine. sin(x²) means square the angle first. sin(sin x) means take the sine twice. They are three different functions with three different derivatives, and the only defence is to read the parentheses before you start.

A mass on a spring, done properly

Attach a mass to a spring, pull it 5 cm from rest and let go. Its displacement is well modelled by x(t) = 5 cos(2t) centimetres, with t in seconds.

1. Velocity: v(t) = x'(t) = 5 · (-sin(2t)) · 2 = -10 sin(2t) cm/s. The chain rule supplied that factor of 2, and dropping it would halve every speed.
2. Acceleration: a(t) = v'(t) = -10 cos(2t) · 2 = -20 cos(2t) cm/s per second.
3. Compare the acceleration with the displacement: a(t) = -4 × 5cos(2t) = -4 x(t).

That relationship, acceleration proportional to displacement and pointing the opposite way, is the definition of simple harmonic motion, and it is why sine and cosine describe pendulums, tides, alternating current and the vibration of a guitar string. Check the instants: at t = 0 the mass is at x = 5 (fully stretched), v = 0 (momentarily still) and a = -20 (yanked hard back toward the middle). At t = π/4, 2t = π/2, so x = 0, v = -10 (moving fastest) and a = 0 (nothing pulling). Fastest where the force is zero, still where the force is greatest. That is the whole behaviour of an oscillator, read off two derivatives.

Why engineers write sine as its own angle

Since d/dx[sin x] = cos x and cos 0 = 1, the tangent line to y = sin x at the origin is simply y = x. That is the small angle approximation, and its accuracy is worth seeing: at x = 0.1 radians, sin x = 0.0998334, so the approximation is off by 0.00017, about one part in six hundred. At x = 0.01 the error is about one part in sixty thousand. Pendulum equations, optics and structural analysis all lean on this, and it is a derivative statement in disguise.

Common misconceptions

• "Degrees and radians are interchangeable if you are careful." Not in calculus. In degrees the derivative of sine is (π/180) cos x, and every subsequent formula carries the extra factor.
• "The derivative of sin(3x) is cos(3x)." Missing the inside derivative. It is 3cos(3x), and in the spring example that missing factor would have halved the velocity.
• "sin²x and sin(x²) are notational variants." They are different functions: 2 sin x cos x against 2x cos(x²).
• "The limit of (sin θ)/θ is 0 because the top goes to 0." The bottom goes to 0 as well, at the same rate, which is exactly why the answer is 1 rather than 0.
• "Differentiating sine four times gets you somewhere new." It cycles: sine, cosine, minus sine, minus cosine, back to sine. That period-four cycle is the reason the fourth derivative of a sine wave is itself, a fact that matters in beam theory.

Where this leaves us

Trigonometric derivatives depend on the angle unit, and radians are the unit that makes the constant 1, which is why calculus never uses degrees. Everything follows from two limits, (sin θ)/θ → 1 proved by squeezing a sector between two triangles, and (cos θ - 1)/θ → 0 proved from the first by the conjugate. With those, the definition of the derivative yields cos x for sine and -sin x for cosine, and the quotient and chain rules produce the other four without a single new limit, with the minus signs all attached to the co-functions. In use they arrive wrapped in products, quotients and compositions, so read the parentheses carefully and never lose the inside derivative. Differentiating a cosine twice returns a negative multiple of itself, which is the mathematical signature of every oscillation in physics.

Sources

1. OpenStax. (2016). 3.5 Derivatives of trigonometric functions. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Derivatives of trig functions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. OpenStax. (2016). 1.3 Trigonometric functions. In Calculus Volume 1. openstax.org
4. MIT OpenCourseWare. (2010). 18.01SC Single Variable Calculus. Massachusetts Institute of Technology. ocw.mit.edu

Bernoulli's absurdly generous bank

In 1683 Jacob Bernoulli asked what happens to compound interest as the compounding gets faster. Put one unit of money in an account paying 100 percent a year. Compounded once, you end with 2. Compounded twice a year at 50 percent each time, you end with (1.5)² = 2.25. Keep going:

Compounding periods per year	Final amount, (1 + 1/n) to the n
1	2
2	2.25
12 (monthly)	2.613035
365 (daily)	2.714567
8760 (hourly)	2.718127

The money does not run away. It converges, on 2.718281828..., a number Euler later called e. Bernoulli could see the limit was between 2 and 3 and could not evaluate it. What matters for us is the reason e is worth a separate letter at all, which is not the bank account. It is that e is the one base for which calculus comes out clean.

Every exponential has a constant attached, and e is where the constant is 1

Start from the definition for f(x) = aₓ, where a is a positive constant.

1. The difference quotient is (aₓ⁺ₕ - aₓ)/h.
2. Factor out aₓ, which does not involve h: aₓ · (aₕ - 1)/h.
3. So f'(x) = aₓ · lim(h → 0) (aₕ - 1)/h. The limit is a number depending only on a, not on x.

Every exponential function is therefore its own derivative times some constant. Which constant? Evaluate the limit numerically at h = 0.001:

Base a	(a to the 0.001, minus 1) divided by 0.001	Recognise it?
2	0.693387	ln 2 = 0.693147
3	1.099215	ln 3 = 1.098612
2.718281828	1.000500	ln e = 1

The constant is ln a, so d/dx [aₓ] = aₓ ln a. And the reason e is called the natural base is the third row: it is the unique base whose constant is exactly 1.

d/dx [eₓ] = eₓ

The core of it: eₓ is its own derivative. Its height and its slope are the same number at every point, and no other function does this except constant multiples of it.

Picture that claim on the graph. At x = 0 the height is 1 and the slope is 1. At x = 2 the height is about 7.389 and the tangent there climbs at 7.389 per unit. The curve is steep exactly in proportion to how high it already is, which is why exponential growth outruns every polynomial eventually.

With the chain rule attached

In practice the exponent is never just x. d/dx [e⁵] = eⁱ · u', where u is whatever sits upstairs.

• y = e³ₓ gives y' = 3e³ₓ.
• y = e raised to x² gives y' = 2x e raised to x².
• y = e raised to -kt gives y' = -k e raised to -kt, the pattern behind every decay law.
• y = 5ₓ gives y' = 5ₓ ln 5, roughly 1.609 × 5ₓ.

The logarithm, from implicit differentiation

Here is where the previous lesson pays for itself. We want d/dx [ln x], and we know nothing about it directly. But the logarithm is defined by the exponential: y = ln x means exactly eₖ = x. Differentiate that equation implicitly.

1. Left side: d/dx[eₖ] = eₖ · (dy/dx), by the chain rule, since y is a function of x.
2. Right side: d/dx[x] = 1.
3. So eₖ (dy/dx) = 1, giving dy/dx = 1/eₖ.
4. But eₖ = x. So dy/dx = 1/x.

d/dx [ln x] = 1/x, for x > 0. Three lines, no new limit, and a surprising answer: differentiating a transcendental function produced a rational one. That single fact is why Module 6 will be able to integrate 1/x at all, and it is the reason the natural logarithm turns up in problems that never mentioned logarithms.

Two companions follow at once. For any base, d/dx [logₐ x] = 1/(x ln a), since logₐ x = (ln x)/(ln a) and the denominator is a constant. And with the chain rule, d/dx [ln u] = u'/u, which is the form you will actually use.

Five worked examples

• y = ln(x² + 1). Inside derivative over inside: y' = 2x/(x² + 1).
• y = x ln x. Product rule: y' = 1 · ln x + x · (1/x) = ln x + 1. Neat enough to be worth remembering; it reappears in Module 6.
• y = eₓ sin x. Product rule: y' = eₓ sin x + eₓ cos x = eₓ(sin x + cos x).
• y = ln(sin x). Chain rule: y' = cos x / sin x = cot x.
• y = (ln x)³. Chain rule, inside is ln x: y' = 3(ln x)² · (1/x).

Watch the difference between the last two patterns. ln(sin x) puts the log outside; (ln x)³ puts the cube outside. Deciding which function is on the outside is the entire chain rule decision, and it is settled by the brackets.

When x is in the base and the exponent at once

Differentiate y = xₓ. The power rule does not apply, because the exponent is not constant. The exponential rule does not apply, because the base is not constant. It is a genuinely new shape, and the tool is logarithmic differentiation: take the natural log of both sides first, which converts the exponent into a multiplier.

1. ln y = ln(xₓ) = x ln x.
2. Differentiate both sides implicitly. Left: (1/y)(dy/dx). Right, by the product rule: ln x + 1.
3. So (1/y)(dy/dx) = ln x + 1.
4. Multiply by y and substitute back: dy/dx = xₓ(ln x + 1).

Check the shape of the answer at x = 1: it gives 1 × (0 + 1) = 1. The function xₓ passes through (1, 1) climbing at slope 1, which is what a table of values shows: at 1.01 the function is about 1.0100, so the rise over the run is about 1.00. Agreement.

The same technique tames ugly products and quotients. For y = (x² + 1)⁵(2x - 1)³, taking logs turns the product into a sum, ln y = 5ln(x² + 1) + 3ln(2x - 1), and differentiating gives (1/y)y' = 10x/(x² + 1) + 6/(2x - 1). Multiply back by y and you are done, without ever expanding an eighth-degree polynomial.

Half-life, and why the derivative is proportional to the amount

Carbon-14 has a half-life of about 5730 years. Model the remaining amount as N(t) = N₀ e to the power -kt. What is k?

1. Half remains at t = 5730: N₀/2 = N₀ e to the power -5730k.
2. Cancel and take logs: ln(1/2) = -5730k, so k = (ln 2)/5730 = 0.693147/5730 = 1.2097 × 10⁻⁴ per year.
3. Now differentiate: N'(t) = -kN₀ e to the power -kt, which is -k N(t).

Read that last line. The rate of decay at any instant is a fixed fraction of what is present, which is what "radioactive" means physically: each atom decays independently with the same probability per unit time, so twice as many atoms means twice the count rate. In a 1 gram sample the rate is enormous in absolute terms and tiny in relative terms, about 0.012 percent of the remaining carbon-14 per year. Module 7 will run this backwards and solve the differential equation N' = -kN to recover the exponential in the first place.

Common misconceptions

• "The derivative of 2ₓ is x times 2 to the x minus 1." That is the power rule applied where it does not belong. The variable is upstairs, so the answer is 2ₓ ln 2.
• "eₓ is its own derivative, so aₓ is its own derivative." Only for a = e. Every other base carries a factor of ln a, which is bigger than 1 above e and smaller below it.
• "The derivative of ln(3x) is 1/(3x)." The chain rule gives 3/(3x) = 1/x. Interestingly the 3 cancels, which is a good sanity check: ln(3x) = ln 3 + ln x differs from ln x by a constant, so the derivatives must be equal.
• "log means log base 10, always." In calculus and in most of science, an unqualified log in this context is the natural log; we write ln here to avoid the ambiguity entirely.
• "Logarithmic differentiation is a special formula." It is two things you already know, log identities and implicit differentiation, used in sequence.

Summing up

Every exponential is its own derivative times a constant, and that constant is ln a; e is defined by the fact that the constant equals 1, so eₓ is its own derivative and nothing else is. With the chain rule, eⁱ differentiates to eⁱu', which is the pattern behind growth and decay everywhere. The natural log's derivative falls out of implicit differentiation of eₖ = x and is 1/x, or u'/u with a chain, which is why a transcendental function ends up governing rational integrals in Module 6. When the variable sits in both the base and the exponent, take logs first, differentiate implicitly, then multiply back; the same trick collapses awkward products and quotients into sums. And a decay law is exactly the statement that the derivative is proportional to the amount present, which is the differential equation this course closes with.

Sources

1. OpenStax. (2016). 3.9 Derivatives of exponential and logarithmic functions. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 1.5 Exponential and logarithmic functions. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). Derivatives of exponential and logarithm functions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. OpenStax. (2016). 6.8 Exponential growth and decay. In Calculus Volume 1. openstax.org

The camera that catches you without seeing you

Average speed enforcement works like this: one camera reads your plate at the start of a stretch of motorway, another reads it 30 km later, and the system divides. If those readings are 15 minutes apart, your average speed was 120 km/h. In a 100 km/h zone that is a ticket, and the legal reasoning is a theorem: since your position was a continuous, differentiable function of time, there must have been at least one instant at which your speedometer read exactly 120.

You cannot escape it by arguing that you were doing 90 for a while and 150 for a while. That only makes it worse, because it means you passed through 120 on the way. The claim that a differentiable function must, at some instant, actually do what it did on average is the mean value theorem, and it turns out to be the tool that converts every intuition about graphs into something provable.

Rolle's theorem: the flat spot

Start with the special case. If f is continuous on [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f'(c) = 0.

The picture is unarguable: if you start and end at the same height, you must have turned around, and at the turn the tangent is horizontal.

Verify it on f(x) = x² - 4x + 3 over [1, 3]. Endpoints: f(1) = 1 - 4 + 3 = 0 and f(3) = 9 - 12 + 3 = 0. Equal, so Rolle applies. Then f'(x) = 2x - 4, which is 0 at x = 2, comfortably inside the interval. The parabola's vertex is exactly where the theorem promised.

The mean value theorem: the tilted version

If f is continuous on [a, b] and differentiable on (a, b), then there is at least one c in (a, b) with

f'(c) = (f(b) - f(a)) / (b - a).

The right side is the slope of the secant line joining the endpoints, which is the average rate of change. The theorem says the instantaneous rate hits the average rate at least once. Rolle is the case where the secant happens to be horizontal, and the general proof is exactly that: tilt the picture by subtracting the secant line from f, apply Rolle to what is left, and tilt back.

Worked example. Take f(x) = x³ on [0, 2] and find every c the theorem guarantees.

1. Average rate: (f(2) - f(0))/(2 - 0) = (8 - 0)/2 = 4.
2. Instantaneous rate: f'(x) = 3x².
3. Set them equal: 3c² = 4, so c² = 4/3 and c = 2/√3 ≈ 1.1547. (The negative root is outside the interval.)
4. Check: 1.1547 does lie strictly between 0 and 2, as promised.

The hypotheses are not ornamental. Take f(x) = |x| on [-1, 1]. The average rate of change is (1 - 1)/2 = 0, so the theorem would demand a point with zero slope. But the derivative is −1 on the left, +1 on the right, and undefined at 0. No contradiction: f is not differentiable on the whole open interval, so the theorem never applied.

What the theorem is actually for

The mean value theorem is rarely the answer to a question. It is the machinery underneath three facts you will use constantly, and here is where they come from.

Fact one: if f' > 0 throughout an interval, f is increasing there. Proof: take any x₁ < x₂ in the interval. By the theorem, f(x₂) - f(x₁) = f'(c)(x₂ - x₁) for some c between them. The second factor is positive because x₂ > x₁, and the first is positive by assumption, so f(x₂) > f(x₁). That is the definition of increasing. Reverse every sign for the decreasing case.

Fact two: if f' = 0 throughout an interval, f is constant there. Same argument with a zero factor: every pair of outputs is equal.

Fact three: if f' = g' throughout an interval, then f = g + C for some constant C. Apply fact two to f - g.

Why this matters: fact three is the reason every indefinite integral in Module 6 ends with "+ C", and the reason that C cannot be dropped. It is not a bookkeeping convention. It is the exact statement of how much a function is undetermined by its derivative.

Critical points

A critical point of f is a number c in the domain of f where f'(c) = 0 or where f'(c) does not exist. These are the only places an interior maximum or minimum can hide, a result usually credited to Fermat: at an interior peak the function stops rising before it starts falling, so if the derivative exists at all, it is zero.

The converse is false and the counterexample matters. f(x) = x³ has f'(0) = 0, so 0 is critical, but f'(x) = 3x² is positive on both sides. The function is increasing through the point, pausing without turning. A critical point is a suspect, not a conviction.

The first derivative test, worked end to end

Analyse f(x) = x³ - 3x² - 9x + 5.

1. Differentiate: f'(x) = 3x² - 6x - 9.
2. Factor: 3(x² - 2x - 3) = 3(x - 3)(x + 1).
3. Critical points where this is zero: x = -1 and x = 3. The derivative is a polynomial, so it is never undefined.
4. Build a sign chart by testing one point in each of the three intervals the critical points create.

Interval	Test point	Sign of f'	Behaviour
x less than −1	x = -2: 3(-5)(-1) = 15	positive	increasing
−1 to 3	x = 0: 3(-3)(1) = -9	negative	decreasing
x greater than 3	x = 4: 3(1)(5) = 15	positive	increasing

Now read the chart. At x = -1 the sign goes from positive to negative, so the function rises then falls: a local maximum, with value f(-1) = -1 - 3 + 9 + 5 = 10. At x = 3 the sign goes from negative to positive: a local minimum, with value f(3) = 27 - 27 - 27 + 5 = -22.

That is the whole test. Positive to negative is a peak, negative to positive is a valley, and no sign change means neither. You never had to plot a point.

A critical point where the derivative does not exist

Take f(x) = x to the power 2/3, whose graph has a sharp cusp at the origin. Its derivative is (2/3)x to the power −1/3, which is undefined at 0, so 0 is a critical point even though the derivative is never zero anywhere. Test the sign: for negative x the cube root is negative, so f' is negative; for positive x it is positive. Negative to positive means a local minimum at x = 0, and the value there is 0. The graph confirms it: the cusp is the lowest point. Skip the undefined-derivative case and you would have found no extremum at all.

Absolute extrema: the candidates test

On a closed interval, the extreme value theorem from Module 1 guarantees an absolute maximum and minimum exist. Finding them takes three steps and no sign chart at all, because the answer must occur at a critical point or an endpoint.

Find the absolute extrema of f(x) = x³ - 3x² - 9x + 5 on [-2, 4].

1. Critical points in the interval: x = -1 and x = 3, both inside.
2. Endpoints: x = -2 and x = 4.
3. Evaluate f at all four and compare.

Candidate	f(x)
x = -2	-8 - 12 + 18 + 5 = 3
x = -1	-1 - 3 + 9 + 5 = 10
x = 3	27 - 27 - 27 + 5 = -22
x = 4	64 - 48 - 36 + 5 = -15

Absolute maximum 10 at x = -1; absolute minimum −22 at x = 3. Note that the right endpoint beat neither, but on a slightly wider interval it would have: extend to [-2, 6] and f(6) = 216 - 108 - 54 + 5 = 59 becomes the maximum. Endpoints are candidates precisely because they can win.

Common misconceptions

• "f'(c) = 0 means c is a maximum or minimum." x³ at 0 says otherwise. You must check the sign change.
• "Critical points only come from setting the derivative to zero." Points where the derivative fails to exist count too, and the cusp of x to the 2/3 is a genuine minimum found only that way.
• "The mean value theorem tells you where c is." It tells you c exists. Finding it means solving f'(c) = the average rate, which may or may not be easy.
• "A function that is increasing must have a positive derivative everywhere." Almost. f(x) = x³ is increasing on every interval yet its derivative is 0 at the origin. Positive derivative implies increasing; increasing implies the derivative is at least zero.
• "Local maximum means highest point on the graph." Local means highest nearby. Here the local maximum is 10 while the function reaches 59 at x = 6.

What to remember

Rolle's theorem says a differentiable function returning to its starting height must level off somewhere in between, and the mean value theorem tilts that into the general statement that the instantaneous rate equals the average rate at least once, which is the argument a speed camera relies on. Its real value is what it proves: a positive derivative forces a function to increase, a zero derivative forces it to be constant, and two functions with the same derivative can differ only by a constant, which is where the "+ C" of integration comes from. Critical points, where the derivative is zero or undefined, are the only interior candidates for extrema, and the first derivative test classifies each by whether the sign of f' flips from positive to negative, negative to positive, or not at all. On a closed interval, compare the values at every critical point and both endpoints, and the largest and smallest of those numbers are the absolute extrema.

Sources

1. OpenStax. (2016). 4.4 The mean value theorem. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 4.3 Maxima and minima. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). The mean value theorem. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Mean value theorem. en.wikipedia.org

Two cones and one steady tap

Stand a cone on its point and fill it from a tap running at a constant 2 litres per second. The water level races up at first, when the cone is narrow, and slows as the cone widens. Now stand an identical cone on its base, point upward, and fill it the same way. The level creeps up at first, in the wide part, then accelerates as the space narrows to the tip.

In both tanks the depth is increasing the whole time, so in both the first derivative is positive. Anyone reading only dh/dt > 0 would call these the same situation. They are not the same at all, and what separates them is whether that rate is itself rising or falling. That is the second derivative, and it is the subject of this lesson.

Key idea: the first derivative says which way the curve is going; the second says which way it is bending.

Concave up, concave down

A curve is concave up on an interval when f''(x) > 0 there, and concave down when f''(x) < 0. The reason is direct: f'' is the derivative of f', so a positive second derivative means the slope is increasing, which bends the curve upward into a shape that would hold water. A negative second derivative means the slope is decreasing and the curve arches over, spilling it.

Two other readings of the same fact are worth carrying:

• Concave up means the curve lies above its tangent lines; concave down means it lies below them.
• Concave up on y = x²: at x = -2 the slope is −4, at 0 it is 0, at 2 it is 4. The slope is climbing even where the function is falling.

Inflection points, and the trap in defining them

An inflection point is where concavity changes, from up to down or down to up. Since f'' is what carries concavity, an inflection point requires f'' to change sign there, which usually means f'' = 0 at that point.

The trap is treating "usually" as "always" in the other direction. f(x) = x⁴ has f''(x) = 12x², which is 0 at the origin. But 12x² is positive on both sides, so the concavity never changes: the curve is concave up throughout and the origin is a flat bottom, not an inflection. A zero of the second derivative is a candidate, and the sign change is the test, exactly as with critical points and the first derivative.

The second derivative test

At a critical point c where f'(c) = 0:

If	Then	Because
f''(c) > 0	local minimum at c	flat and concave up is the bottom of a bowl
f''(c) < 0	local maximum at c	flat and concave down is the top of a dome
f''(c) = 0	no conclusion	use the first derivative test instead

Try it on the function from the previous lesson, f(x) = x³ - 3x² - 9x + 5, whose critical points were −1 and 3.

1. f'(x) = 3x² - 6x - 9, so f''(x) = 6x - 6.
2. At x = -1: f''(-1) = -12, negative, so local maximum. The sign chart said maximum too.
3. At x = 3: f''(3) = 12, positive, so local minimum. Again the two methods agree.

The second derivative test is faster when the second derivative is easy to compute, since it needs one number rather than a whole sign chart. It is useless when it returns zero, and this is not a rare inconvenience: x⁴ (minimum), -x⁴ (maximum) and x³ (neither) all have first and second derivatives equal to zero at the origin, so no amount of second derivative information could tell them apart. When it comes back zero, go back to the sign of f'.

The four combinations, and how to read them out loud

f'	f''	Shape	Said plainly
positive	positive	rising, steepening	growing, and growth is accelerating
positive	negative	rising, levelling off	still growing, but the growth is slowing
negative	negative	falling, steepening	declining, and the decline is worsening
negative	positive	falling, levelling off	still declining, but the decline is easing

The second row is the one people misread in public argument. A quantity whose second derivative has just turned negative is still increasing; nothing has got smaller. What has changed is the rate. Reporting "the rise is slowing" as "the numbers are falling" confuses f' with f'', and the distinction is worth being able to state precisely.

A complete sketch, start to finish

Sketch f(x) = x⁴ - 4x³ using nothing but calculus.

Step 1, intercepts. Factor: x³(x - 4). Zeros at x = 0 and x = 4. The y-intercept is f(0) = 0.

Step 2, end behaviour. The leading term is x⁴, an even power with a positive coefficient, so the curve rises to positive infinity on both sides.

Step 3, first derivative. f'(x) = 4x³ - 12x² = 4x²(x - 3). Critical points at x = 0 and x = 3.

Interval	Sign of 4x²	Sign of (x - 3)	Sign of f'	Behaviour
x less than 0	+	−	−	decreasing
0 to 3	+	−	−	decreasing
x greater than 3	+	+	+	increasing

Two things to notice. At x = 3 the derivative flips from negative to positive: a local minimum, with f(3) = 81 - 108 = -27. At x = 0 the derivative is zero but does not change sign, because the 4x² factor is positive on both sides. So the curve flattens at the origin and keeps going down, with no extremum there at all.

Step 4, second derivative. f''(x) = 12x² - 24x = 12x(x - 2). Zeros at x = 0 and x = 2.

Interval	Sign of f''	Concavity
x less than 0	+	concave up
0 to 2	−	concave down
x greater than 2	+	concave up

The sign changes at both zeros, so both are genuine inflection points: (0, 0) and (2, -16).

Step 5, assemble. Coming in from the far left the curve is falling steeply and concave up. It flattens momentarily at the origin, which is simultaneously a critical point and an inflection, then continues downward while concave down. It bottoms out at (3, -27), turns, and climbs away to the right, crossing the axis again at x = 4, concave up from x = 2 onward.

Every feature of that description came from two derivatives and a factorisation. You could now plot four points and draw a curve that is correct in every qualitative detail.

One with an asymptote

Sketch f(x) = x/(x² + 1). The denominator is never zero, so there are no vertical asymptotes, and by the degree comparison from Module 1 the function tends to 0 in both directions, giving a horizontal asymptote at y = 0.

1. Quotient rule: f'(x) = [(1)(x² + 1) - x(2x)] / (x² + 1)² = (1 - x²)/(x² + 1)².
2. The denominator is always positive, so the sign is the sign of 1 - x²: negative outside [-1, 1], positive inside.
3. Critical points at x = -1 and x = 1. Negative to positive at −1 makes it a minimum, f(-1) = -1/2; positive to negative at 1 makes it a maximum, f(1) = 1/2.

So this curve climbs from the asymptote on the far left, dips to -1/2, rises through the origin to 1/2, then settles back toward zero. It crosses its own horizontal asymptote at the origin, which functions are perfectly entitled to do.

Common misconceptions

• "f'' = 0 means there is an inflection point." x⁴ at the origin refutes this. The second derivative must change sign, not merely touch zero.
• "Concave up means increasing." They are independent. y = x² for negative x is decreasing and concave up at the same time.
• "If the second derivative test fails, the point is not an extremum." It fails by returning no information. x⁴ has a genuine minimum at a point where the test is silent.
• "Every critical point is an extremum, and every extremum is a smooth turn." The origin in x⁴ - 4x³ is a critical point that is not an extremum, and it is an inflection point as well.
• "Growth slowing means the total is going down." A negative second derivative with a positive first derivative means still rising, just less steeply. Nothing has decreased.

The takeaway

The second derivative measures bending: positive means concave up, with the curve above its tangents and its slopes increasing, and negative means concave down. Inflection points are where that sign changes, so a zero of f'' is only a candidate and the sign change is the test, exactly as it is for critical points. At a point where f' = 0, the sign of f'' classifies it immediately, a minimum if positive and a maximum if negative, but a zero second derivative tells you nothing and sends you back to the first derivative test. Combining the two sign charts with intercepts and end behaviour produces a complete sketch: for x⁴ - 4x³ that meant a minimum at (3, -27), inflections at (0, 0) and (2, -16), and a flat spot at the origin that is not an extremum. Four combinations of signs describe every stretch of every curve, and being able to say which one you are in is what separates reading a graph from guessing at one.

Sources

1. OpenStax. (2016). 4.5 Derivatives and the shape of a graph. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). The shape of a graph, part II. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Dawkins, P. (n.d.). The shape of a graph, part I. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Second derivative. en.wikipedia.org

The drinks can that mathematics says is the wrong shape

A standard aluminium drinks can holds 355 millilitres. Ask calculus for the cheapest such can, meaning the one using the least metal, and it returns a definite answer: radius 3.84 cm, height 7.67 cm. The height should be exactly twice the radius, so the ideal can is as tall as it is wide, and it uses about 278 square centimetres of aluminium.

Real 355 mL cans are about 6.6 cm across and 12.2 cm tall. They are visibly taller and thinner than the optimum, and they use a few percent more metal than they need to. The manufacturers are not bad at calculus. The lid is made of thicker, more expensive alloy than the wall, so shrinking the lid is worth more than the model says; cans must stack, ship and fit a hand. What the calculation gives you is the answer to the question you actually asked, which is why stating the question precisely is most of the work.

This lesson is that process, run slowly, four times.

The five steps

1. Name and draw. Assign a letter to every varying quantity and sketch the situation. Label the sketch.
2. Write the objective. One equation for the quantity being maximised or minimised.
3. Use the constraint to eliminate a variable. The objective usually has two unknowns; the constraint given in the problem lets you write one in terms of the other. State the domain of what is left, since a negative length is not a candidate.
4. Differentiate and solve. Set the derivative to zero and find the critical points.
5. Justify and answer. Show it really is a maximum or minimum, then answer the question that was asked, with units.

The upshot: steps 2 and 3 are where problems are won or lost. Step 4 is the easy part.

Problem one: the field on the river

A farmer has 2400 m of fencing and wants to enclose a rectangular field beside a straight river. No fence is needed along the river. What is the largest area she can enclose?

1. Name and draw. Let x be the length of each of the two sides perpendicular to the river, and y the length of the side parallel to it. Only three sides are fenced.
2. Objective. Area A = xy.
3. Constraint. The fencing used is 2x + y = 2400, so y = 2400 - 2x. Substituting, A(x) = x(2400 - 2x) = 2400x - 2x². Domain: x must be at least 0, and y must be at least 0, so 0 ≤ x ≤ 1200.
4. Differentiate. A'(x) = 2400 - 4x, which is zero at x = 600.
5. Justify. A''(x) = -4, negative everywhere, so the curve is concave down and this critical point is the absolute maximum. Alternatively, check the candidates: A(0) = 0, A(1200) = 0, A(600) = 600 × 1200 = 720000.

Answer: 600 m out from the river, 1200 m along it, enclosing 720,000 square metres. Notice the shape of the answer: the side parallel to the river is exactly half the total fencing. That is not a coincidence and it recurs in every problem of this family, which is a useful check when you meet the next one.

Problem two: the can, all the way down

Find the dimensions of a closed cylindrical can of volume 355 cm³ using the least material.

1. Name. Radius r, height h, both in centimetres.
2. Objective. Surface area of a closed cylinder: two circular ends plus the wall, S = 2πr² + 2πrh.
3. Constraint. πr²h = 355, so h = 355/(πr²). Substitute: S(r) = 2πr² + 2πr · 355/(πr²) = 2πr² + 710/r. Domain: r > 0, an open interval, which is why the second derivative test will be needed rather than a candidates list.
4. Differentiate. Write 710/r as 710r⁻¹ before differentiating. S'(r) = 4πr - 710/r². Setting that to zero gives 4πr = 710/r², so r³ = 710/(4π) = 56.5 and r = 3.837 cm.
5. Justify. S''(r) = 4π + 1420/r³, which is positive for every positive r, so S is concave up throughout and this is the absolute minimum.

Now finish the question. h = 355/(π × 3.837²) = 355/46.25 = 7.676 cm, which is 2r to three figures. And the material used is S = 2π(14.72) + 710/3.837 = 92.5 + 185.0 = 277.5 square centimetres.

The relationship h = 2r is exact, and you can see why algebraically: substituting r³ = 710/(4π) back into the constraint always produces a height equal to the diameter. Every optimal closed cylinder, at any volume, is as tall as it is wide.

Problem three: the open box

A rectangular sheet of card measures 20 cm by 30 cm. Equal squares are cut from the four corners and the flaps folded up to make an open box. What cut gives the largest volume?

1. Name. Let x be the side of each cut square, in centimetres. After folding, the box is x tall, 20 - 2x wide and 30 - 2x long, since a square is removed from each end of both dimensions.
2. Objective. V(x) = x(20 - 2x)(30 - 2x).
3. Domain. The cut must be positive and cannot exceed half the shorter side, so 0 < x < 10. At both ends the volume is 0, which is a good sign that a maximum lives inside.
4. Expand, then differentiate. (20 - 2x)(30 - 2x) = 600 - 100x + 4x², so V(x) = 600x - 100x² + 4x³ and V'(x) = 600 - 200x + 12x².
5. Solve. Divide by 4: 3x² - 50x + 150 = 0. The quadratic formula gives x = (50 ± √(2500 - 1800))/6 = (50 ± √700)/6, so x = 3.924 or x = 12.743.
6. Choose and justify. The second root is outside the domain, so discard it. On (0, 10), V' is positive before 3.924 and negative after, so it is a maximum.

Answer: cut squares of about 3.92 cm. The box is then 3.92 by 12.15 by 22.15 cm, with volume about 1056 cm³. Discarding the second root is not a technicality: a cut of 12.74 cm would remove more card than exists, and the algebra has no way of knowing that. The domain is what protects you.

Problem four, and a trick worth owning

Which point on the parabola y = x² is closest to (0, 3)?

The distance is D = √(x² + (x² - 3)²), and differentiating that root is unpleasant. Here is the move: minimising a distance is the same as minimising its square, because squaring is an increasing function on non-negative numbers, so both are smallest at the same x.

1. Minimise f(x) = x² + (x² - 3)² instead.
2. Let u = x², so f = u + (u - 3)² = u² - 5u + 9.
3. df/du = 2u - 5 = 0 gives u = 2.5, so x = ±√2.5 ≈ ±1.581.
4. Then f = 6.25 - 12.5 + 9 = 2.75, so the distance is √2.75 ≈ 1.658.

Two closest points, symmetric about the axis, at roughly (±1.581, 2.5). Note that the closest point is not the vertex (0, 0), which is 3 units away.

Why the justification step is not optional

A critical point can be a maximum, a minimum, or neither, and the problem never tells you which. Every year students find x = 600 for the fence, write down 720,000, and lose marks for never showing it was a maximum. Two acceptable justifications:

• Second derivative test, when the second derivative is easy and has a constant sign on the domain, as in the fence and the can.
• Candidates test, when the domain is a closed interval: evaluate at the critical points and both endpoints and compare.

When the domain is open, as with the can, the candidates test is unavailable and you must argue from concavity or from the sign of the first derivative on either side.

Common misconceptions

• "Set the derivative to zero and you are finished." You have found a candidate. Whether it is the answer, and whether it lies in the domain, are separate questions.
• "The constraint and the objective are interchangeable." They are not. The objective is what you differentiate; the constraint is what you substitute. Differentiating the constraint by mistake is the most common wrong turn in this topic.
• "Domains are a formality." The box problem produces a root of 12.74 cm from a 20 cm sheet. Only the domain rejects it.
• "The answer is the critical point." The question usually asks for dimensions, an area or a cost. Finish the arithmetic and attach units.
• "The closest point on a curve is directly below the target." Not generally. For the parabola and (0, 3), the vertex is a critical point of the distance function, but it is a local maximum of the distance rather than a minimum: it sits 3 units away while the true closest points, off to either side, sit about 1.66 away.

Recap

Optimisation is a translation problem wearing a calculus hat. Name the variables and draw the picture, write one equation for the quantity you want to make largest or smallest, use the constraint to reduce it to a single variable, state the domain, then differentiate and solve. Justify what you found: the second derivative when its sign is constant, the candidates test when the domain is closed. The fence answer, 600 m by 1200 m for 720,000 square metres, has the parallel side equal to half the fencing; the optimal closed can is exactly as tall as it is wide, at radius 3.84 cm for 355 mL; the best cut from a 20 by 30 sheet is about 3.92 cm, and the domain is what rejects the second root of the same quadratic. When a distance is involved, minimise its square. And finish by answering the question that was actually asked, in the units it was asked in.

Sources

1. OpenStax. (2016). 4.7 Applied optimization problems. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Optimization. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. OpenStax. (2016). 4.3 Maxima and minima. In Calculus Volume 1. openstax.org
4. Wikipedia. (n.d.). Mathematical optimization. en.wikipedia.org

The ladder that ends up moving infinitely fast

A 5 m ladder leans against a vertical wall. Someone drags the foot of it away from the wall at a steady 0.6 m/s. How fast is the top sliding down?

The answer is not a single number, because it changes as the ladder falls. Let x be the distance from the wall to the foot and y the height of the top. Pythagoras gives x² + y² = 25 at every instant. Differentiate both sides with respect to time, remembering that x and y are both functions of t:

2x (dx/dt) + 2y (dy/dt) = 0, so dy/dt = -(x/y)(dx/dt).

When the foot is 3 m out, the top is at y = √(25 - 9) = 4 m, and dy/dt = -(3/4)(0.6) = -0.45 m/s. The minus sign is the answer telling you the top is descending, which is worth noticing rather than discarding.

Now push the model. When the foot is 4.9 m out, y = √(25 - 24.01) = 0.995 m, and the top is dropping at -(4.9/0.995)(0.6) = -2.95 m/s. As y approaches 0 the ratio x/y runs to infinity, so the model says the top hits the floor at infinite speed. Physically that never happens, because a real ladder loses contact with the wall well before then and the assumption that the top stays touching stops being true. A model is a set of assumptions, and knowing where they fail is part of using it honestly.

Bottom line: a related rates problem is one equation, differentiated with respect to time, with the chain rule applied to every variable.

The procedure, and the one step people get backwards

1. Draw and label. Give a letter to every quantity that changes. Give a number only to what never changes.
2. Write the rates down in symbols. What you are given, and what you want, each as a derivative with respect to t: dx/dt = 0.6, find dy/dt.
3. Find one equation relating the variables. Usually Pythagoras, a volume or area formula, or a trigonometric ratio.
4. Differentiate the whole equation with respect to t. Every variable picks up its own rate factor.
5. Substitute the instantaneous values last, then solve for the unknown rate.
6. Answer with units and interpret the sign.

Step 5 is the one that goes wrong. Suppose you had substituted x = 3 and y = 4 into x² + y² = 25 before differentiating. You would be looking at 9 + 16 = 25, an equation with no variables in it, and differentiating it gives 0 = 0. The instant values are true at one moment; the relationship is true at every moment, and only something true at every moment has a derivative worth taking.

The salt pile

A council depot dumps road salt at 3 cubic metres per minute onto a conical pile whose height always stays equal to the diameter of its base. How fast is the pile getting taller when it is 4 m high?

1. Variables. Height h, base radius r, volume V, all functions of t. Given dV/dt = 3 m³/min, want dh/dt when h = 4.
2. Relation. A cone has V = (1/3)πr²h, but that is two variables. The condition "height equals diameter" says h = 2r, so r = h/2. Substitute: V = (1/3)π(h²/4)h = πh³/12.
3. Differentiate. dV/dt = (π/12)(3h²)(dh/dt) = (πh²/4)(dh/dt).
4. Substitute now. 3 = (π × 16/4)(dh/dt) = 4π(dh/dt).
5. Solve. dh/dt = 3/(4π) = 0.2387 m/min.

Using the constraint to eliminate r in step 2 is essential. Differentiating V = (1/3)πr²h as it stands produces an equation containing dr/dt, which you were never told and do not want.

Sanity check the size of the answer. The pile is 4 m tall, so its base radius is 2 m and its cross-sectional area at the base is about 12.6 square metres. Adding 3 cubic metres a minute over roughly that footprint should raise things by a few tens of centimetres a minute, and 24 cm is exactly that order.

The ripple

Drop a stone in still water. The circular ripple's radius grows at 0.5 m/s. How fast is the enclosed area growing when the radius is 4 m?

1. A = πr².
2. Differentiate: dA/dt = 2πr (dr/dt).
3. Substitute: dA/dt = 2π(4)(0.5) = 4π ≈ 12.57 square metres per second.

The interesting part is what the formula says about later instants. At r = 8 the same radial speed gives dA/dt = 8π, twice as fast. The edge is moving at a constant rate but there is more edge, so area accumulates faster and faster. Related rates problems are often worth reading for this kind of structural fact, not only for the number.

Two cars leaving one intersection

Car A leaves an intersection heading north at 60 km/h. At the same moment car B leaves heading east at 80 km/h. How fast is the distance between them increasing two hours later?

1. Variables. a is A's distance north, b is B's distance east, z is the separation. da/dt = 60, db/dt = 80, want dz/dt.
2. Relation. z² = a² + b².
3. Differentiate. 2z (dz/dt) = 2a (da/dt) + 2b (db/dt), so z (dz/dt) = a (da/dt) + b (db/dt).
4. Instant values. After 2 hours, a = 120, b = 160, and z = √(14400 + 25600) = √40000 = 200.
5. Solve. 200(dz/dt) = 120(60) + 160(80) = 7200 + 12800 = 20000, so dz/dt = 100 km/h.

That answer has a tidy interpretation: 100 is exactly √(60² + 80²). For two objects moving in fixed perpendicular directions from a common start, the separation grows at the magnitude of the combined velocity, and it does so at every instant, not just at two hours. Getting a clean number is often a hint that something structural is going on.

The tracking camera

A camera sits 900 m from a launch pad and tilts to keep a rising rocket in frame. The rocket is climbing at 300 m/s. How fast must the camera rotate when the rocket is 1200 m up?

1. Variables. h is the rocket's height, θ the camera's elevation angle. The 900 m is fixed, so it is a number, not a variable.
2. Relation. tan θ = h/900.
3. Differentiate with respect to t. The left side needs the chain rule: sec²θ (dθ/dt) = (1/900)(dh/dt).
4. Instant values. When h = 1200, the camera-to-rocket distance is √(900² + 1200²) = 1500, so sec θ = 1500/900 = 5/3 and sec²θ = 25/9.
5. Solve. (25/9)(dθ/dt) = 300/900 = 1/3, so dθ/dt = (9/25)(1/3) = 0.12 radians per second, about 6.9 degrees per second.

Radians again. If you had worked the trigonometry in degrees, the derivative of the tangent would have carried a factor of π/180 and the answer would have been wrong by a factor of about 57.

Reading the answer

Two habits separate a finished related rates solution from an unfinished one.

• Units. They come from the quantities: a volume rate is cubic metres per minute, an angular rate is radians per second. If your units do not come out right, an equation was set up wrongly.
• Sign. Negative means the quantity is decreasing. The ladder's top had dy/dt = -0.45 m/s; reporting 0.45 m/s without the sign discards the fact that it is falling.

Common misconceptions

• "Put the numbers in first, it saves work." It destroys the problem. Substituting the instantaneous values before differentiating turns a relationship into an arithmetic identity whose derivative is 0 = 0.
• "Every length in the picture is a variable." Only the ones that change. The camera's 900 m stand-off is fixed, and treating it as a variable adds a rate that does not exist.
• "The cone formula can be differentiated as it stands." Not unless you know dr/dt. Use the given shape constraint to reduce to one variable first, exactly as in an optimisation problem.
• "A constant rate in one quantity means a constant rate in another." The ripple's radius grows steadily while its area accelerates, and the ladder's foot moves steadily while its top accelerates downward without bound.
• "The model is the situation." The ladder result is genuinely infinite in the limit and genuinely false in a workshop. Knowing which assumption breaks, here that the top stays in contact with the wall, is part of the answer.

Looking back

Related rates problems all have the same skeleton: find one equation that ties the changing quantities together, differentiate it with respect to time so that every variable contributes its own rate through the chain rule, and only then put in the values that hold at the instant of interest. Reversing those last two steps is the classic failure, because instantaneous values are true at a moment and only a permanent relationship can be differentiated. Use any given shape constraint to reduce to a single variable before differentiating, as with the cone whose height equalled its diameter. The answers carry units and signs that mean something: 0.2387 m/min for a rising salt pile, 4π square metres per second for a spreading ripple, 100 km/h for two cars at right angles, 0.12 radians per second for a tracking camera. And when a model reports an infinite speed, the useful response is to ask which of its assumptions has just stopped being true.

Sources

1. OpenStax. (2016). 4.1 Related rates. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Related rates. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. OpenStax. (2016). 3.6 The chain rule. In Calculus Volume 1. openstax.org
4. Wikipedia. (n.d.). Related rates. en.wikipedia.org

Reading a car's speed off the skid marks

A car brakes hard and leaves 40 m of skid marks before stopping. On that road surface the deceleration is about 6 m/s². How fast was it going when the driver hit the brakes? Accident investigators answer this question routinely, and the reasoning is pure calculus run backwards.

1. Acceleration is known: a(t) = -6.
2. Velocity is a function whose derivative is -6. One such function is -6t, but so is -6t + 25 and -6t - 3. Write v(t) = -6t + v₀, where v₀ is the unknown initial speed.
3. Position is a function whose derivative is v: s(t) = -3t² + v₀t, taking s(0) = 0 at the start of the skid.
4. The car stops when v = 0, at t = v₀/6. Its distance then is -3(v₀²/36) + v₀²/6 = v₀²/12.
5. Set that to 40: v₀² = 480, so v₀ = 21.9 m/s, about 79 km/h.

Two things happened there. We reversed differentiation twice, and each time an unknown constant appeared which extra information later pinned down. Both are the subject of this lesson.

The reverse question, and its unavoidable ambiguity

An antiderivative of f is any function F with F'(x) = f(x). Antiderivatives are never unique, because adding a constant does not change a derivative: x², x² + 7 and x² - 1.5 all differentiate to 2x.

How much ambiguity is there? Exactly one constant's worth, and that is a theorem rather than an observation. Fact three from the mean value theorem lesson says that two functions with the same derivative on an interval differ by a constant. So if F'(x) = f(x), then every antiderivative of f has the form F(x) + C, and no others exist.

So what?: the "+ C" is not a formality. It is the precise statement of how much information about a function is lost when you differentiate it, and recovering that information is what an initial condition does.

The notation

We write

∫ f(x) dx = F(x) + C

and call it the indefinite integral of f. The elongated S was Leibniz's, chosen because the definite version in two lessons' time is a sum. The dx names the variable you are reversing with respect to, which matters as soon as more than one letter is in play: ∫ 3t dx = 3tx + C, treating t as a constant, while ∫ 3t dt = 1.5t² + C.

The table, which is the derivative table read right to left

Integral	Result	Because the derivative of the result is
∫ xⁿ dx, for n ≠ -1	xⁿ⁺¹/(n + 1) + C	xⁿ
∫ (1/x) dx	ln|x| + C	1/x
∫ eₓ dx	eₓ + C	eₓ
∫ cos x dx	sin x + C	cos x
∫ sin x dx	-cos x + C	sin x, since the minus signs cancel
∫ sec²x dx	tan x + C	sec²x
∫ sec x tan x dx	sec x + C	sec x tan x
∫ 1/(1 + x²) dx	arctan x + C	1/(1 + x²)

The reversed power rule is worth saying in words, because the order of operations flips: add one to the exponent, then divide by the new exponent. Differentiating multiplies by the old exponent and subtracts one; integrating adds one and divides by the new one.

Every line of that table can be checked in five seconds by differentiating the right-hand column, and you should form the habit. An antiderivative is the one kind of answer in mathematics that verifies itself.

Why 1/x is the exception

Apply the reversed power rule to x⁻¹ and you are asked to add one to the exponent, giving x⁰, and then divide by the new exponent, which is 0. That is why the rule carries the condition n ≠ -1. The gap is filled by a function from a completely different family: since d/dx[ln x] = 1/x, the natural logarithm is the missing antiderivative.

The absolute value in ln|x| is doing real work. The function 1/x is defined for negative x, but ln x is not, so without the absolute value the formula would cover only half the domain. Check the negative side: for x < 0, ln|x| = ln(-x), whose derivative by the chain rule is (-1)/(-x) = 1/x. It works.

What integration does and does not distribute over

Integration is linear: ∫ (f + g) dx = ∫ f dx + ∫ g dx, and constants pull out, ∫ cf dx = c ∫ f dx. That is all.

There is no product rule for integrals and no quotient rule, and the failure is not subtle. Take ∫ x · x² dx. The correct answer is ∫ x³ dx = x⁴/4 + C. Multiplying the separate integrals gives (x²/2)(x³/3) = x⁵/6, a different function of a different degree. When a product will not yield, the tools are algebraic rewriting, then substitution in Module 7, and beyond this course a technique called integration by parts.

Worked examples

• ∫ (6x² - 4x + 5) dx = 2x³ - 2x² + 5x + C. Check by differentiating: 6x² - 4x + 5.
• ∫ (3/x + 2eₓ) dx = 3ln|x| + 2eₓ + C.
• ∫ (√x + 1/x²) dx. Rewrite as ∫ (x to the power 1/2 + x⁻²) dx = (2/3)x to the power 3/2 - x⁻¹ + C.
• ∫ (5cos x - 2sec²x) dx = 5 sin x - 2 tan x + C.
• ∫ (x³ + 2x)/x dx. Divide first: ∫ (x² + 2) dx = x³/3 + 2x + C. Trying to integrate the quotient as it stands leads nowhere.
• ∫ tan²x dx. There is no entry for this, but the identity tan²x = sec²x - 1 converts it into two entries there are: tan x - x + C.

The last two are the pattern to internalise: when an integral is not in the table, the first question is whether algebra or an identity can move it into the table.

Pinning down C: initial value problems

One condition, one constant. Find f given f'(x) = 6x - 4 and f(2) = 5.

1. Integrate: f(x) = 3x² - 4x + C.
2. Impose the condition: f(2) = 12 - 8 + C = 5, so C = 1.
3. Answer: f(x) = 3x² - 4x + 1.

Two integrations, two constants. A ball is thrown upward at 20 m/s from a rooftop 45 m above the ground. Gravity gives a(t) = -9.8 m/s². When does it land, and how high does it get?

1. Integrate acceleration: v(t) = -9.8t + C₁. The condition v(0) = 20 gives C₁ = 20, so v(t) = -9.8t + 20.
2. Integrate velocity: s(t) = -4.9t² + 20t + C₂. The condition s(0) = 45 gives C₂ = 45.
3. Highest point is where v = 0: t = 20/9.8 = 2.041 s. Then s = -4.9(4.165) + 40.82 + 45 = 65.4 m above the ground.
4. Lands when s = 0: solve 4.9t² - 20t - 45 = 0, giving t = (20 + √1282)/9.8 = 5.69 s. The negative root is discarded, since the throw happened at t = 0.

Notice how the two constants entered: the first carried the initial velocity, the second the initial height. Each integration loses one piece of information and each condition restores one.

Common misconceptions

• "The + C is a convention you can drop." Drop it in an initial value problem and you will get the wrong function. It represents an entire family of curves, all parallel, and the condition selects one.
• "The antiderivative of 1/x is x⁰/0." The reversed power rule genuinely fails there, and the answer comes from a different family entirely: ln|x|.
• "The integral of a product is the product of the integrals." ∫ x · x² dx is x⁴/4, not x⁵/6.
• "Integrating is harder than differentiating because the rules are harder." The rules are the same rules. What is harder is that there is no procedure guaranteed to work, so integration is a matter of recognising forms, while differentiation is an algorithm.
• "The dx at the end is decoration." It names the variable. It also becomes essential machinery in substitution, where you will manipulate it directly.

The short version

An antiderivative of f is any function whose derivative is f, and the mean value theorem guarantees that all of them differ by a constant, which is exactly what the "+ C" in ∫ f(x) dx = F(x) + C records. The antiderivative table is the derivative table read backwards, with the reversed power rule saying to add one to the exponent and divide by the new one, and with 1/x as the single exception, filled by ln|x|. Integration is linear and nothing more: there is no product or quotient rule, so an integral that is not in the table must first be rewritten by algebra or a trigonometric identity. An initial condition pins down each constant, one per integration, which is how an accelerometer reading turns into a velocity and then into a position, and how 40 m of skid marks turns into a speed of 79 km/h.

Sources

1. OpenStax. (2016). 4.10 Antiderivatives. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Indefinite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. OpenStax. (2016). 5.4 Integration formulas and the net change theorem. In Calculus Volume 1. openstax.org
4. Wikipedia. (n.d.). Antiderivative. en.wikipedia.org

How far did the car travel?

A driver accelerates from rest onto a motorway slip road and a passenger reads the speedometer every ten seconds:

Time (s)	0	10	20	30	40	50	60
Speed (m/s)	0	8	15	19	22	24	25

How far did the car go in that minute? There is no formula for the speed, only seven readings, and speed multiplied by time gives distance only when the speed is constant. But it is nearly constant over ten seconds, so estimate.

• Assume the speed at the start of each interval held throughout it. Distance is about (0 + 8 + 15 + 19 + 22 + 24) × 10 = 880 m.
• Assume the speed at the end of each interval held throughout it. Distance is about (8 + 15 + 19 + 22 + 24 + 25) × 10 = 1130 m.

Because the car is speeding up, the first estimate is too small and the second too large, so the true distance is between 880 m and 1130 m. Averaging gives 1005 m, which is what the trapezoid estimate amounts to. And there is a clean way to make the bracket tighter: read the speedometer more often. That single sentence is the whole idea of integration.

Worth holding on to: multiplying a rate by a time and adding up the pieces gives a total, and the answer becomes exact as the pieces shrink.

The same picture as area

Plot speed against time and each of those products is the area of a rectangle: height is the speed, width is the ten seconds. The left estimate is a staircase of rectangles hugging the curve from below, the right estimate a staircase from above, and the true distance is the area under the speed graph.

Generalise. To find the area under y = f(x) from x = a to x = b:

1. Cut the interval into n equal pieces, each of width Δx = (b - a)/n.
2. On each piece choose a sample point: its left end, its right end, or its midpoint.
3. Build a rectangle of that height and width Δx.
4. Add the areas. That total is a Riemann sum, written Σ f(xᵢ) Δx with the sum running from i = 1 to n.

For an increasing function, left sums underestimate and right sums overestimate; for a decreasing function it is the other way round. Midpoint sums are usually much better than either, which is why they are worth the extra arithmetic.

Estimating, then computing exactly

Take f(x) = x² on [0, 3] with n = 6, so Δx = 0.5.

• Right endpoints are 0.5, 1, 1.5, 2, 2.5, 3, whose squares are 0.25, 1, 2.25, 4, 6.25, 9. The sum is 22.75, and multiplying by 0.5 gives 11.375.
• Left endpoints are 0, 0.5, 1, 1.5, 2, 2.5, whose squares total 13.75, giving 6.875.

So the area is between 6.875 and 11.375, a bracket 4.5 wide. Now do it properly by letting n run to infinity.

1. With n rectangles, Δx = 3/n and the right endpoint of the i-th is xᵢ = 3i/n.
2. The Riemann sum is Σ (3i/n)² (3/n) = (27/n³) Σ i², with the sum from 1 to n.
3. Use the standard formula Σ i² = n(n + 1)(2n + 1)/6.
4. Substitute: (27/n³) · n(n + 1)(2n + 1)/6 = 27(2n² + 3n + 1)/(6n²) = 4.5(2 + 3/n + 1/n²).
5. Let n → ∞. The 3/n and 1/n² terms vanish, exactly as in the limits-at-infinity work of Module 1, leaving 4.5 × 2 = 9.

The exact area is 9. Notice it sits between the two estimates, closer to the average of them, and notice how much work that was. Getting a single number required a summation identity and a limit. In two lessons the fundamental theorem will produce the same 9 in one line, which is the reason it is called fundamental.

The definition

The definite integral of f from a to b is the limit of its Riemann sums as the rectangles shrink:

∫(a to b) f(x) dx = lim(n → ∞) Σ f(xᵢ) Δx

It is a number, not a family of functions, so there is no constant of integration attached to it. If f is continuous on [a, b] the limit is guaranteed to exist and the choice of sample points makes no difference to it.

Area with a sign

The definite integral computes signed area. Where f is below the axis, f(xᵢ) is negative, so those rectangles contribute negative amounts.

∫(0 to 2π) sin x dx = 0, because the hump above the axis from 0 to π exactly cancels the identical hump below it from π to 2π. The area enclosed is not zero, it is 4, but the signed area is.

This is a feature, not a defect, and the speedometer example says why. If velocity is negative the car is reversing, and the integral of velocity gives displacement, the net change in position. To get the total distance driven you would integrate the speed, which is the absolute value of the velocity. Net and total are different questions, and the integral answers the net one unless you tell it otherwise.

The properties you will actually use

Property	Statement	Why
Zero width	∫(a to a) f = 0	no interval, no area
Reversal	∫(b to a) f = -∫(a to b) f	the widths change sign
Splitting	∫(a to c) f = ∫(a to b) f + ∫(b to c) f	areas of adjacent pieces add
Linearity	∫ (cf + g) = c∫ f + ∫ g	inherited from sums
Comparison	if f ≤ g on [a, b] then ∫ f ≤ ∫ g	shorter rectangles, smaller total

Two integrals you can do with geometry alone

Since the integral is area, any region whose area you already know can be integrated without calculus.

A semicircle. ∫(-3 to 3) √(9 - x²) dx. The curve y = √(9 - x²) is the upper half of the circle x² + y² = 9, so the region is a semicircle of radius 3. Its area is (1/2)π(3²) = 4.5π ≈ 14.14. Antiderivatives for that integrand are unpleasant; geometry takes ten seconds.

A trapezoid. ∫(0 to 4) (2x + 1) dx. The region under a straight line is a trapezoid with parallel sides f(0) = 1 and f(4) = 9 and width 4, so its area is ((1 + 9)/2)(4) = 20. Check it later against the antiderivative x² + x, which gives (16 + 4) - 0 = 20. The two methods must agree, and they do.

Archimedes got there first, without limits

Around 250 BCE, in Quadrature of the Parabola, Archimedes found the exact area of a parabolic segment: it is four thirds of the area of the largest triangle inscribed in it. His method was to fill the segment with infinitely many triangles and sum the resulting geometric series, an argument he stated with the care of someone who knew he was working without a theory of limits. It took nineteen centuries to build the theory that made his method routine, and the modern definition above is the finished product.

Common misconceptions

• "A definite integral is a function." An indefinite integral is a family of functions; a definite integral is a single number.
• "The integral gives the area, so it cannot be negative." It gives signed area. Regions below the axis count negatively, which is exactly what makes it the right tool for net change.
• "Left and right sums bracket the answer for every function." Only for monotonic ones. For a function that rises then falls, neither is guaranteed to be an underestimate.
• "More rectangles is just more accuracy." The limit is not an approximation getting better; it is the definition of the exact value.
• "You need an antiderivative to compute a definite integral." Not when the region is a shape whose area you know, such as the semicircle above.

Putting it together

A total accumulated from a varying rate is estimated by chopping the interval into pieces, multiplying the rate on each piece by its width, and adding. That is a Riemann sum, and geometrically it is a staircase of rectangles under a curve. For a monotonic function the left and right sums bracket the truth, and midpoints do better than both. Letting the number of rectangles run to infinity turns the estimate into the definite integral, a single number defined as that limit, which for x² on [0, 3] is exactly 9 by way of a summation formula and a limit at infinity. The integral measures signed area, so it gives displacement rather than distance when the rate goes negative, and it obeys splitting, reversal, linearity and comparison rules that all come straight from the sums. When the region is a triangle, trapezoid or semicircle, geometry evaluates it faster than any calculus will.

Sources

1. OpenStax. (2016). 5.1 Approximating areas. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 5.2 The definite integral. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). The definition of the definite integral. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Riemann sum. en.wikipedia.org

Nine, the hard way and the easy way

Last lesson, finding the area under y = x² from 0 to 3 took a partition, a summation identity for Σi², an algebraic simplification and a limit at infinity. The answer was 9.

Here it is again. An antiderivative of x² is x³/3. Evaluate it at the two ends and subtract: 27/3 - 0 = 9.

One line, no rectangles, no limit. Nothing about the two calculations looks related: one is about slopes and the other about areas, and they were pursued as separate subjects for two thousand years. The theorem that connects them is the reason calculus is a single subject rather than two, and this lesson is that theorem in both of its halves.

Part one: accumulate, then differentiate, and you are back where you started

Fix a continuous function f and a starting point a, and define the accumulation function

A(x) = ∫(a to x) f(t) dt

which is the signed area gathered from a up to a movable right edge at x. Note the variable of integration is written t so it cannot be confused with the limit x.

Part 1 of the fundamental theorem: A'(x) = f(x). The rate at which area accumulates is exactly the height of the curve at the moving edge.

Why? Push the edge from x to x + h. The extra area is a thin strip of width h whose height is about f(x), so A(x + h) - A(x) ≈ f(x) h. Divide by h and the strip's area becomes f(x); take the limit as the strip narrows and the approximation becomes equality, because f is continuous and so cannot change much across a shrinking strip.

A water meter makes it concrete. The meter's reading is total litres delivered, an accumulation. Its rate of change is the flow at that instant. Read it as a sentence and it is nearly obvious; write it as an equation and it is a theorem that took centuries.

Remember: differentiating an integral with a variable upper limit hands the integrand straight back.

Two worked cases:

• d/dx ∫(1 to x) √(t³ + 1) dt = √(x³ + 1). Nobody can write an elementary antiderivative for that integrand, and nobody needs to.
• d/dx ∫(0 to x²) sin t dt. The upper limit is a function, so this is a composition and the chain rule applies: the answer is sin(x²) · 2x.

Part two: how to actually evaluate an integral

Part 2 of the fundamental theorem: if F is any antiderivative of a continuous f on [a, b], then

∫(a to b) f(x) dx = F(b) - F(a).

The usual shorthand writes the antiderivative in brackets with the limits attached, and the phrase "evaluate from a to b" means substitute the top, substitute the bottom, subtract.

Two features are worth noticing. First, the constant of integration is irrelevant here: using F + C instead of F gives (F(b) + C) - (F(a) + C), and the C cancels. That is why definite integrals never carry a "+ C". Second, the word "any" is doing work: you may pick whichever antiderivative is most convenient, because they all give the same difference.

Five evaluations

Integral	Antiderivative	Value
∫(0 to 3) x² dx	x³/3	9 - 0 = 9
∫(1 to 4) (2x + 1) dx	x² + x	20 - 2 = 18
∫(0 to π) sin x dx	-cos x	1 - (-1) = 2
∫(1 to e) (1/x) dx	ln|x|	1 - 0 = 1
∫(0 to 1) eₓ dx	eₓ	e - 1 ≈ 1.718

The fourth row is worth a second look: the area under the reciprocal curve from 1 to e is exactly 1. That is not a coincidence, it is a definition in disguise, since e is precisely the number that makes it true.

Why "fundamental" is the right word

Two questions grew up separately. The tangent problem asked for the slope of a curve at a point, and Fermat and Descartes were working on it in the 1630s. The area problem asked for the region under a curve, and Archimedes had been working on it in the third century BCE. Isaac Barrow, Newton's teacher at Cambridge, saw in the 1660s that the two were geometrically linked. Newton and Leibniz turned that observation into a calculus, an actual set of rules that solved both problems at once.

What the theorem buys is not elegance but tractability. Computing a definite integral from the definition means evaluating a limit of sums, which for anything more complicated than x² is brutal or impossible. Part 2 converts it into finding one antiderivative and doing one subtraction. Every applied integral in the rest of this course, and in physics and statistics and engineering, is computed that way.

The net change theorem

Read Part 2 with f = F' and it says something you can use without ever thinking about area:

∫(a to b) F'(x) dx = F(b) - F(a)

The integral of a rate of change gives the net change. If r(t) is the flow into a tank in litres per minute, then ∫ r dt is litres. If C'(q) is marginal cost, then integrating it over a production range gives the additional cost.

Worked example. Water flows into a tank at r(t) = 20 - 0.5t litres per minute for the first 20 minutes. How much water enters?

1. An antiderivative of 20 - 0.5t is 20t - 0.25t².
2. At t = 20: 400 - 100 = 300. At t = 0: 0.
3. Net change: 300 litres.

Sanity check: the flow starts at 20 L/min and falls linearly to 10 L/min, averaging 15 L/min over 20 minutes, which is 300 litres. The integral of a linear rate is a trapezoid, and the two agree.

Displacement against distance, in full

A particle moves with velocity v(t) = t² - 4t metres per second for 0 ≤ t ≤ 5.

Displacement is the integral of velocity:

1. Antiderivative: t³/3 - 2t².
2. At 5: 125/3 - 50 = 41.667 - 50 = -8.333. At 0: 0.
3. Displacement is -8.333 m: the particle finishes 8.33 m behind where it started.

Total distance requires knowing where the velocity changes sign. Factor: v(t) = t(t - 4), negative on (0, 4) and positive on (4, 5). So split the integral at 4.

1. From 0 to 4: 64/3 - 32 = -10.667. The particle travelled 10.667 m backwards.
2. From 4 to 5: -8.333 - (-10.667) = 2.333 m forwards.
3. Total distance: 10.667 + 2.333 = 13 m.

Displacement −8.33 m, distance travelled 13 m. Both are correct answers to different questions, and the split at the sign change is what separates them.

Average value

The average value of f on [a, b] is

favg = (1/(b - a)) ∫(a to b) f(x) dx

which is just the total divided by the width, exactly as an average of numbers is their total divided by how many there are. For x² on [0, 3] the average value is 9/3 = 3.

There is a companion theorem: for a continuous function, the average value is actually attained somewhere. Here we need c² = 3, so c = √3 ≈ 1.732, which lies inside [0, 3]. Geometrically, the rectangle of height 3 over [0, 3] has exactly the same area, 9, as the region under the parabola.

Common misconceptions

• "The two parts are the same statement." Part 1 differentiates an integral and returns the integrand; Part 2 evaluates an integral using an antiderivative. One says the operations undo each other in one order, the other in the other order.
• "A definite integral needs a + C." It cancels in the subtraction. Only indefinite integrals carry it.
• "The integral of velocity is the distance travelled." It is the displacement. Distance requires splitting at every sign change and adding the magnitudes, which is why the particle above has a displacement of −8.33 m and a distance of 13 m.
• "You must use the simplest antiderivative." Any one works, because they differ by a constant that vanishes on subtraction.
• "Part 1 is useless because you can just integrate first." Often you cannot. There is no elementary antiderivative for √(t³ + 1), and Part 1 still hands you the derivative of its accumulation function instantly.

What you now know

Part 1 of the fundamental theorem says that differentiating an accumulation function returns the integrand, so the rate at which area piles up is the height of the curve at the moving edge; with a function in the upper limit, the chain rule supplies an extra factor. Part 2 says that any antiderivative evaluates a definite integral by a single subtraction, which is why the constant of integration disappears and why an infinite limiting process collapses into arithmetic. Together they say differentiation and integration are inverse operations, joining two problems that had been separate since Archimedes. Read as the net change theorem, Part 2 turns any rate into a total: 300 litres from a declining flow, a displacement of −8.33 m from a velocity that changes sign, and, when the sign changes are handled separately, a distance of 13 m instead. Dividing an integral by the width of the interval gives the average value of the function, which a continuous function is guaranteed to attain somewhere inside.

Sources

1. OpenStax. (2016). 5.3 The fundamental theorem of calculus. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 5.4 Integration formulas and the net change theorem. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). Computing definite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Fundamental theorem of calculus. en.wikipedia.org

Reading the chain rule backwards

Differentiate sin(x²). The chain rule gives cos(x²) · 2x. So without doing any work at all we know

∫ 2x cos(x²) dx = sin(x²) + C.

Now cover that up and look only at the integral. Could you have found the answer? Nothing in the antiderivative table matches 2x cos(x²). What you would have to notice is that the integrand contains a composition, cos of something, multiplied by the derivative of that something. That pattern is the signature of a chain rule, and substitution is the technique for undoing it.

In short: substitution is the chain rule run in reverse, and the skill is spotting the inside function whose derivative is already sitting in the integrand.

The rule

If u = g(x), then du = g'(x) dx, and

∫ f(g(x)) g'(x) dx = ∫ f(u) du.

That statement about du is where the dx stops being decoration. Treating du/dx = g'(x) as though you could multiply through by dx is not literally division, but it is a legitimate and provable manoeuvre, and it is what makes the bookkeeping work.

The procedure, every time:

1. Choose u: usually the inside of a composition, the thing under a root, the exponent, or the denominator.
2. Compute du = g'(x) dx.
3. Replace every x in the integral, including the dx. If any x survives, either fix it or change the substitution.
4. Integrate in u.
5. Substitute back to x, unless it is a definite integral whose limits you converted.

Five, worked

One: the clean case. ∫ 2x cos(x²) dx. Let u = x², so du = 2x dx. The integral becomes ∫ cos u du = sin u + C = sin(x²) + C.

Two: a power. ∫ 3x²(x³ + 5)⁴ dx. Let u = x³ + 5, so du = 3x² dx. Then ∫ u⁴ du = u⁵/5 + C = (x³ + 5)⁵/5 + C. Verify by differentiating: 5(x³ + 5)⁴(3x²)/5 = 3x²(x³ + 5)⁴. Correct.

Three: adjusting a constant. ∫ x/(x² + 1) dx. Let u = x² + 1, so du = 2x dx. The integrand has x dx, not 2x dx, but the difference is only a constant: x dx = du/2. So the integral is (1/2)∫ du/u = (1/2)ln|u| + C = (1/2)ln(x² + 1) + C. The absolute value can be dropped here because x² + 1 is always positive.

Four: a linear inside. ∫ e⁵ₓ dx. Let u = 5x, so du = 5 dx and dx = du/5. The integral is (1/5)∫ eⁱ du = (1/5)e⁵ₓ + C. Linear substitutions are so common that many people do them by inspection, but write them out until the constant never goes astray.

Five: one you should know. ∫ tan x dx. Write it as ∫ (sin x)/(cos x) dx and let u = cos x, so du = -sin x dx, which means sin x dx = -du. The integral becomes -∫ du/u = -ln|u| + C = -ln|cos x| + C, usually written ln|sec x| + C. The choice of the denominator as u is the one to remember: when a fraction has the derivative of its bottom sitting on top, a logarithm is coming.

When a stray x refuses to leave

Not every substitution clears every x, and the fix is to solve for x and substitute that too.

Evaluate ∫ x²√(x + 1) dx. Let u = x + 1, so du = dx and, crucially, x = u - 1.

1. Rewrite: ∫ (u - 1)² √u du.
2. Expand: (u² - 2u + 1)u to the power 1/2, which is u to the 5/2 minus 2u to the 3/2 plus u to the 1/2.
3. Integrate term by term with the reversed power rule: (2/7)u to the 7/2, minus (4/5)u to the 5/2, plus (2/3)u to the 3/2.
4. Back-substitute u = x + 1.

Nothing subtle happened there, but it is worth seeing once so that a leftover x does not look like a dead end. If the substitution can be inverted, use the inverse.

Definite integrals: change the limits or change back

Evaluate ∫(0 to 2) x e to the power x², dx.

Method A, change the limits. Let u = x², so du = 2x dx and x dx = du/2. When x = 0, u = 0; when x = 2, u = 4. The integral becomes

(1/2)∫(0 to 4) eⁱ du = (1/2)(e⁴ - e⁰) = (1/2)(54.598 - 1) = 26.799.

Method B, back-substitute first. Find the antiderivative (1/2)e to the x², then evaluate from 0 to 2, giving the same number.

Either is fine. What is never fine is changing the variable and keeping the old limits: writing (1/2)∫(0 to 2) eⁱ du would evaluate to (1/2)(e² - 1) = 3.19, which is not the answer to anything. The limits belong to the variable.

When substitution simply does not work

Substitution needs the derivative of the inside function to be present, up to a constant factor. Compare:

Integral	Works?	Why
∫ x cos(x²) dx	yes	u = x² needs x dx, which is there
∫ cos(x²) dx	no	the missing factor of x cannot be manufactured; this integral has no elementary antiderivative at all
∫ x² e to the x³ dx	yes	u = x³ needs x² dx, which is there
∫ x e to the x³ dx	no	the wrong power of x is present

Missing a constant is fine, because constants move in and out of integrals freely. Missing a variable factor is fatal, because x cannot be pulled out of an integral. That distinction is the whole test, and it is worth applying before you start rather than after three lines of algebra.

A symmetry shortcut worth having

If f is odd, meaning f(-x) = -f(x), then ∫(-a to a) f(x) dx = 0, because the signed area on the left cancels the signed area on the right. So ∫(-2 to 2) x³ dx = 0 without any computation. If f is even, meaning f(-x) = f(x), then ∫(-a to a) f = 2∫(0 to a) f, which halves the work. Checking symmetry before integrating occasionally saves the whole problem.

Common misconceptions

• "You can substitute for anything." Only when the derivative of your chosen inside is present as a factor, up to a constant.
• "Constants and variables can both be adjusted." A missing 2 is fixed by writing du/2. A missing x cannot be fixed at all, because variables do not pass through an integral sign.
• "Keep the original limits after substituting." The limits are values of the variable. Change the variable, change the limits, or convert back before evaluating.
• "Forgetting to convert dx is a small slip." It changes the integral. Every dx must become a du.
• "If substitution fails, the integral is unsolvable." Sometimes true, as with cos(x²), and sometimes not: another substitution, an identity, or a technique beyond this course may work.

Pulling it together

Substitution reverses the chain rule. Choose u as the inside of a composition, compute du = g'(x)dx, replace every x including the dx, integrate in u, and convert back. A missing constant factor is adjusted freely, a missing variable factor cannot be, and that single test tells you in advance whether the method applies: ∫ x cos(x²)dx yields at once while ∫ cos(x²)dx has no elementary antiderivative. When a stray x survives, invert the substitution and replace it too. For a definite integral, either change the limits along with the variable or convert back before evaluating, but never mix the two. And check for odd or even symmetry first, since an odd function over a symmetric interval integrates to zero before you write anything down.

Sources

1. OpenStax. (2016). 5.5 Substitution. In Calculus Volume 1. openstax.org
2. Dawkins, P. (n.d.). Substitution rule for indefinite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. OpenStax. (2016). 5.6 Integrals involving exponential and logarithmic functions. In Calculus Volume 1. openstax.org
4. Wikipedia. (n.d.). Integration by substitution. en.wikipedia.org

Kepler and the wine merchant

In 1615 Johannes Kepler published Nova stereometria doliorum vinariorum, a book on measuring the volume of wine casks. He had bought wine for his wedding and watched the merchant gauge the barrel by pushing a rod in diagonally through the bung hole and reading off a mark. Kepler wanted to know whether that could possibly be right, and to find out he cut the barrel, on paper, into thin circular slices and added up their volumes.

That is the whole method of this lesson. A hard area becomes a sum of thin strips; a hard volume becomes a sum of thin slices; and in both cases the sum becomes an integral. Kepler had to invent his own arguments for each shape. With the fundamental theorem, we get the general procedure and one formula.

What matters here: do not memorise four formulas. Write down the area or volume of one thin slice, then integrate it.

Area between two curves

If f(x) ≥ g(x) throughout [a, b], the area of the region between them is

A = ∫(a to b) [f(x) - g(x)] dx.

Each thin vertical strip has height f(x) - g(x), the top curve minus the bottom curve, and width dx. Notice this handles regions below the axis with no special care: if both curves are negative, the difference of the heights is still the true height of the strip.

Worked example. Find the area enclosed by y = x² and y = x + 2.

1. Find the intersections first. Set x² = x + 2, so x² - x - 2 = 0, which factors as (x - 2)(x + 1) = 0. The limits are x = -1 and x = 2. Without this step you have no interval.
2. Decide which is on top. Test a point inside, say x = 0: the line gives 2, the parabola gives 0. The line is above.
3. Integrate the difference. A = ∫(-1 to 2) (x + 2 - x²) dx = [x²/2 + 2x - x³/3] evaluated from −1 to 2.
4. At 2: 2 + 4 - 8/3 = 3.333. At −1: 0.5 - 2 + 1/3 = -1.167. Subtract: 3.333 + 1.167 = 4.5.

The area is 9/2. A rough check: the region is roughly 3 wide and averages perhaps 1.5 tall, so 4.5 is the right size.

When the curves swap places

If the curves cross inside the interval, the top curve is not the top curve everywhere, and integrating a single difference would let the two parts cancel. Split at every crossing.

Find the total area between y = x³ and y = x on [-1, 1]. They meet where x³ = x, that is at −1, 0 and 1.

• On (0, 1), test x = 0.5: x = 0.5 beats x³ = 0.125, so the line is on top. The integral of x - x³ from 0 to 1 is 1/2 - 1/4 = 1/4.
• On (-1, 0), test x = -0.5: x³ = -0.125 beats x = -0.5, so the cubic is on top. By symmetry that piece is also 1/4.

Total area 1/2. Had you integrated x - x³ straight across from −1 to 1 you would have got 0, because the two pieces have opposite signs. The integral was not wrong; it answered a different question, the signed one.

Turning the picture on its side

Sometimes horizontal strips are far easier than vertical ones. Find the area bounded by x = y² and x = y + 2.

Solved for y, the first curve is two branches and you would need two integrals. Left as functions of y, it is one. Intersections: y² = y + 2 gives y = -1 and y = 2. For y between them the line is to the right, so the strip width is right minus left:

A = ∫(-1 to 2) (y + 2 - y²) dy = 4.5,

the same arithmetic as before because it is the same integral with a different letter. The rule of thumb: if the region is bounded above and below by single curves, integrate in x; if it is bounded left and right, integrate in y.

Volumes by revolution: the disk method

Take a region under y = f(x) and spin it around the x-axis. Each thin vertical strip sweeps out a thin disk: a circle of radius f(x) and thickness dx, so of volume π[f(x)]² dx. Add them up:

V = π ∫(a to b) [f(x)]² dx.

Worked example. Rotate the region under y = √x from 0 to 4 about the x-axis. V = π∫(0 to 4) x dx = π[x²/2] = π(8) = 8π ≈ 25.13.

Check the method against something known. Rotate the line y = (r/h)x from x = 0 to x = h about the x-axis and you should get a cone of radius r and height h.

V = π∫(0 to h) (r²/h²)x² dx = π(r²/h²)(h³/3) = (1/3)πr²h.

That is exactly the cone formula from geometry, derived rather than remembered. A method that reproduces the answers you already trust is a method you can use on shapes you do not.

The washer method

When the region does not touch the axis of rotation, each slice is an annulus, a disk with a hole. Subtract the hole:

V = π ∫(a to b) ([R(x)]² - [r(x)]²) dx

where R is the distance from the axis to the far edge and r the distance to the near edge.

Worked example. Rotate the region between y = 2x and y = x² about the x-axis.

1. Intersections: x² = 2x gives x = 0 and x = 2.
2. Between them the line is above the parabola (test x = 1: 2 against 1), so R = 2x and r = x².
3. V = π∫(0 to 2) (4x² - x⁴) dx = π[4x³/3 - x⁵/5] from 0 to 2.
4. At 2: 32/3 - 32/5 = (160 - 96)/15 = 64/15. So V = 64π/15 ≈ 13.40.

The single most common error here is computing π∫(R - r)² dx instead of π∫(R² - r²) dx. Those are different: the first squares a difference, the second subtracts two squares. Only the second matches the geometry, because you are subtracting the area of the inner circle from the area of the outer one.

Rotating about some other line

The formulas never change; only the radii do. The radius is always the distance from the axis of rotation to the edge of the region.

Rotate the region under y = √x from 0 to 4 about the line y = -1. Every point has moved one unit further from the axis, so the outer radius is √x + 1 and the inner radius is 1, the gap between the axis and the x-axis.

1. V = π∫(0 to 4) [(√x + 1)² - 1²] dx.
2. Expand: (x + 2√x + 1) - 1 = x + 2√x.
3. V = π[x²/2 + (4/3)x to the power 3/2 ] from 0 to 4 = π(8 + 32/3) = 56π/3 ≈ 58.64.

Compare with the 8π ≈ 25.13 from rotating the same region about the x-axis. Moving the axis further away more than doubled the volume, which is what you would expect, since volume depends on the square of the radius.

Slices that are not circles

Nothing in the slicing argument requires a solid of revolution. If you know the area A(x) of the cross-section at each x, the volume is ∫ A(x) dx.

Worked example. A solid has as its base the disk x² + y² = 4, and every cross-section perpendicular to the x-axis is a square.

1. At position x, the base runs from y = -√(4 - x²) to y = +√(4 - x²), so the side of the square is 2√(4 - x²).
2. Cross-sectional area: [2√(4 - x²)]² = 4(4 - x²).
3. V = ∫(-2 to 2) 4(4 - x²) dx = 4[4x - x³/3] from −2 to 2.
4. At 2: 8 - 8/3 = 16/3. At −2: -16/3. Difference 32/3, times 4 gives 128/3 ≈ 42.67.

No circles appear anywhere in that answer, and no π either, which is a good sign that the setup was right.

Common misconceptions

• "Integrate the difference across the whole interval and you get the area." Only if one curve stays on top. Otherwise the pieces cancel, as with x and x³ on [-1, 1], which gives 0 instead of 1/2.
• "You can skip finding the intersections." They are the limits of integration. Without them there is no integral to write.
• "The washer integrand is (R - r) squared." It is R² - r². The two agree only when r is zero.
• "Regions below the axis need absolute values." Not for area between curves. The top curve minus the bottom curve is positive wherever the labelling is right, even if both are negative.
• "Rotating about a different line needs a different formula." Same formula, new radii, measured from the new axis.

What to carry forward

Every problem in this lesson is the same move: slice the region, write down the area or volume of one thin slice, and integrate. For area between curves that slice is a strip of height top minus bottom, so find the intersections first, decide which curve is above by testing an interior point, and split the integral wherever they trade places. Choose horizontal strips and integrate in y when the region is bounded left and right rather than above and below. For a solid of revolution the slice is a disk of volume πR²dx, or a washer of volume π(R² - r²)dx when the region stands off the axis, and the method proves the cone formula as a special case. Rotating about a line other than an axis changes only the radii. And when the cross-sections are squares or triangles rather than circles, the same argument applies with a different area formula, which is where Kepler's barrels came from.

Sources

1. OpenStax. (2016). 6.1 Areas between curves. In Calculus Volume 1. openstax.org
2. OpenStax. (2016). 6.2 Determining volumes by slicing. In Calculus Volume 1. openstax.org
3. Dawkins, P. (n.d.). Area between curves. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Dawkins, P. (n.d.). Volumes of solids of revolution: rings. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Wikipedia. (n.d.). Solid of revolution. en.wikipedia.org

A body in the ice

In September 1991 two hikers found a corpse emerging from a melting glacier in the Oetztal Alps, on the border between Austria and Italy. The body turned out to be about 5,300 years old. The dating came from carbon-14, and the arithmetic is short: with a half-life of 5730 years, the decay constant is k = (ln 2)/5730 = 1.2097 × 10⁻⁴ per year, so after 5300 years the surviving fraction is

e to the power (-1.2097 × 10⁻⁴ × 5300) = e to the power -0.6411 = 0.527.

About 53 percent of the original carbon-14 is still there. Run it backwards and a measured fraction gives an age: t = -ln(0.527)/k = 0.6407/0.00012097 = 5296 years.

Everything in that calculation rests on the formula N = N₀e to the power -kt. Where does that formula come from? Not from a table. It is the solution of a differential equation, and this lesson derives it.

What a differential equation is

A differential equation is an equation containing a derivative, and a solution is a function, not a number. That difference is worth pausing on. Solving x² = 9 gives you two numbers; solving dy/dx = 2y gives you a family of functions.

Checking a proposed solution is always available and always cheap: substitute. Does y = 3e²ₓ solve y' = 2y? Differentiate the candidate: y' = 6e²ₓ. Compute twice the candidate: 2y = 6e²ₓ. They match, so yes. And so does y = 7e²ₓ, and y = -0.4e²ₓ, which is the family.

Bottom line: a differential equation describes how something changes; solving it recovers the thing itself, up to constants that an initial condition supplies.

Separable equations

An equation is separable if it can be arranged so that every y sits on one side with the dy and every x on the other with the dx:

dy/dx = f(x)g(y) becomes dy/g(y) = f(x) dx.

Then integrate both sides. Manipulating dy and dx like that is a shorthand for a substitution argument, and it is legitimate; if it feels like sleight of hand, notice that every answer it produces can be checked by differentiating, and they all check out.

Worked example one. Solve dy/dx = 2xy.

1. Separate: dy/y = 2x dx.
2. Integrate both sides: ln|y| = x² + C.
3. Exponentiate: |y| = e to the power x² + C, which is eᶜ times e to the power x².
4. Absorb the constant: write A for ±eᶜ, giving y = A e to the power x².

Check by differentiating: y' = A e to the x² times 2x = 2xy. Correct. Note that A = 0, meaning y = 0, also solves the original equation, and it slipped through the division in step 1. Dividing by y assumed y was not zero, so constant-zero solutions must be checked separately.

Worked example two: where the exponential law comes from

Solve dy/dt = ky, the statement that a quantity changes at a rate proportional to how much of it there is.

1. Separate: dy/y = k dt.
2. Integrate: ln|y| = kt + C.
3. Exponentiate: y = A eᵏᵗ.
4. Apply the initial condition y(0) = y₀: at t = 0 the exponential is 1, so A = y₀.

y = y₀eᵏᵗ. Positive k is growth, negative k is decay, and the carbon-14 formula is this with k = -(ln 2)/5730. The exponential function was not chosen because it fits the data. It is forced: it is the only shape a quantity can take if its rate is proportional to its size.

Worked example three: an initial value problem

Solve dy/dx = x/y with y(0) = 3.

1. Separate: y dy = x dx.
2. Integrate: y²/2 = x²/2 + C.
3. Multiply by 2 and rename the constant: y² = x² + K.
4. Apply the condition: at x = 0, y = 3, so 9 = 0 + K and K = 9.
5. Solve for y, taking the positive root because the initial value is positive: y = √(x² + 9).

The general solution y² - x² = K is a family of hyperbolas; the initial condition picks one branch of one of them. That is the usual shape of these problems: a family from the integration, a single member from the condition.

Newton's law of cooling

A hot object loses heat at a rate proportional to how much hotter it is than its surroundings:

dT/dt = -k(T - Tₛ), where Tₛ is the surrounding temperature.

Substituting u = T - Tₛ turns this into du/dt = -ku, which we have already solved. So u = u₀e to the power -kt, and

T(t) = Tₛ + (T₀ - Tₛ)e to the power -kt.

Worked example. Coffee at 90 °C is left in a 20 °C room and cools to 60 °C in ten minutes. What is its temperature after twenty minutes?

1. Excess over room temperature: initially 90 - 20 = 70, after ten minutes 60 - 20 = 40.
2. So 40 = 70 e to the power -10k, giving e to the power -10k = 4/7 = 0.5714.
3. Take logs: -10k = ln(0.5714) = -0.5596, so k = 0.05596 per minute.
4. At twenty minutes the excess is 70 e to the power -1.1192 = 70 × 0.3265 = 22.86.
5. Temperature: 20 + 22.86 = 42.9 °C.

There is a shortcut worth noticing. Each ten-minute interval multiplies the excess by 4/7, so after twenty minutes the excess is 70 × (4/7)² = 70 × 16/49 = 22.86, the same answer with no logarithms at all. Exponential decay always works that way: equal time intervals multiply by equal factors.

Half-life and doubling time

Setting y = 2y₀ in y = y₀eᵏᵗ gives eᵏᵗ = 2, so the doubling time is t = (ln 2)/k. The same algebra with 1/2 gives the half-life, which is why the two formulas look identical.

Worked example. A bacterial culture doubles every 3 hours. By what factor does it grow in 12 hours?

1. k = (ln 2)/3 = 0.2310 per hour.
2. Growth factor over 12 hours: e to the power (0.2310 × 12) = e to the power 2.7726 = 16.
3. And 16 is 2⁴, which is exactly four doublings in twelve hours. The calculus agrees with the counting, as it must.

Reading the equation without solving it

A differential equation assigns a slope to every point of the plane. Sketch short segments with those slopes and you have a slope field; any solution curve must run tangent to the segments it passes through. For dy/dx = 2xy, the slope is 0 all along both axes and steepens away from them, and the solution curves are the bell-like shapes y = Ae to the x² flaring upward and downward according to the sign of A. You can read the qualitative behaviour off the field before doing any integration, which is often the fastest way to sanity-check an answer.

Where exponential models stop being true

The bacterial culture above would reach a mass of 16 in twelve hours, 256 in a day, and roughly ten million times its starting size in three days. No culture does that, because nutrients run out. The honest fix replaces dP/dt = kP with the logistic equation dP/dt = kP(1 - P/M), where M is the carrying capacity. When P is small the bracket is close to 1 and growth is nearly exponential; as P approaches M the bracket approaches 0 and growth stalls. The solution is an S-shaped curve rather than a runaway one.

Exponential growth is the correct model for the early phase of almost everything and the wrong model for the late phase of almost everything. Knowing which phase you are in is the modelling judgement that the mathematics cannot make for you.

Common misconceptions

• "The solution of a differential equation is a number." It is a function, or a family of them, and an initial condition selects one member.
• "Every differential equation is separable." Most are not. dy/dx = x + y cannot be separated, and needs a method beyond this course.
• "You only need one constant of integration, on one side." One is enough, because a constant on each side can be combined into a single one, but it must not be forgotten entirely.
• "Dividing by y is harmless." It discards the constant solution y = 0. Check for such solutions separately.
• "Exponential growth means fast growth." It means proportional growth. With a small k it is slow for a long time and then abruptly enormous, which is precisely what makes it hard to see coming.

Where this leaves us

A differential equation is a statement about a rate, and its solution is a function. Separable equations are the ones that can be arranged with all the y and dy on one side and all the x and dx on the other, and integrating both sides then produces a family of solutions from which an initial condition selects one. Applying that method to dy/dt = ky derives the exponential law y = y₀eᵏᵗ rather than assuming it, and shows that proportional rate of change and exponential shape are the same statement. From there, radiocarbon dating is one substitution, Newton's law of cooling is one change of variable, and half-lives and doubling times are both (ln 2)/k. Slope fields let you see the family before you compute it. And when growth is bounded by resources, the exponential model must give way to the logistic one, which is a modelling decision rather than a mathematical one.

Sources

1. OpenStax. (2016). 4.3 Separable equations. In Calculus Volume 2. openstax.org
2. OpenStax. (2016). 4.1 Basics of differential equations. In Calculus Volume 2. openstax.org
3. OpenStax. (2016). 6.8 Exponential growth and decay. In Calculus Volume 1. openstax.org
4. Dawkins, P. (n.d.). Separable differential equations. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Wikipedia. (n.d.). Radiocarbon dating. en.wikipedia.org