Lesson 12: The Chain Rule

(a) Opening Question

You already know that d/dx[x⁵] = 5x⁴. Now try this one:

Find d/dx[(x² + 1)⁵].

Tempting move: "It's something to the 5th power, so the derivative is 5(x² + 1)⁴." Write that down — and then test it. If that were the whole answer, then squaring-then-raising would behave exactly like raising a bare variable. But (x² + 1)⁵ is a function inside another function: you take x, run it through g(x) = x² + 1, and then run that result through the 5th-power machine.

Multiply out a smaller version, (x² + 1)²= x⁴ + 2x² + 1, and differentiate honestly: 4x³ + 4x = (2)(x² + 1)(2x). Notice the extra factor of 2x — exactly the derivative of the inside. That stray factor is the entire point of this lesson. The 5th-power version will need it too: the answer is 5(x² + 1)⁴ · 2x. Today you'll learn the rule that produces that inside factor automatically, every time.

(b) Core Concepts

The composition you're differentiating

A composite function is a function plugged into another function, written f(g(x)) or (f ∘ g)(x). Here g is the inner (inside) function — it touches x first — and f is the outer (outside) function — it acts on whatever g produces. For (x² + 1)⁵, the inside is g(x) = x² + 1 and the outside is f(u) = u⁵.

The rule

Chain Rule. If f and g are differentiable, then d/dx[f(g(x))] = f'(g(x)) · g'(x).

In words: the derivative of the outside (with the inside left alone) times the derivative of the inside. People also call this the outside-inside method. In Leibniz notation, if y = f(u) and u = g(x), then

dy/dx = dy/du · du/dx

and the du's "cancel" as a memory aid (they aren't really fractions, but the cancellation pattern is reliable). The single most important habit: after differentiating the outer function, you are not done — you must multiply by the derivative of the inner function. Forgetting that inside factor is the #1 derivative error on the AP exam.

Layer-peeling: the method that never fails

Treat a composite like an onion and peel one layer at a time, working outside in:

1. Identify the outermost operation — the last thing you'd do if you evaluated by hand.
2. Differentiate that outer operation, leaving everything inside it untouched.
3. Multiply by the derivative of what was inside.
4. If the inside is itself a composite, repeat on it.

Quick example. For sin(x²): the last thing you do to x is take the sine, so the outer function is sine. d/dx[sin(▢)] = cos(▢) · (▢)', giving cos(x²) · 2x = 2x cos(x²). The 2x is the derivative of the inside x² — don't lose it.

The general power rule

The most common chain-rule pattern combines it with the power rule. Memorize this special case:

General Power Rule. d/dx[(g(x))ⁿ] = n·(g(x))ⁿ⁻¹ · g'(x).

Multiply down the exponent, drop the power by one, then multiply by the derivative of the base. So d/dx[(x² + 1)⁵] = 5(x² + 1)⁴ · 2x = 10x(x² + 1)⁴. The final · 2x is the inside derivative the Opening Question warned you about. This pattern also covers roots and reciprocals once you rewrite them as powers: √(x² + 1) = (x² + 1)^(1/2) and 1/(x² + 1)³ = (x² + 1)⁻³.

Chain rule with trig functions

Each trig derivative gets a chain-rule companion. The pattern is always (derivative of the trig function, evaluated at the inside) times (derivative of the inside):

d/dx[sin(g)] = cos(g)·g'       d/dx[cos(g)] = −sin(g)·g'
d/dx[tan(g)] = sec²(g)·g'      d/dx[sec(g)] = sec(g)tan(g)·g'

Watch the notation trap: sin²x means (sin x)², an outer squaring of an inner sine, while sin(x²) is an outer sine of an inner squaring. They differentiate completely differently — see Common Mistake 4.

Worked example: a multi-layer (3-function) composite

Differentiate y = sin(√(x² + 1)). Peel from the outside in — there are three layers: sine (outer), square root (middle), and x² + 1 (inner).

• Outer is sine: cos(√(x² + 1)), leaving the inside alone, times the derivative of √(x² + 1).
• That derivative is itself a general-power-rule job: (x² + 1)^(1/2) differentiates to ½(x² + 1)^(−1/2) · 2x = x/√(x² + 1).

Multiplying the layers:

y' = cos(√(x² + 1)) · x/√(x² + 1)

Every nested layer contributes one factor. (sympy confirms y' = x·cos(√(x²+1))/√(x²+1).)

Worked example: a trig composite

Differentiate y = cos³x. Rewrite to see the layers: y = (cos x)³, an outer cube of an inner cosine. General power rule: 3(cos x)² · (cos x)' = 3cos²x · (−sin x) = −3 sin x cos²x.

Worked example: chain rule from a table of values

You will not always have a formula. The AP exam loves giving a table and asking for a composite's derivative at a point.

Let h(x) = f(g(x)). Find h'(1).

By the chain rule, h'(1) = f'(g(1)) · g'(1). Read the table: g(1) = 2, so f'(g(1)) = f'(2) = −3; and g'(1) = 4. Therefore

h'(1) = f'(g(1))·g'(1) = f'(2)·g'(1) = (−3)(4) = −12.

The discipline that earns the point: evaluate the inside first (g(1) = 2), use it as the input to f', then multiply by g'(1). Mixing up which value feeds which function is the classic table-problem error.

(c) Worked Examples

Example 1 — General power rule [NO CALC]

Problem. Differentiate y = (3x² − 5x + 1)⁴.

Strategy. Outer = 4th power, inner = 3x² − 5x + 1. Use the general power rule.

Solution.

y' = 4(3x² − 5x + 1)³ · (6x − 5)

Justification / check. The factor (6x − 5) is the derivative of the inside; without it the answer is wrong. (sympy: (24x − 20)(3x² − 5x + 1)³, which is 4(6x−5)(3x²−5x+1)³ factored the same way. ✓)

Example 2 — Trig composite [NO CALC]

Problem. Differentiate y = sin(3x²).

Strategy. Outer = sine, inner = 3x². Use d/dx[sin g] = cos g · g'.

Solution.

y' = cos(3x²) · 6x = 6x cos(3x²)

Justification / check. Inside derivative d/dx[3x²] = 6x. (sympy: 6x cos(3x²). ✓)

Example 3 — Product rule combined with chain rule [NO CALC]

Problem. Differentiate y = (2x + 1)³(x − 4)².

Strategy. This is a product of two functions, each of which needs the chain rule (general power rule). Product rule first: (uv)' = u'v + uv', where u = (2x+1)³ and v = (x−4)².

Solution. Each factor's derivative uses the general power rule:

• u' = 3(2x+1)² · 2 = 6(2x+1)²
• v' = 2(x−4) · 1 = 2(x−4)

y' = 6(2x+1)²(x−4)² + (2x+1)³ · 2(x−4)

Factor out the common 2(2x+1)²(x−4):

y' = 2(2x+1)²(x−4)[3(x−4) + (2x+1)]
   = 2(2x+1)²(x−4)(5x − 11)

Justification / check. (sympy: 2(x−4)(2x+1)²(5x−11). ✓) Both inside derivatives — the ·2 on u' and the ·1 on v' — are required.

Example 4 — Table-of-values composite [NO CALC]

Problem. Using the table, let m(x) = g(f(x)). Find m'(2).

Strategy. Chain rule: m'(x) = g'(f(x)) · f'(x).

Solution.

m'(2) = g'(f(2)) · f'(2) = g'(1) · f'(2) = (−1)(−2) = 2

Justification / check. f(2) = 1 feeds g', giving g'(1) = −1; multiply by f'(2) = −2. Result 2. ✓

(d) Common Mistakes

Forgetting the inside derivative (THE #1 error).

Students write d/dx[(x² + 1)⁵] = 5(x² + 1)⁴ and stop. Why it's wrong: you differentiated the outer power but never multiplied by g'(x) = 2x. Fix: end every chain-rule line by literally asking "times the derivative of the inside" — here · 2x, giving 10x(x² + 1)⁴.

Stopping after one layer in a multi-layer composite.

For sin(√(x² + 1)), a student peels the sine, then the root, but forgets the innermost 2x. Why it's wrong: every layer contributes a factor. Fix: keep peeling until the inside is just x. The chain has as many factors as it has nested functions.

Applying the chain rule to a non-composite.

On y = x³ + 5x, a student "multiplies by the derivative of the inside" and invents extra factors. Why it's wrong: x³ is not a composite — x is not "inside" anything. Fix: the chain rule applies only when a whole function sits inside another. A bare power of x uses the ordinary power rule with no extra factor.

Mixing up which function is inside (sin²x vs. sin(x²)).

sin²x = (sin x)² has derivative 2 sin x · cos x (outer square, inner sine). sin(x²) has derivative cos(x²) · 2x (outer sine, inner square). Fix: ask "what is the last operation I'd perform?" — that's the outer function.

Plugging the wrong value into a table problem.

For h'(1) = f'(g(1))·g'(1), students compute f'(1) instead of f'(g(1)). Fix: evaluate the inside (g(1)) first and use it as the input to f'.

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## (e) Practice Problems

Differentiate y = sin(√x). (Short answer.)

Differentiate y = (4x − 1)³(x + 2)² and factor your answer. (Short answer.)

(Justify) A student claims d/dx[cos(2x)] = −sin(2x). State whether this is correct, and justify your answer with a correct computation.

(Justify) Explain, using the chain rule, why d/dx[(g(x))ⁿ] requires the factor g'(x) but d/dx[xⁿ] does not. Reference the idea of a composite function in your explanation.

Table for Problems 11–13.

| x | f(x) | f'(x) | g(x) | g'(x) |

|---|---|---|---|---|

| 1 | 4 | 2 | 3 | −1 |

| 2 | 1 | 5 | 2 | 3 |

| 3 | 2 | −4 | 2 | 6 |

If k(x) = g(f(x)), find k'(3). (Short answer.)

Let p(x) = f(x²). Find p'(1). (Short answer — calculator optional; the work is by hand from the table.)

Find an equation of the tangent line to y = (x² − 3)³ at x = 2. (Short answer.)

Differentiate y = sin(cos x). (Short answer.)

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## (f) AP Exam Focus — Free Response

> The functions f and g are differentiable for all real numbers. Selected values of f, f', g, and g' are given in the table below.

| x | f(x) | f'(x) | g(x) | g'(x) |

|---|---|---|---|---|

| 1 | 3 | 5 | 2 | −1 |

| 2 | 4 | −2 | 4 | 2 |

| 3 | 1 | 1 | 1 | −4 |

| 4 | 2 | 3 | 3 | 3 |

Let k be the function defined by k(x) = f(g(x)).

(a) (2 points) Find k'(3). Show the computation that leads to your answer.

(b) (3 points) Write an equation of the line tangent to the graph of k at x = 3.

(c) (2 points) Let m be the function defined by m(x) = g(f(x)). Find m'(2). Show the work that leads to your answer.

(d) (2 points) A student computes k'(3) by evaluating f'(3)·g'(3). Explain why this computation is incorrect, and identify the correct quantity that should be evaluated in place of f'(3).

Total: 9 points

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### Model Solution

(a) By the chain rule, k'(x) = f'(g(x))·g'(x), so

k'(3) = f'(g(3))·g'(3) = f'(1)·g'(3) = (5)(−4) = −20.

Since g(3) = 1, the inside value 1 is used as the input to f', giving f'(1) = 5. k'(3) = −20.

(b) The line tangent to k at x = 3 passes through (3, k(3)) with slope k'(3).

• Point: k(3) = f(g(3)) = f(1) = 3, so the point is (3, 3).
• Slope: k'(3) = −20 from part (a).

y − 3 = −20(x − 3),   i.e.   y = −20x + 63.

(c) By the chain rule, m'(x) = g'(f(x))·f'(x), so

m'(2) = g'(f(2))·f'(2) = g'(4)·f'(2) = (3)(−2) = −6.

Here f(2) = 4, so the inside value 4 feeds g', giving g'(4) = 3. m'(2) = −6.

(d) The chain rule gives k'(3) = f'(g(3))·g'(3). The outer derivative f' must be evaluated at the inner output g(3) = 1, not at 3. The student evaluated f'(3), which uses the wrong input to f'. The correct quantity is f'(g(3)) = f'(1) = 5. (The student's answer f'(3)·g'(3) = (1)(−4) = −4 is therefore wrong; the correct value is −20.)

Scoring Commentary

Where students lose points:

• Writing f'(3) instead of f'(g(3)) — the most common table error (part a).
• In (b), using k(3) = 3·1 = 3? No — k(3) = f(g(3)), a composition, not a product. Students who multiply table entries lose the point.
• "Linkage": (b) awards the slope point even if (a) was wrong, provided the student uses their own k'(3) consistently. But an unsupported slope earns nothing.
• In (d), vague answers ("he used the wrong number") earn 0; the response must name which function is evaluated at which input.

y' = 12(4x−1)²(x+2)² + (4x−1)³·2(x+2)
   = 2(4x−1)²(x+2)[6(x+2) + (4x−1)]
   = 2(4x−1)²(x+2)(10x + 11)