The Chain Rule

Master the technique for differentiating composite functions: the single most powerful and frequently used derivative rule in calculus.

AP Calculus BC

What Is a Composite Function?

Before learning the Chain Rule, you need to understand the functions it applies to. A composite function is a function built by plugging one function inside another:

$$ h(x) = f\!\bigl(g(x)\bigr) $$

Outer function $ f $

The function applied last: it wraps around everything.

Inner function $ g $

The function applied first: it lives inside the outer function.

Visualizing Composition as a Pipeline

Think of composition as an assembly line: $x$ enters, $g$ transforms it into $u = g(x)$, then $f$ transforms $u$ into the final output $f(g(x))$. The Chain Rule tells us how the rate of change flows through this pipeline.

Why the Chain Rule Matters

Most real-world functions are compositions. Without the Chain Rule, you cannot differentiate the vast majority of functions you will encounter in calculus and beyond.

Trig Compositions

$ \sin(3x) $, $ \cos(x^2) $, $ \tan(e^x) $

Exponential/Log

$ e^{5x} $, $ \ln(\sin x) $, $ 2^{x^3} $

Powers

$ (3x+1)^7 $, $ \sqrt{x^2+1} $

The Chain Rule: Formal Statement

$$ h'(x) = f'\!\bigl(g(x)\bigr) \cdot g'(x) $$

In words: differentiate the outer function (keeping the inner function unchanged), then multiply by the derivative of the inner function.

Chain Rule Diagram

Chain Rule vs Product Rule

Students often confuse when to use each rule. The distinction is simple:

	Chain Rule	Product Rule
Structure	Function inside function: $ f(g(x)) $	Function times function: $ f(x) \cdot g(x) $
Keyword	"of" : $ \sin $ of $ x^2 $	"times" : $ x^2 $ times $ \sin x $
Formula	$ f'(g(x))\cdot g'(x) $	$ f'g + fg' $
Example	$ \sin(x^2) $	$ x^2 \sin x $

How to Apply the Chain Rule

1

Identify the composite. Ask: "Is there a function inside another function?" If yes, the Chain Rule applies.

2

Label outer and inner. Write $ f(u) $ for the outer function and $ u = g(x) $ for the inner function.

3

Differentiate the outer. Find $ f'(u) $, leaving $ u $ in place (do not substitute back yet).

4

Differentiate the inner. Find $ g'(x) $.

5

Multiply and substitute. Write $ f'(g(x)) \cdot g'(x) $, then simplify.

Example 1: Basic Chain Rule

Differentiate $ h(x) = (3x + 1)^5 $.

1

Outer: $ f(u) = u^5 $. Inner: $ u = g(x) = 3x + 1 $.

2

Outer derivative: $ f'(u) = 5u^4 $.

3

Inner derivative: $ g'(x) = 3 $.

4

Combine: $ h'(x) = 5(3x+1)^4 \cdot 3 = 15(3x+1)^4 $.

Example 2: Trigonometric Composition

Differentiate $ y = \cos(x^3) $.

1

Outer: $ f(u) = \cos u $. Inner: $ u = x^3 $.

2

Outer derivative: $ f'(u) = -\sin u $.

3

Inner derivative: $ g'(x) = 3x^2 $.

4

Combine: $ \dfrac{dy}{dx} = -\sin(x^3) \cdot 3x^2 = -3x^2\sin(x^3) $.

Example 3: Exponential Composition

Differentiate $ y = e^{\sin x} $.

1

Outer: $ f(u) = e^u $. Inner: $ u = \sin x $.

2

Outer derivative: $ f'(u) = e^u $ (exponential is its own derivative).

3

Inner derivative: $ g'(x) = \cos x $.

4

Combine: $ \dfrac{dy}{dx} = e^{\sin x} \cdot \cos x $.

Example 4: Double Chain Rule

Differentiate $ y = \sin^2(4x) = \bigl[\sin(4x)\bigr]^2 $.

1

Outermost: $ (\;\cdot\;)^2 $. Middle: $ \sin(\cdot) $. Innermost: $ 4x $.

2

Differentiate layer by layer (outside in):

3

$ \dfrac{dy}{dx} = 2\bigl[\sin(4x)\bigr]^1 \cdot \cos(4x) \cdot 4 $

4

Simplify: $ y' = 8\sin(4x)\cos(4x) = 4\sin(8x) $.

Graphical Interpretation

Compare $ y = \sin x $ and $ y = \sin(2x) $. The Chain Rule predicts that the derivative of $ \sin(2x) $ is twice as steep.

Common Mistakes

✕ Forgetting the inner derivative

Writing $ \frac{d}{dx}[\sin(x^2)] = \cos(x^2) $ without multiplying by $ 2x $. This is the most common error.

✕ Replacing the inner function

Writing $ \cos(2x) $ instead of $ \cos(x^2) $. You must keep $g(x)$ inside $f'$, not substitute $g'(x)$.

✕ Using the Product Rule instead

Treating $ \sin(x^2) $ as $ \sin(x) \cdot x^2 $. Composition ($ f \circ g $) is not multiplication ($ f \cdot g $).

Chain Rule Quick Reference

Function	Derivative
$ \bigl[g(x)\bigr]^n $	$ n\bigl[g(x)\bigr]^{n-1} \cdot g'(x) $
$ \sin\!\bigl(g(x)\bigr) $	$ \cos\!\bigl(g(x)\bigr) \cdot g'(x) $
$ \cos\!\bigl(g(x)\bigr) $	$ -\sin\!\bigl(g(x)\bigr) \cdot g'(x) $
$ e^{g(x)} $	$ e^{g(x)} \cdot g'(x) $
$ \ln\!\bigl(g(x)\bigr) $	$ \dfrac{g'(x)}{g(x)} $
$ \tan\!\bigl(g(x)\bigr) $	$ \sec^2\!\bigl(g(x)\bigr) \cdot g'(x) $

Question 1 of 5

What is the derivative of $ f(x) = (2x+5)^3 $?

By the Chain Rule: $ f'(x) = 3(2x+5)^2 \cdot 2 = 6(2x+5)^2 $. The factor of 2 is the derivative of the inner function $ 2x+5 $.

Question 2 of 5

Find $ \dfrac{d}{dx}\bigl[e^{3x^2}\bigr] $.

The outer function is $ e^u $ (derivative: $ e^u $). The inner function is $ u = 3x^2 $ (derivative: $ 6x $). So $ \frac{d}{dx}[e^{3x^2}] = e^{3x^2} \cdot 6x $.

Question 3 of 5

What is $ \dfrac{d}{dx}\bigl[\ln(\cos x)\bigr] $?

Using $ \frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)} $: the inner derivative is $ g'(x) = -\sin x $, so the answer is $ \frac{-\sin x}{\cos x} = -\tan x $.

Question 4 of 5

If $ h(x) = \sqrt{x^3 + 7} $, what is $ h'(x) $?

Rewrite as $ (x^3+7)^{1/2} $. Outer derivative: $ \frac{1}{2}(x^3+7)^{-1/2} $. Inner derivative: $ 3x^2 $. Combined: $ \frac{3x^2}{2\sqrt{x^3+7}} $.

Question 5 of 5

Find the derivative of $ y = \sin^3(2x) $.

This requires a double Chain Rule. Outermost: $ u^3 \to 3u^2 $. Middle: $ \sin(v) \to \cos(v) $. Innermost: $ 2x \to 2 $. Combined: $ 3\sin^2(2x) \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x) $.

	Chain Rule	Product Rule
Structure	Function inside function: \( f(g(x)) \)	Function times function: \( f(x) \cdot g(x) \)
Keyword	"of" : \( \sin \) of \( x^2 \)	"times" : \( x^2 \) times \( \sin x \)
Formula	\( f'(g(x))\cdot g'(x) \)	\( f'g + fg' \)
Example	\( \sin(x^2) \)	\( x^2 \sin x \)

Function	Derivative
\( \bigl[g(x)\bigr]^n \)	\( n\bigl[g(x)\bigr]^{n-1} \cdot g'(x) \)
\( \sin\!\bigl(g(x)\bigr) \)	\( \cos\!\bigl(g(x)\bigr) \cdot g'(x) \)
\( \cos\!\bigl(g(x)\bigr) \)	\( -\sin\!\bigl(g(x)\bigr) \cdot g'(x) \)
\( e^{g(x)} \)	\( e^{g(x)} \cdot g'(x) \)
\( \ln\!\bigl(g(x)\bigr) \)	\( \dfrac{g'(x)}{g(x)} \)
\( \tan\!\bigl(g(x)\bigr) \)	\( \sec^2\!\bigl(g(x)\bigr) \cdot g'(x) \)