Chain Rule

What Is a Composite Function?

A composite function is a function applied to another function. If $y = f(g(x))$, we call $g(x)$ the inner function and $f$ the outer function.

Examples:

• $y = (3x + 1)^4$: outer = $(\cdot)^4$, inner = $3x + 1$
• $y = \sin(x^2)$: outer = $\sin(\cdot)$, inner = $x^2$
• $y = e^{5x}$: outer = $e^{(\cdot)}$, inner = $5x$
• $y = \ln(\cos x)$: outer = $\ln(\cdot)$, inner = $\cos x$

The Chain Rule Formula

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

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

In words: derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.

In Leibniz notation, if $y = f(u)$ and $u = g(x)$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

This looks like the $du$'s "cancel," which is a useful mnemonic (though not rigorous).

The Generalized Power Rule

The most common application of the chain rule is with powers:

$\frac{d}{dx}[u^n] = n \cdot u^{n-1} \cdot \frac{du}{dx}$

where $u$ is any function of $x$. For example:

$\frac{d}{dx}[(x^2 + 3x)^7] = 7(x^2 + 3x)^6 \cdot (2x + 3)$

Chain Rule with Trigonometric Functions

Every trig derivative gains a factor of the inner derivative:

$\frac{d}{dx}[\sin(u)] = \cos(u) \cdot u'$

$\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot u'$

$\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot u'$

Examples:

• $\frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot 3 = 3\cos(3x)$
• $\frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot 2x = -2x\sin(x^2)$
• $\frac{d}{dx}[\tan(5x+1)] = \sec^2(5x+1) \cdot 5 = 5\sec^2(5x+1)$

Chain Rule with Exponentials and Logarithms

$\frac{d}{dx}[e^u] = e^u \cdot u'$

$\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot u' = \frac{u'}{u}$

Examples:

• $\frac{d}{dx}[e^{3x^2}] = e^{3x^2} \cdot 6x = 6xe^{3x^2}$
• $\frac{d}{dx}[\ln(x^2 + 1)] = \frac{2x}{x^2 + 1}$

Nested Chain Rule (Multiple Layers)

Sometimes functions have more than two layers. Apply the chain rule from the outside in, multiplying each derivative:

$\frac{d}{dx}[\sin^3(2x)] = \frac{d}{dx}[(\sin(2x))^3]$

Three layers: outer = $(\cdot)^3$, middle = $\sin(\cdot)$, innermost = $2x$.

$= 3[\sin(2x)]^2 \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x)$

Chain Rule Combined with Product Rule

Many AP problems require both rules simultaneously. Differentiate the product using the product rule, and apply the chain rule whenever you differentiate a composite factor.

Example: $\frac{d}{dx}[x^2 \sin(3x)]$

Product rule: $2x \cdot \sin(3x) + x^2 \cdot \cos(3x) \cdot 3$

$= 2x\sin(3x) + 3x^2\cos(3x)$

Chain Rule Combined with Quotient Rule

Example: $\frac{d}{dx}\left[\frac{e^{2x}}{x+1}\right]$

Quotient rule, with chain rule on $e^{2x}$:

$= \frac{2e^{2x}(x+1) - e^{2x}(1)}{(x+1)^2} = \frac{e^{2x}(2x + 2 - 1)}{(x+1)^2} = \frac{e^{2x}(2x+1)}{(x+1)^2}$

Implicit Differentiation and the Chain Rule

Implicit differentiation is the chain rule applied to $y$ as a function of $x$. Whenever you differentiate a term involving $y$, multiply by $\frac{dy}{dx}$:

$\frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx}$

$\frac{d}{dx}[\sin(y)] = \cos(y) \cdot \frac{dy}{dx}$

This connection to implicit differentiation makes the chain rule essential for related rates problems.

Tip 1: Identify the Layers Before You Start

Before writing anything, pause and identify the outer and inner functions. For a triple composition like $e^{\sin(x^2)}$, list the layers: outer = $e^{(\cdot)}$, middle = $\sin(\cdot)$, inner = $x^2$.

Tip 2: Work from the Outside In

Always differentiate the outermost function first, keeping the inner function unchanged. Then multiply by the derivative of the inner function. For multiple layers, repeat.

Tip 3: Don't Forget the Inner Derivative

The most common chain rule error is forgetting to multiply by the derivative of the inner function. Every time you differentiate a composite, ask yourself: "Did I multiply by the inside derivative?"

Tip 4: Use Leibniz Notation for Complex Problems

If a problem is complicated, introduce a substitution: let $u = \text{inner function}$, find $\frac{dy}{du}$ and $\frac{du}{dx}$ separately, then multiply. This breaks the problem into manageable steps.

Tip 5: Recognize Chain Rule in Reverse (for Integration)

Understanding the chain rule well prepares you for $u$-substitution in integration. If $\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)$, then $\int F'(g(x)) \cdot g'(x)\,dx = F(g(x)) + C$.