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The Chain Rule

Reference: Stewart 2.5  •  Chapter: 2  •  Section: 5

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When Derivatives Get Layered

How do you differentiate $\sqrt{x^2 + 1}$? The power rule doesn't directly apply because the exponent sits on top of an expression, not just $x$. This function has layers: a square root wrapped around $(x^2 + 1)$.

Think of it like peeling an onion. The outer layer is the square root. The inner layer is $x^2 + 1$. The Chain Rule tells us how to differentiate functions that are built from layers: one function applied to another.

The key insight: if the inner part changes, it affects how fast the outer part changes. If $u$ changes twice as fast as $x$, and $y$ changes three times as fast as $u$, then $y$ changes six times as fast as $x$. Rates of change multiply.

Prerequisite Map

PrerequisitesThe Derivative DefinitionFunction Composition
This skillThe Chain Rule
Leads toNested Chain RuleImplicit Differentiation

Quick Reference

Property Value
Concept Differentiation Rules
Chapter Ch 3, §4
Difficulty Intermediate
Time ~20 minutes

Key Concepts

The Chain Rule Formula

If $F(x) = f(g(x))$ is a composite function where $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then:

$$\boxed{F'(x) = f'(g(x)) \cdot g'(x)}$$

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

Leibniz Notation

If $y = f(u)$ and $u = g(x)$ are both differentiable, then:

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

This notation is memorable because the $du$ terms appear to "cancel," though this is just a helpful mnemonic, not rigorous mathematics.

Identifying Inner and Outer Functions

Function: √(x² + 1)

Outer function: √(□) = (□)^(1/2)    ← What operation is applied last?
Inner function: x² + 1              ← What's inside the outer operation?

The outer function is applied last when you evaluate the expression. If you were computing $\sqrt{x^2 + 1}$ for a specific $x$:

  1. First compute $x^2 + 1$ (inner)
  2. Then take the square root (outer)

The Chain Rule Process

  1. Identify the outer function $f$ and inner function $g$
  2. Differentiate the outer function, keeping the inner function unchanged: $f'(g(x))$
  3. Multiply by the derivative of the inner function: $g'(x)$

Visual Representation

                    Chain Rule

  y = f(g(x))   →   dy/dx = f'(g(x)) · g'(x)
      ↓                     ↓           ↓
  ┌───────┐           ┌─────────┐   ┌─────┐
  │ outer │           │ d(outer)│ × │d(in)│
  │  f    │           │  at     │   │     │
  │───────│           │ inner   │   │     │
  │ inner │           └─────────┘   └─────┘
  │  g(x) │
  └───────┘

Common Chain Rule Patterns

Composite Form Derivative
$\sin(g(x))$ $\cos(g(x)) \cdot g'(x)$
$\cos(g(x))$ $-\sin(g(x)) \cdot g'(x)$
$\tan(g(x))$ $\sec^2(g(x)) \cdot g'(x)$
$e^{g(x)}$ $e^{g(x)} \cdot g'(x)$
$\ln(g(x))$ $\frac{1}{g(x)} \cdot g'(x)$
$[g(x)]^n$ $n[g(x)]^{n-1} \cdot g'(x)$

Practice Problems

Level 1 Identifying Inner and Outer Functions

For each composite function, identify the inner function $g(x)$ and outer function $f(u)$.

  1. $h(x) = \sin(3x)$
  2. $h(x) = (5x - 2)^7$
  3. $h(x) = \sqrt{x^3 + 4x}$
Thought Process

Ask yourself: "What operation is performed last when I evaluate this function?"

  • In $\sin(3x)$, you first compute $3x$, then take the sine
  • In $(5x - 2)^7$, you first compute $5x - 2$, then raise to the 7th power
  • In $\sqrt{x^3 + 4x}$, you first compute $x^3 + 4x$, then take the square root
Show Answer

(a) $h(x) = \sin(3x)$

  • Inner: $g(x) = 3x$
  • Outer: $f(u) = \sin u$

(b) $h(x) = (5x - 2)^7$

  • Inner: $g(x) = 5x - 2$
  • Outer: $f(u) = u^7$

(c) $h(x) = \sqrt{x^3 + 4x}$

  • Inner: $g(x) = x^3 + 4x$
  • Outer: $f(u) = \sqrt{u} = u^{1/2}$
Level 2 Basic Chain Rule Application

Find the derivative of $f(x) = \cos(4x)$.

Thought Process
  1. Identify: outer = $\cos$, inner = $4x$
  2. Derivative of outer at inner: $-\sin(4x)$
  3. Derivative of inner: $\frac{d}{dx}(4x) = 4$
  4. Multiply: $-\sin(4x) \cdot 4$
Show Answer

Let $u = 4x$ (inner function).

$$\frac{d}{dx}\cos(4x) = -\sin(4x) \cdot \frac{d}{dx}(4x)$$

$$= -\sin(4x) \cdot 4 = -4\sin(4x)$$

Level 3 Chain Rule with Polynomial Inner Functions

Differentiate $y = \sin(x^2 - 3x + 1)$.

Thought Process
  1. Outer function: $\sin(\square)$
  2. Inner function: $x^2 - 3x + 1$
  3. Apply chain rule: $\cos(\text{inner}) \cdot (\text{derivative of inner})$
  4. The derivative of $x^2 - 3x + 1$ is $2x - 3$
Show Answer

$$\frac{dy}{dx} = \cos(x^2 - 3x + 1) \cdot \frac{d}{dx}(x^2 - 3x + 1)$$

$$= \cos(x^2 - 3x + 1) \cdot (2x - 3)$$

$$= (2x - 3)\cos(x^2 - 3x + 1)$$

Level 4 Conceptual: Rate of Change Interpretation

A particle moves so that its position at time $t$ is $s(t) = \sin(\theta(t))$, where $\theta(t)$ gives the angle (in radians) as a function of time.

  1. Find an expression for the velocity $v(t) = s'(t)$ in terms of $\theta(t)$ and $\theta'(t)$.
  2. If at $t = 2$ seconds, $\theta(2) = \pi/3$ and the angle is changing at 0.5 rad/sec, what is the particle's velocity at $t = 2$?
Thought Process

This problem connects the chain rule to physics. The position depends on an angle that changes over time, so:

  • $s = \sin(\theta)$ where $\theta$ depends on $t$
  • By chain rule: $\frac{ds}{dt} = \frac{ds}{d\theta} \cdot \frac{d\theta}{dt}$

For part (b), plug in the given values: $\theta = \pi/3$ means $\cos(\pi/3) = 1/2$, and $\theta' = 0.5$.

Show Answer

(a) Using the chain rule:

$$v(t) = s'(t) = \frac{d}{dt}\sin(\theta(t)) = \cos(\theta(t)) \cdot \theta'(t)$$

(b) At $t = 2$:

  • $\theta(2) = \pi/3$
  • $\theta'(2) = 0.5$ rad/sec

$$v(2) = \cos(\pi/3) \cdot 0.5 = \frac{1}{2} \cdot 0.5 = 0.25 \text{ units/sec}$$

Level 5 Proving a Differentiation Identity

Use the chain rule to prove: if $f$ is an even function (meaning $f(-x) = f(x)$ for all $x$), then $f'$ is an odd function (meaning $f'(-x) = -f'(x)$).

Thought Process

Start with what we know: $f(-x) = f(x)$.

To find $f'(-x)$, differentiate both sides of $f(-x) = f(x)$ with respect to $x$.

The left side requires the chain rule since $-x$ is the inner function.

Show Answer

Given: $f(-x) = f(x)$ for all $x$.

Differentiate both sides with respect to $x$:

Left side (using chain rule with inner function $u = -x$): $$\frac{d}{dx}f(-x) = f'(-x) \cdot \frac{d}{dx}(-x) = f'(-x) \cdot (-1) = -f'(-x)$$

Right side: $$\frac{d}{dx}f(x) = f'(x)$$

Setting them equal: $$-f'(-x) = f'(x)$$

Multiply both sides by $-1$: $$f'(-x) = -f'(x)$$

This is exactly the definition of an odd function. $\square$

CCI-Style Conceptual Questions

Conceptual Understanding Rate Multiplication

If $y = f(u)$ and $u = g(x)$, and at a particular point:

  • $u$ is changing 3 times as fast as $x$ (i.e., $\frac{du}{dx} = 3$)
  • $y$ is changing 4 times as fast as $u$ (i.e., $\frac{dy}{du} = 4$)

How fast is $y$ changing with respect to $x$ at that point?

  1. $\frac{dy}{dx} = 7$
  2. $\frac{dy}{dx} = 12$
  3. $\frac{dy}{dx} = 1$
  4. $\frac{dy}{dx} = \frac{4}{3}$
Thought Process

The chain rule says rates of change multiply, not add. Think of it physically: if an intermediate quantity doubles, and the output triples for each unit of that intermediate quantity, the overall effect is $2 \times 3 = 6$.

Show Answer

(B) $\frac{dy}{dx} = 12$

By the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4 \cdot 3 = 12$

Mastery Checklist

Mental Model

The Gear Train:

Think of the chain rule like connected gears. If the small gear (inner function) turns twice as fast as the input shaft, and the large gear (outer function) turns three times as fast as the small gear, then the output shaft turns $2 \times 3 = 6$ times as fast as the input. Rates multiply through the chain.

Connections

Looking back:

Looking ahead:

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Product & Quotient Rules Skills Index Generalized Power Rule

Last updated: 2026-01-22