The Chain Rule
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
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 $x$:
- First compute $x^2 + 1$ (inner)
- Then take the square root (outer)
The Chain Rule Process
- Identify the outer function $f$ and inner function $g$
- Differentiate the outer function, keeping the inner function unchanged: $f'(g(x))$
- 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
For each composite function, identify the inner function $g(x)$ and outer function $f(u)$.
- $h(x) = \sin(3x)$
- $h(x) = (5x - 2)^7$
- $h(x) = \sqrt{x^3 + 4x}$
Find the derivative of $f(x) = \cos(4x)$.
Differentiate $y = \sin(x^2 - 3x + 1)$.
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.
- Find an expression for the velocity $v(t) = s'(t)$ in terms of $\theta(t)$ and $\theta'(t)$.
- 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 $t = 2$?
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 $f'(-x) = -f'(x)$).
CCI-Style Conceptual Questions
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?
- $\frac{dy}{dx} = 7$
- $\frac{dy}{dx} = 12$
- $\frac{dy}{dx} = 1$
- $\frac{dy}{dx} = \frac{4}{3}$
Common Misconceptions
the derivative of f(g(x)) is f’(x) times g’(x).
This is the composition-is-not-chaining error. The outer derivative must be evaluated at the inner function, not at x. For example, $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$, not $\cos(x) \cdot 2x$. Substituting $x = 1$: the correct answer gives $\cos(1) \cdot 2 \approx 1.08$, while the wrong formula gives $\cos(1) \cdot 2 \approx$ the same numerical coincidence only because the inner function happens to equal $x$ at $x = 1$ -- but at $x = 2$ the correct answer is $\cos(4) \cdot 4 \approx -2.61$ and the wrong formula gives $\cos(2) \cdot 4 \approx -1.66$, which differ.
the factor g’(x) is a single constant that can be looked up once.
This is the rate-as-fixed-number error. The inner derivative $g'(x)$ is itself a function of $x$, not a single number. For $g(x) = x^2$, the factor $g'(x) = 2x$ equals 2 at $x = 1$, equals 4 at $x = 2$, and equals 0 at $x = 0$. Writing a fixed number in place of $g'(x)$ produces an incorrect derivative at every point except one.
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:
- The chain rule builds on all basic derivative rules (power, trig, exponential)
- Function composition from precalculus is essential for identifying the structure
Looking ahead:
- The Generalized Power Rule is the most common chain rule application
- Implicit differentiation is essentially the chain rule applied to $y(x)$
- Related rates problems chain together multiple rates of change
| Previous | Up | Next |
|---|---|---|
| Product & Quotient Rules | Skills Index | Generalized Power Rule |
Last updated: 2026-01-22