The Chain Rule is arguably the most important differentiation rule in calculus. It tells you how to differentiate composite functions — functions inside other functions. Without the Chain Rule, you'd be stuck with functions like $\sin(3x)$, $(x^2+1)^{10}$, or $\ln(2x)$.
🎯 In this section you will learn
- The Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
- How to identify the "inner" and "outer" functions
- Applying the Chain Rule with power rule, trig functions, exponentials, and logs
- The "outside-inside" method (Leibniz notation: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$)
📌 The Chain Rule
If $y = f(u)$ and $u = g(x)$, then:
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$
$$ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) $$
💡 Memory Trick
"Derivative of the outside, times derivative of the inside."
"Outside stays the same (with inside plugged in), then multiply by derivative of inside."
🔍 Identifying Inner and Outer Functions
Before using the Chain Rule, you need to identify the inner function (what's inside) and the outer function (what's applied to it).
| Composite Function | Inner $g(x)$ | Outer $f(u)$ |
|---|
| $\sin(3x)$ | $3x$ | $\sin u$ |
| $(x^2+1)^{10}$ | $x^2+1$ | $u^{10}$ |
| $\ln(2x)$ | $2x$ | $\ln u$ |
| $e^{\sin x}$ | $\sin x$ | $e^u$ |
| $\sqrt{x^2+1}$ | $x^2+1$ | $\sqrt{u} = u^{1/2}$ |
Find $\frac{d}{dx}(3x+1)^5$.
①
Identify inner and outer
Inner: $g(x)=3x+1$, Outer: $f(u)=u^5$
②
Derivative of outer
$f'(u)=5u^4$, so $f'(g(x))=5(3x+1)^4$
③
Derivative of inner
$g'(x)=3$
④
Multiply
$\frac{d}{dx}(3x+1)^5 = 5(3x+1)^4 \cdot 3 = 15(3x+1)^4$
Find $\frac{d}{dx}\sin(5x)$.
①
Identify
Inner: $5x$, Outer: $\sin u$
②
Derivative of outer
$\frac{d}{du}\sin u = \cos u$, so $\cos(5x)$
③
Derivative of inner
$\frac{d}{dx}(5x) = 5$
④
Multiply
$\frac{d}{dx}\sin(5x) = \cos(5x) \cdot 5 = 5\cos(5x)$
Find $\frac{d}{dx}e^{\sin x}$.
- Inner: $\sin x$, Outer: $e^u$
- Derivative of outer: $e^{\sin x}$
- Derivative of inner: $\cos x$
- $\frac{d}{dx}e^{\sin x} = e^{\sin x} \cdot \cos x$
Find $\frac{d}{dx}\ln(x^2+1)$.
- Inner: $x^2+1$, Outer: $\ln u$
- Derivative of outer: $\frac{1}{x^2+1}$
- Derivative of inner: $2x$
- $\frac{d}{dx}\ln(x^2+1) = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$
Find $\frac{d}{dx}\sqrt{4x-3}$.
- Rewrite: $(4x-3)^{1/2}$
- Inner: $4x-3$, Outer: $u^{1/2}$
- Derivative of outer: $\frac{1}{2}(4x-3)^{-1/2} = \frac{1}{2\sqrt{4x-3}}$
- Derivative of inner: $4$
- $\frac{d}{dx}\sqrt{4x-3} = \frac{1}{2\sqrt{4x-3}} \cdot 4 = \frac{4}{2\sqrt{4x-3}} = \frac{2}{\sqrt{4x-3}}$
📌 Multiple Layers (Nested Functions)
Sometimes you have more than two layers — functions inside functions inside functions. Just apply the Chain Rule repeatedly:
$$ \frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) $$
Find $\frac{d}{dx}\sin^2(3x)$.
①
Rewrite
$\sin^2(3x) = [\sin(3x)]^2$
②
Outer: $u^2$, Middle: $\sin v$, Inner: $3x$
Derivative: $2[\sin(3x)] \cdot \cos(3x) \cdot 3$
③
Simplify
$\frac{d}{dx}\sin^2(3x) = 6\sin(3x)\cos(3x) = 3\sin(6x)$ (using double-angle)
⚠️ Common Mistakes
- Forgetting the inner derivative: Many students stop after finding $f'(g(x))$ and forget to multiply by $g'(x)$.
- Power Rule on the outside without Chain Rule: $\frac{d}{dx}(x^2+1)^5 \neq 5(x^2+1)^4$ — that's missing the $2x$ from the inside!
- Misidentifying inside vs outside: In $\sin(2x)$, the inside is $2x$, not $\sin x$.
- Applying Chain Rule when it's just a constant multiple: For $2e^x$, use constant multiple rule, not chain rule.
🔍 Key Takeaways
- Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
- Identify the inner function first, then the outer function.
- Derivative = (derivative of outer, with inner plugged in) × (derivative of inner).
- For multiple layers, apply the Chain Rule repeatedly from outside to inside.
- The Chain Rule works with every type of function — power, trig, exponential, log.
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