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Calculus AB · Unit 3 · 3.1
AP Calculus · Differentiation: Composite, Implicit, and Inverse Functions
The Chain Rule
Differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
key idea$\frac{d}{dx}f\bigl(g(x)\bigr)=f'\bigl(g(x)\bigr)\cdot g'(x)$
The rule
A composition is a function inside another function. Differentiate the outer one leaving the inside alone, then multiply by the derivative of the inside. That trailing factor is the whole rule — and the thing everyone forgets.
| Expression | Outer $f$ | Inner $g$ | Derivative |
|---|
| $(3x+1)^{5}$ | $u^{5}$ | $3x+1$ | $5(3x+1)^{4}\cdot\mathbf{3}$ |
| $\sin(x^{2})$ | $\sin u$ | $x^{2}$ | $\cos(x^{2})\cdot\mathbf{2x}$ |
| $e^{4x}$ | $e^{u}$ | $4x$ | $e^{4x}\cdot\mathbf{4}$ |
| $\ln(5x-2)$ | $\ln u$ | $5x-2$ | $\dfrac{1}{5x-2}\cdot\mathbf{5}$ |
| $\sqrt{x^{2}+9}$ | $\sqrt{u}$ | $x^{2}+9$ | $\dfrac{1}{2\sqrt{x^{2}+9}}\cdot\mathbf{2x}$ |
| $\tan^{3}x$ | $u^{3}$ | $\tan x$ | $3\tan^{2}x\cdot\mathbf{\sec^{2}x}$ |
The bold factor is the inner derivative. Every error on this topic is a missing or wrong bold factor.
identifying the layers
- Ask “what is the last operation I would perform?” — that is the outer function.
- Differentiate outer, keep the inside unchanged, then multiply by the inner derivative.
- $\tan^{3}x$ means $(\tan x)^{3}$ — the power is outer, the trig is inner.
- If the inside is just $x$, the inner derivative is $1$ and the rule reduces to the basic one.
Pattern recognition
Power outside$\tan^{3}x=(\tan x)^{3}$ — inner is the trig function
Power inside$\sin(x^{2})$ — inner is $x^{2}$, outer is $\sin$
Count the layerseach layer contributes exactly one factor
Worked examples
EX 1
One layer
$(3x+1)^{5}$Differentiate $y=(3x+1)^{5}$.
1
Name the layers
Last operation is raising to the 5th power, so outer is $u^{5}$ and inner is $u=3x+1$.
2
Differentiate the outer
$\dfrac{d}{du}\left[u^{5}\right]=5u^{4}$, so we get $5(3x+1)^{4}$ — inside untouched.
3
Differentiate the inner
$\dfrac{d}{dx}[3x+1]=3$.
4
Multiply
$y'=5(3x+1)^{4}\cdot3=15(3x+1)^{4}$.
5
Sanity check
Expanding $(3x+1)^{5}$ would give a leading term $243x^{5}$, whose derivative is $1215x^{4}$. And $15\cdot3^{4}=1215$. ✓
EX 2
Two layers
$\sin^{2}(4x)$Differentiate $y=\sin^{2}(4x)$.
1
Peel from the outside
Last operation is squaring: outer $u^{2}$, inner $\sin(4x)$.
2
First application
$2\sin(4x)\cdot\dfrac{d}{dx}\left[\sin(4x)\right]$.
3
The inner is itself a composition
$\dfrac{d}{dx}\left[\sin(4x)\right]=\cos(4x)\cdot4$.
4
Combine all three factors
$y'=2\sin(4x)\cdot\cos(4x)\cdot4=8\sin(4x)\cos(4x)$.
5
Note the structure
Three layers give three factors. Peel one at a time and nothing is dropped.
EX 3
Chain inside a product
$x^{2}e^{-3x}$Differentiate $y=x^{2}e^{-3x}$.
1
Identify the top-level structure
A product of $x^{2}$ and $e^{-3x}$, so the product rule leads.
2
Differentiate each factor
$\dfrac{d}{dx}\left[x^{2}\right]=2x$. For $e^{-3x}$ the chain rule gives $e^{-3x}\cdot(-3)$.
3
Apply the product rule
$y'=2x\,e^{-3x}+x^{2}\left(-3e^{-3x}\right)$.
4
Factor
$y'=x\,e^{-3x}\left(2-3x\right)$.
5
Where the rules sit
Product rule outside, chain rule inside one of its factors. They compose; neither replaces the other.
EX 4
From a table
structuralGiven $g(2)=5$, $g'(2)=-3$, $f'(5)=4$, find $\dfrac{d}{dx}\left[f(g(x))\right]$ at $x=2$.
1
Write the rule
$\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$.
2
Substitute $x=2$
$=f'(g(2))\cdot g'(2)=f'(5)\cdot(-3)$.
3
Use the given values
$=4\cdot(-3)=-12$.
4
Note which value of $f'$ was needed
$f'(5)$, not $f'(2)$ — the outer derivative is evaluated at the inner output. That substitution is what this question tests.
Watch out
The missing inner derivative
$\dfrac{d}{dx}\left[\sin(4x)\right]$ is $4\cos(4x)$, not $\cos(4x)$. If your answer would be unchanged when the inner function changes, you dropped the factor.
Evaluate the outer derivative at the inner value
In Example 4 you need $f'(5)$ because $g(2)=5$. Using $f'(2)$ is the classic table-question error.
Do not change the inside while differentiating the outside
$\dfrac{d}{dx}\left[(3x+1)^{5}\right]$ begins $5(3x+1)^{4}$ — the bracket keeps its contents. Writing $5(3)^{4}$ discards the function.
$\sin^{2}x$ and $\sin(x^{2})$ are different
The first squares the output, the second squares the input. They have different outer functions and different derivatives.