3.1 - The Chain Rule

Composite Functions Review

A composite function is a function inside another function. Before you can use the chain rule, you have to identify the layers.

Example 1:

f(x) = \cos(x^3)

Outer: $f(u) = \cos u$

Inner: $g(x) = x^3$

Example 2:

f(x) = \sqrt{\ln x}

Outer: $f(u) = \sqrt{u}$

Inner: $g(x) = \ln x$

Example 3:

f(x) = \sin(\cos(2x))

Outer: $f(u) = \sin u$

Middle: $g(u) = \cos u$

Inner: $h(x) = 2x$

That one has three layers, like an onion. You peel away one layer at a time when you differentiate.

The Formula

For a composite function

f(g(x))

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

In words: take the derivative of the outer function (leaving the inside alone), then multiply by the derivative of the inside.

The “onion rule”: peel one layer at a time, taking derivatives as you go from outside to inside.

Leibniz form. If

y = f(u)

and

u = g(x)

, the chain rule says:

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

Same rule, different notation: the rate of change of the outer function times the rate of change of the inner function. The

du

’s look like they cancel, which makes it easy to remember (they aren’t really fractions, but they behave like them). In Example 1 below,

y = u^5

and

u = 3x^2 - 7

, so

\frac{dy}{du} = 5u^4

and

\frac{du}{dx} = 6x

This form is handy for word problems because the units chain together too. Say the air temperature

T

(in °C) depends on the height

h

of a weather balloon (in meters), and the height depends on time

t

(in minutes). If

\frac{dT}{dh} = -0.0065

°C per meter and the balloon rises at

\frac{dh}{dt} = 40

meters per minute, then:

\frac{dT}{dt} = \frac{dT}{dh} \cdot \frac{dh}{dt} = (-0.0065)(40) = -0.26 \ {}^\circ\text{C per minute}

Check the units:

\frac{{}^\circ\text{C}}{\text{min}} = \frac{{}^\circ\text{C}}{\text{m}} \cdot \frac{\text{m}}{\text{min}}

. The meters cancel, just like the

dh

’s.

Basic Examples

Example 1: Power Function Composition

Find

\frac{d}{dx}[(3x^2 - 7)^5]

Outer: $u^5$ , derivative $5u^4$

Inner: $3x^2 - 7$ , derivative $6x$

\frac{d}{dx}[(3x^2 - 7)^5] = 5(3x^2 - 7)^4 \cdot 6x = 30x(3x^2 - 7)^4

Without the chain rule you’d have to fully expand

(3x^2 - 7)^5

, which would be miserable.

Example 2: Square Root Composition

Find

\frac{d}{dx}[\sqrt{5x + 2}]

Recall

\frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}}

Outer: $\sqrt{u}$ , derivative $\frac{1}{2\sqrt{u}}$

Inner: $5x + 2$ , derivative $5$

\frac{d}{dx}[\sqrt{5x + 2}] = \frac{1}{2\sqrt{5x + 2}} \cdot 5 = \frac{5}{2\sqrt{5x + 2}}

Example 3: Squared Trig Function (common error)

Find

\frac{d}{dx}[\cos^2(3x)]

First, understand that

\cos^2(3x) = [\cos(3x)]^2

. That gives you three layers:

Outer: $u^2$

Middle: $\cos u$

Inner: $3x$

Peel them off:

Layer 1: bring down the 2, giving $2[\cos(3x)]^1$

Layer 2: derivative of cosine, giving $-\sin(3x)$

Layer 3: derivative of $3x$ , giving $3$

\frac{d}{dx}[\cos^2(3x)] = 2\cos(3x) \cdot (-\sin(3x)) \cdot 3 = -6\cos(3x)\sin(3x)

Common mistake: forgetting that the square is its own layer. People often write

-\sin^2(3x) \cdot 3

, which is wrong.

Logarithmic Functions with the Chain Rule

Method 1: Direct Chain Rule

Find

\frac{d}{dx}[\ln(x^4)]

\frac{d}{dx}[\ln(x^4)] = \frac{1}{x^4} \cdot 4x^3 = \frac{4x^3}{x^4} = \frac{4}{x}

Method 2: Use Log Properties First

Using the power rule for logs,

\ln(x^4) = 4\ln x

\frac{d}{dx}[4\ln x] = 4 \cdot \frac{1}{x} = \frac{4}{x}

Same answer. Method 2 is faster whenever you just have a variable raised to a power inside the log.

Exponential and Trig Functions with the Chain Rule

Every basic derivative from Unit 2 works the same way inside the chain rule: take the usual derivative, leave the inside alone, then multiply by the derivative of the inside.

\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x) \qquad \frac{d}{dx}[a^{g(x)}] = a^{g(x)} \ln(a) \cdot g'(x)

\frac{d}{dx}[\sin(g(x))] = \cos(g(x)) \cdot g'(x) \qquad \frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}

Example 1: Exponential with a Polynomial Exponent

Find

\frac{d}{dx}[e^{x^2 - 4x}]

Outer: $e^u$ , derivative $e^u$ (it stays the same)

Inner: $x^2 - 4x$ , derivative $2x - 4$

\frac{d}{dx}[e^{x^2 - 4x}] = e^{x^2 - 4x} \cdot (2x - 4) = (2x - 4)e^{x^2 - 4x}

Common mistake: using the power rule and writing

(x^2 - 4x)e^{x^2 - 4x - 1}

. The variable is in the exponent, so the power rule doesn’t apply. The other classic error is stopping at

e^{x^2 - 4x}

and forgetting to multiply by

2x - 4

Example 2: A Base Other Than e

Find

\frac{d}{dx}[5^{3x}]

\frac{d}{dx}[5^{3x}] = 5^{3x} \ln(5) \cdot 3 = 3\ln(5) \cdot 5^{3x}

Example 3: Trig with a Polynomial Inside

Find

\frac{d}{dx}[\tan(4x^3)]

Outer: $\tan u$ , derivative $\sec^2 u$

Inner: $4x^3$ , derivative $12x^2$

\frac{d}{dx}[\tan(4x^3)] = \sec^2(4x^3) \cdot 12x^2 = 12x^2\sec^2(4x^3)

Combining the Chain Rule with Other Rules

Chain Rule + Quotient Rule

Find

f'(t)

f(t) = \left[\dfrac{t^2 - 3}{3t + 1}\right]^4

Outer: $u^4$ , derivative $4u^3$

Inner: $\frac{t^2 - 3}{3t + 1}$ , which needs the quotient rule

f'(t) = 4\left[\frac{t^2 - 3}{3t + 1}\right]^3 \cdot \frac{d}{dt}\left[\frac{t^2 - 3}{3t + 1}\right]

Quotient rule on the inside:

\frac{d}{dt}\left[\frac{t^2 - 3}{3t + 1}\right] = \frac{(2t)(3t + 1) - (t^2 - 3)(3)}{(3t + 1)^2} = \frac{3t^2 + 2t + 9}{(3t + 1)^2}

So:

f'(t) = 4\left[\frac{t^2 - 3}{3t + 1}\right]^3 \cdot \frac{3t^2 + 2t + 9}{(3t + 1)^2}

Simplified:

f'(t) = \frac{4(t^2 - 3)^3(3t^2 + 2t + 9)}{(3t + 1)^5}

Multiple forms are correct, and you may need to manipulate yours to match multiple choice options.

Evaluating Derivatives at Specific Points

Chain Rule + Product Rule with Evaluation

Find

h'(2)

h(x) = 3x\sqrt{2 - x}

Step 1: Product rule.

h'(x) = 3\sqrt{2 - x} + 3x \cdot \frac{1}{2\sqrt{2 - x}} \cdot (-1) = 3\sqrt{2 - x} - \frac{3x}{2\sqrt{2 - x}}

Step 2: Plug in

x = 2

right away.

h'(2) = 3\sqrt{0} - \frac{6}{2\sqrt{0}}

That’s division by zero, so

h'(2)

is undefined.

A Valid Evaluation Point

Find

g'(-1)

g(x) = 4x\sqrt{3 - x}

Step 1: Find

g'(x)

g'(x) = 4\sqrt{3 - x} + 4x \cdot \frac{1}{2\sqrt{3 - x}} \cdot (-1) = 4\sqrt{3 - x} - \frac{2x}{\sqrt{3 - x}}

Step 2: Plug in

x = -1

g'(-1) = 4\sqrt{4} + \frac{2}{\sqrt{4}} = 4(2) + \frac{2}{2} = 8 + 1 = 9

Tip: plug in values as soon as you have the derivative. It cuts down on algebra errors.

Using Tables of Values

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
2	5	3	4	-1
4	8	6	2	4

Problem 1: Simple Chain Rule

Find

h'(2)

h(x) = [f(x)]^3

h'(x) = 3[f(x)]^2 \cdot f'(x)

h'(2) = 3[f(2)]^2 \cdot f'(2) = 3(5)^2 \cdot 3 = 3(25)(3) = 225

Problem 2: Square Root Chain Rule

Find

k'(4)

k(x) = \sqrt{g(x)}

k'(x) = \frac{1}{2\sqrt{g(x)}} \cdot g'(x)

k'(4) = \frac{1}{2\sqrt{2}} \cdot 4 = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

Problem 3: Nested Composition

Find

p'(2)

p(x) = f(g(x))

p'(x) = f'(g(x)) \cdot g'(x)

From the table,

g(2) = 4

and

g'(2) = -1

p'(2) = f'(4) \cdot (-1) = 6 \cdot (-1) = -6

Key Takeaways

Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

The “onion rule”: work from outside to inside, one layer at a time

Common error: squared trig functions like $\sin^2(3x)$ have three layers, not two (the square, the trig function, and the inner argument)

Log shortcut: $\frac{d}{dx}[\ln(x^n)] = \frac{n}{x}$ by using log properties first

Exponentials: $\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)$ , so the exponent stays the same and you multiply by its derivative

Leibniz form: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ , and the units chain together the same way

The chain rule often pairs with the product rule, quotient rule, or trig derivatives

Plug in numerical values right after finding the derivative to minimize errors

Answers can look different depending on how you simplify, so know how to manipulate your result

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