The Chain Rule (College Board AP® Calculus AB)

Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Derivatives of composite functions

How do I differentiate composite functions?

Composite functions, of the form $f (g (x))$ can be differentiated using the chain rule
The chain rule states that, if $y = f (u)$ and $u = g (x)$ (i.e. $y$ is a function of $u$ , and $u$ is a function of $x$ ), then
- $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
Or in function notation, if $h (x) = f (g (x))$
- $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$
It is called the chain rule as this idea can be extended in a longer 'chain'
- E.g. $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d r} \times \frac{d r}{d x}$
The terms are not really fractions (they are derivatives), but we can treat them in a similar way in this context
- You should be able to spot how the terms on the right cancel to get $\frac{d y}{d x}$

How do I use the chain rule?

This is easiest to demonstrate through an example
Using the chain rule in the form $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ to differentiate $y = {(2 x^{3} + 4 x)}^{6}$
Substitute the "inside" function for $u$
- $u = 2 x^{3} + 4 x$
- So now $y = u^{6}$
Differentiate $u$ with respect to $x$
- $\frac{d u}{d x} = 6 x^{2} + 4$
Differentiate $y$ with respect to $u$
- $\frac{d y}{d u} = 6 u^{5}$
Apply the chain rule which states $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
- $\frac{d y}{d x} = 6 u^{5} \times (6 x^{2} + 4)$
Substitute for $u$ using the same substitution that was made at the start
- $\frac{d y}{d x} = 6 {(2 x^{3} + 4 x)}^{5} (6 x^{2} + 4)$
After plenty of practice you may be able to skip several of these steps
- E.g. To differentiate $y = {(2 x^{3} + 4 x)}^{6}$ ,
  - Use the power as a coefficient, and reduce the power by 1 (as you would for powers of $x$ )
  - Then multiply by the derivative of the inside function
  - $\frac{d y}{d x} = 6 {(2 x^{3} + 4 x)}^{5} (6 x^{2} + 4)$

Worked Example

Differentiate the following functions.

(a) $f (x) = \sin (3 x^{4} + 5 x)$

Answer:

Substitute the inside function for u

$u = 3 x^{4} + 5 x$

$y = \sin u$

Differentiate $u$ with respect to $x$ and differentiate $y$ with respect to $u$

$\frac{d u}{d x} = 12 x^{3} + 5$

$\frac{d y}{d u} = \cos u$

Apply the chain rule which states $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

$\frac{d y}{d x} = \cos u \cdot (12 x^{3} + 5)$

Substitute for $u$ using the same substitution that was made at the start

$\frac{d y}{d x} = \cos (3 x^{4} + 5 x) \cdot (12 x^{3} + 5)$

$f^{'} (x) = (12 x^{3} + 5) \cos (3 x^{4} + 5 x)$

(b) $g (x) = e^{\tan x}$

Answer:

Substitute the inside function for u (note here that $\tan x$ is 'inside' the exponential function)

$u = \tan x$

$y = e^{u}$

Differentiate $u$ with respect to $x$ and differentiate $y$ with respect to $u$

$\frac{d u}{d x} = \sec^{2} x$

$\frac{d y}{d u} = e^{u}$

Apply the chain rule which states $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

$\frac{d y}{d x} = e^{u} \cdot \sec^{2} x$

Substitute for $u$ using the same substitution that was made at the start

$\frac{d y}{d x} = e^{\tan x} \cdot \sec^{2} x$

$f^{'} (x) = e^{\tan x} \sec^{2} x$

How might chain rule questions be made more difficult?

The chain rule may be combined with other skills in more complex problems
- When applying the quotient rule or product rule, one or both of the expressions may require the chain rule
- Alternatively, when applying the chain rule, you may need to use the product or quotient rule
You may also encounter a function within a function, within another function
- E.g. $\sin (e^{3 x^{2} + 2})$ or $\ln (\sin (2 x^{3} + 3 x))$
- In this scenario you would have to apply the chain rule twice
- This is the same as making two substitutions, for $u$ and $v$ , and using $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d v} \times \frac{d v}{d x}$

Worked Example

Differentiate the following functions.

(a) $f (x) = \cos x \cdot e^{4 x^{3} + 3 x^{2} + 2}$

Answer:

Applying the product rule

Let $u = \cos x$ and $v = e^{4 x^{3} + 3 x^{2} + 2}$

Differentiate $u$

$u^{'} = - \sin x$

To differentiate $v$ , use the chain rule
It is a good idea to use a different letter for the substitution if you have already used $u$

Let $w = 4 x^{3} + 3 x^{2} + 2$ , so $v = e^{w}$

$\frac{d w}{d x} = 12 x^{2} + 6 x$ and $\frac{d v}{d w} = e^{w}$

$\begin{array}{rcl} \frac{d v}{d x} & = & \frac{d v}{d w} \times \frac{d w}{d x} \\ \frac{d v}{d x} & = & e^{w} \cdot (12 x^{2} + 6 x) \\ v^{'} & = & (12 x^{2} + 6 x) e^{4 x^{3} + 3 x^{2} + 2} \end{array}$

This might be a chain rule application that you do not need to work out formally
The exponential term is simply multiplied by the derivative of the power

Apply the product rule, $u' v + u v'$

$f^{'} (x) = - \sin x \cdot e^{4 x^{3} + 3 x^{2} + 2} + \cos x \cdot (12 x^{2} + 6 x) e^{4 x^{3} + 3 x^{2} + 2}$

This could be factorized

$f^{'} (x) = e^{4 x^{3} + 3 x^{2} + 2} ((12 x^{2} + 6 x) \cos x - \sin x)$

(b) $g (x) = \sin ({(3 x^{4} + 5 x^{2})}^{5})$

Answer:

Make a substitution for the function inside the sin function

$u = {(3 x^{4} + 5 x^{2})}^{5}$ so $y = \sin u$

The function $u$ requires the chain rule to be differentiate it
This might be a chain rule application that you do not need to work out formally, but we will show the full working for this example

Make another substitution, with a different letter, for the function inside $u$

$v = 3 x^{4} + 5 x^{2}$ so $u = v^{5}$

Find the derivatives of the expressions which no longer require the chain rule

$\frac{d y}{d u} = \cos u$

$\frac{d u}{d v} = 5 v^{4}$

$\frac{d v}{d x} = 12 x^{3} + 10 x$

Find an expression for $\frac{d y}{d x}$ involving these three derivatives

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d v} \times \frac{d v}{d x}$

Fill in the derivatives

$\frac{d y}{d x} = \cos u \cdot 5 v^{4} \cdot (12 x^{3} + 10 x)$

Substitute in $u$ and $v$

$\frac{d y}{d x} = \cos ({(3 x^{4} + 5 x^{2})}^{5}) \cdot 5 {(3 x^{4} + 5 x^{2})}^{4} \cdot (12 x^{3} + 10 x)$

$g^{'} (x) = 5 {(3 x^{4} + 5 x^{2})}^{4} (12 x^{3} + 10 x) \cos ({(3 x^{4} + 5 x^{2})}^{5})$

Last updated: 8 August 2024

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Test yourself

Did this page help you?

Previous:Derivatives of Tangent and Reciprocal Trig FunctionsNext:The Inverse Function Theorem

The Chain Rule (College Board AP® Calculus AB)

Study Guide

Derivatives of composite functions

How do I differentiate composite functions?

How do I use the chain rule?

How might chain rule questions be made more difficult?

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Continuous Functions

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Implicit Differentiation

Implicit Differentiation

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules

Constant of Integration

Riemann Sums & Definite Integrals