The Chain Rule (College Board AP® Calculus AB)

Revision Note

Author

Jamie Wood

Expertise

Maths

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 $v$

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

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

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The Chain Rule (College Board AP® Calculus AB)

Revision Note

Derivatives of composite functions

How do I differentiate composite functions?

How do I use the chain rule?

Worked Example

How might chain rule questions be made more difficult?

Worked Example

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

Author: Jamie Wood