Chain Rule (Cambridge (CIE) IGCSE Additional Maths): Revision Note

Exam code: 606

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Amber

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27 June 2025

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Chain rule

What is the chain rule?

The chain rule states if $y$ is a function of $u$ and $u$ is a function of $x$ then

$y = f (u (x))$

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

In function notation this could be written

$y = f (g (x))$

$\frac{d y}{d x} = f' (g (x)) g' (x)$

How do I know when to use the chain rule?

The chain rule is used when we are trying to differentiate composite functions
- “function of a function”
- these can be identified as the variable (usually $x$ ) does not ‘appear alone’
  - $\sin x$ – not a composite function, $x$ ‘appears alone’
  - $\sin (3 x + 2)$ is a composite function; $x$ is tripled and has 2 added to it before the sine function is applied

How do I use the chain rule?

STEP 1

Identify the two functions

Rewrite $y$ as a function of $u$ ; $y = f (u)$

Write $u$ as a function of $x$ ; $u = g (x)$

STEP 2

Differentiate $y$ with respect to $u$ to get $\frac{d y}{d u}$ Differentiate $u$ with respect to $x$ to get $\frac{d u}{d x}$

STEP 3

Obtain $\frac{d y}{d x}$ by applying the formula $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ and substitute $u$ back in for $g (x)$

In trickier problems chain rule may have to be applied more than once

How do I differentiate (ax + b)ⁿ?

For n = 2 you will most likely expand the brackets and differentiate each term separately
If n > 2 this becomes time-consuming and if n is not a positive integer we need a different method completely
The chain rule allows us to use substitution to differentiate any function in the form y = (ax + b)ⁿ
- Let u = ax + b, then y = uⁿ
- Differentiate both parts separately
  - $\frac{d u}{d x} = a$ and $\frac{d y}{d u} = n u^{n - 1}$
- Put both parts into the chain rule
  - $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} = a \times n u^{n - 1} = a n u^{n - 1}$
- Substitute u = ax + b back into your answer
  - $\frac{d y}{d x} = a n {(a x + b)}^{n - 1}$

How do I differentiate √(ax+b)?

The chain rule allows us to use substitution to differentiate any function in the form $y = \sqrt{a x + b}$
Rewrite $\sqrt{a x + b} = {(a x + b)}^{\frac{1}{2}}$
- Let u = ax + b, then y = u^½
- Differentiate both parts separately
  - $\frac{d u}{d x} = a$ and $\frac{d y}{d u} = \frac{1}{2} u^{- \frac{1}{2}}$
- Put both parts into the chain rule
  - $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} = a \times \frac{1}{2} u^{- \frac{1}{2}} = \frac{a}{2} u^{- \frac{1}{2}}$
- Substitute u = ax + b back into your answer
  - $\frac{d y}{d x} = \frac{a}{2} {(a x + b)}^{- \frac{1}{2}} = \frac{a}{2 \sqrt{a x + b}}$
This method can be used for any fractional power of any linear or non-linear expression
- Provided you know how to differentiate the non-linear expression

Are there any standard results for using chain rule?

The following general results are particularly useful
- If $y = {(f (x))}^{n}$ then $\frac{d y}{d x} = n f^{'} (x) f {(x)}^{n - 1}$

If $y = e^{f (x)}$ then $\frac{d y}{d x} = f^{'} (x) e^{f (x)}$
- If $y = \ln (f (x))$ then $\frac{d y}{d x} = \frac{f^{'} (x)}{f (x)}$
- If $y = \sin (f (x))$ then $\frac{d y}{d x} = f^{'} (x) \cos (f (x))$
- If $y = \cos (f (x))$ then $\frac{d y}{d x} = - f^{'} (x) \sin (f (x))$
- If $y = \tan (f (x))$ then $\frac{d y}{d x} = f^{'} (x) \sec^{2} (f (x))$

Examiner Tips and Tricks

You should aim to be able to spot and carry out the chain rule mentally (rather than use substitution)
- every time you use it, say it to yourself in your head “differentiate the first function ignoring the second, then multiply by the derivative of the second function"

Worked Example

a) Find the derivative of $y = {(x^{2} - 5 x + 7)}^{7}$ .

STEP 1 Identify the two functions and rewrite

$y = u^{7}$ $u = x^{2} - 5 x + 7$

STEP 2 Find $\frac{d y}{d u}$ and $\frac{d u}{d x}$

$\frac{d y}{d u} = 7 u^{6}$ $\frac{d u}{d x} = 2 x - 5$

STEP 3 Apply the chain rule, $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

$\frac{d y}{d x} = 7 u^{6} (2 x - 5)$

Substitute $u$ in terms of $x$ back in

$= 7 {(x^{2} - 5 x + 7)}^{6} (2 x - 5)$

$\frac{d y}{d x} = 7 (2 x - 5) {(x^{2} - 5 x + 7)}^{6}$

b) Find the derivative of $y = \sin (e^{2 x})$ .

(In this solution, we will be applying the mental method discussed in the Exam Tip above)

"... differentiate $\sin$ , ignore $e^{2 x}$ "

$\frac{d y}{d x} = \cos (e^{2 x}) \times$

"... multiply by the derivative of $e^{2 x}$ ": differentiate $e^{2 x}$ using the result "if $y = e^{f (x)}$ , then $\frac{d y}{d x} = f^{'} (x) e^{f (x)}$ "

$\frac{d y}{d x} = \cos (e^{2 x}) \times 2 e^{2 x}$

$\frac{d y}{d x} = 2 e^{2 x} \cos (e^{2 x})$

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Chain Rule (Cambridge (CIE) IGCSE Additional Maths): Revision Note

Chain rule

What is the chain rule?

How do I know when to use the chain rule?

How do I use the chain rule?

How do I differentiate √(ax+b)?

Are there any standard results for using chain rule?

Unlock more, it's free!

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

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