Techniques of Differentiation (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Paul

Reviewed by: Dan Finlay

Product Rule

What is the product rule?

The product rule states that if $y = f (x) g (x)$ is the product of two functions $f (x)$ and $g (x)$ then
- $\frac{d y}{d x} = f' (x) g (x) + f (x) g' (x)$

This is not given on the exam formula sheet, so you need to remember it
This is sometimes written as $y = u v$ where $u$ and $v$ are both functions of $x$
- Then $y' = u' v + u v'$
  - where $y' = \frac{d y}{d x}$ , $u' = \frac{d u}{d x}$ and $v' = \frac{d v}{d x}$

For your final answer make sure you match the notation used in the question

How do I know when to use the product rule?

The product rule is used to differentiate a product of two functions
- This can easily be confused with a 'function of a function' (see the Chain Rule note)
  - $\sin (\cos x)$ is a function of a function, “sin of cos of $x$ ”
  - $\sin x \cos x$ is a product of two functions, “sin x times cos $x$ ”

How do I use the product rule?

To differentiate $y = f (x) g (x)$
- you'll need to make it clear what $f (x), g (x), f^{'} (x)$ and $g^{'} (x)$ are
  - Arranging them in a square can help
STEP 1
Identify the two functions, $f (x)$ and $g (x)$
- Then differentiate each one with respect to $x$ to find $f^{'} (x)$ and $g^{'} (x)$
STEP 2
Obtain $\frac{d y}{d x}$ by applying the product rule formula
- If $y = f (x) g (x)$ , then $\frac{d y}{d x} = f^{'} (x) g (x) + f (x) g^{'} (x)$
- Simplify the answer if
  - it is straightforward to do so
  - or if the question requires a particular form

In trickier problems chain rule may have to be used when finding $u'$ and $v'$

Examiner Tips and Tricks

Using $u, v, u'$ and $v'$ can save time writing
- lay them out in a 2x2 'square' to help keep which is which straight
For trickier functions chain rule may be required along with product rule
- i.e. either $u$ and/or $v$ could be a 'function of a function'
- So chain rule needed to find $u'$ and $v'$

Worked Example

Find the derivative of $y = 5 x^{2} \cos 3 x$ .

We'll use $y = u v$ form

Identify the functions $u$ and $v$

$u = 5 x^{2} v = \cos 3 x$

Differentiate those to find $u'$ and $v'$

$u' = 10 x v' = - 3 \sin 3 x$

Put the pieces together using $y' = u' v + u v'$

$\frac{d y}{d x} = (10 x) (\cos 3 x) + (5 x^{2}) (- 3 \sin 3 x)$

Expand the brackets

$\frac{d y}{d x} = 10 x \cos 3 x - 15 x^{2} \sin 3 x$

Quotient Rule

What is the quotient rule?

The quotient rule states if $y = \frac{f (x)}{g (x)}$ is the quotient of two functions $f (x)$ and $g (x)$ then
- $\frac{d y}{d x} = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{{[g (x)]}^{2}}$
  - This is given on the exam formula sheet, so you don't need to remember it
This is sometimes written as $y = \frac{u}{v}$ where $u$ and $v$ are both functions of $x$
- Then $y' = \frac{u' v - u v'}{v^{2}}$
  - where $y' = \frac{d y}{d x}$ , $u' = \frac{d u}{d x}$ and $v' = \frac{d v}{d x}$

For your final answer make sure you match the notation used in the question

How do I know when to use the quotient rule?

The quotient rule is used to differentiate a quotient of two functions
- If the numerator is a constant, negative powers can be used
  - e.g $\frac{k}{g (x)} = k {(g (x))}^{- 1}$
  - The chain rule can be used here
  - Note that ${(g (x))}^{- 1}$ is different from the inverse function $g^{- 1} (x)$
- If the denominator is a constant, treat it as a factor of the expression
  - $\frac{f (x)}{k} = \frac{1}{k} f (x)$
- The quotient rule will still work for both those cases
  - But it might not be the quickest method

How do I use the quotient rule?

To differentiate $y = \frac{f (x)}{g (x)}$
- you'll need to make it clear what $f (x), g (x), f^{'} (x)$ and $g^{'} (x)$ are
  - Arranging them in a square can help
STEP 1
Identify the two functions, $f (x)$ and $g (x)$
- Then differentiate each one with respect to $x$ to find $f^{'} (x)$ and $g^{'} (x)$
STEP 2
Obtain $\frac{d y}{d x}$ by applying the quotient rule formula
- If $y = \frac{f (x)}{g (x)}$ , then $\frac{d y}{d x} = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{{[g (x)]}^{2}}$
- Simplify the answer if
  - it is straightforward to do so
  - or if the question requires a particular form

In trickier problems chain rule may have to be used when finding $u'$ and $v'$

Examiner Tips and Tricks

For trickier functions chain rule may be required along with product rule
- i.e. either $u$ and/or $v$ could be a 'function of a function'
- So chain rule needed to find $u'$ and $v'$
Look out for functions of the form $y = f (x) (g {(x))}^{- 1}$
- These can be differentiated using a combination of chain rule and product rule
  - It would be good practice to try this sometime!
- But it's probably easier to use laws of indices to rewrite as $y = \frac{f (x)}{g (x)}$
  - and then use the quotient rule

Worked Example

Given the function $f (x) = \frac{\cos 2 x}{3 x + 2}$ , find $f^{'} (x)$ .

We'll use $y = \frac{f (x)}{g (x)}$ form

Identify the functions $f (x)$ and $g (x)$

$f (x) = \cos 2 x g (x) = 3 x + 2$

Differentiate those to find $f' (x)$ and $g^{'} (x)$

$f^{'} (x) = - 2 \sin 2 x g^{'} (x) = 3$

Put the pieces together using $\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{{[g (x)]}^{2}}$ from the exam formula sheet

$f' (x) = \frac{(- 2 \sin 2 x) (3 x + 2) - (\cos 2 x) (3)}{{(3 x + 2)}^{2}}$

Simplify the numerator
(The denominator is simplest left as it is)

$f^{'} (x) = \frac{- 2 (3 x + 2) \sin 2 x - 3 \cos 2 x}{{(3 x + 2)}^{2}}$

Chain Rule

What is the chain rule?

The chain rule is used to differentiate a composite function
- A function of a function
The chain rule is given by the formula
- $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
  - where $y$ is a function of $u$
  - and $u$ is a function of $x$
  - (Of course this ultimately makes $y$ a function of $x$ as well!)

It can also be written in function notation as
- If $y = f (g (x))$ , then $\frac{d y}{d x} = f' (g (x)) g' (x)$
The chain rule makes it possible to differentiate a function of a function
- For example $y = {(5 x^{4} - 2 x)}^{7}$
  - $y = u^{7}$
  - $u = 5 x^{4} - 2 x$
- Or $y = \cos (3 x^{2} - 1)$
  - $y = \cos u$
  - $u = 3 x^{2} - 1$

How can I use the chain rule to differentiate the power of a function?

The chain rule can be used to differentiate a 'power of a function'
- This can save you a lot of work
  - e.g. finding the derivative of $y = {(x^{2} - 5 x + 7)}^{7}$
  - The other option would require trying to expand ${(x^{2} - 5 x + 7)}^{7}$ first!
- You may need to use laws of indices
  - e.g. finding the derivative of $y = \frac{1}{\sqrt{2 x - 3}}$
  - Rewrite first as $y = {(2 x - 3)}^{- \frac{1}{2}}$

A power of a function can be differentiated using this special case of the chain rule:
- If $y = {[f (x)]}^{n}$
  - i.e. if $y$ is the function $f (x)$ raised to the power $n$
- then $\frac{d y}{d x} = n f^{'} (x) {[f (x)]}^{n - 1}$
  - i.e. $n$ times the derivative of $f (x)$ , times $f (x)$ to the power of $n - 1$
  - compare $\frac{d}{d x} (x^{n}) = n x^{n - 1}$
- $n$ can be any power (including fractional and negative powers)
- This formula is not on the exam formula sheet, so you need to remember it
The power of a function may also be differentiated using the general chain rule method in the next section
- But remembering the 'special case' formula is a lot quicker

How can I use the general chain rule to differentiate a function?

STEP 1
Identify $y$ and $u$
- e.g. $y = \cos (3 x^{2} - 1)$
  - $y = \cos u$
  - $u = 3 x^{2} - 1$
STEP 2
Find $\frac{d y}{d u}$ and $\frac{d u}{d x}$
- $\frac{d y}{d u} = - \sin u$
- $\frac{d u}{d x} = 6 x$
STEP 3
Substitute into $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
- $\frac{d y}{d x} = (- \sin u) \times (6 x)$
STEP 4
Substitute the expression for $u$ back in
- This gets $\frac{d y}{d x}$ entirely in terms of $x$
  - $\frac{d y}{d x} = (- \sin (3 x^{2} - 1)) \times (6 x) = - 6 x \sin (3 x^{2} - 1)$

Examiner Tips and Tricks

When asked to differentiate a 'power of a function', the chain rule is usually your best option
- Look out for 'hidden powers'
  - e.g. square roots (fractional powers)
  - or functions in a denominator (negative powers)
To integrate a general 'function of a function' the chain rule is your only option!

Worked Example

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

This is $y = {[f (x)]}^{n}$ with $n = 7$ and $f (x) = x^{2} - 5 x + 7$

Start by finding the derivative of $f (x)$

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

Now use $\frac{d y}{d x} = n f^{'} (x) {[f (x)]}^{n - 1}$

$\frac{d y}{d x} = 7 (2 x - 5) {(x^{2} - 5 x + 7)}^{7 - 1}$

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

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

First use laws of indices to write as a power

$y = {(\sin x)}^{\frac{1}{2}}$

This is $y = {[f (x)]}^{n}$ with $n = \frac{1}{2}$ and $f (x) = \sin x$

Find the derivative of $f (x)$

$\frac{d}{d x} (\sin x) = \cos x$

Now use $\frac{d y}{d x} = n f^{'} (x) {[f (x)]}^{n - 1}$

$\begin{array}{rcl} \frac{d y}{d x} & = & (\frac{1}{2}) (\cos x) {(\sin x)}^{\frac{1}{2} - 1} \\ = & \frac{\cos x}{2} {(\sin x)}^{- \frac{1}{2}} \end{array}$

And by laws of indices ${(\sin x)}^{- \frac{1}{2}} = \frac{1}{{(\sin x)}^{\frac{1}{2}}} = \frac{1}{\sqrt{\sin x}}$

$\frac{d y}{d x} = \frac{\cos x}{2 \sqrt{\sin x}}$

This is not a 'power of a function', so we need to use the general chain rule method

Start by identifying $y$ and $u$

$y = e^{u} u = \cos x$

Differentiate

$\frac{d y}{d u} = e^{u} \frac{d u}{d x} = - \sin x$

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

$\frac{d y}{d x} = (e^{u}) \times (- \sin x)$

Substitute the expression for $u$ back in

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

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

Last updated: 20 October 2024

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Previous: Differentiating Basic FunctionsNext:Finding Gradients

Techniques of Differentiation (Edexcel IGCSE Further Pure Maths)

Revision Note

Product Rule

What is the product rule?

How do I know when to use the product rule?

How do I use the product rule?

Examiner Tips and Tricks

Worked Example

Quotient Rule

What is the quotient rule?

How do I know when to use the quotient rule?

How do I use the quotient rule?

Examiner Tips and Tricks

Worked Example

Chain Rule

What is the chain rule?

How can I use the chain rule to differentiate the power of a function?

How can I use the general chain rule to differentiate a function?

Examiner Tips and Tricks

Worked Example

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Author: Paul

Author: Dan Finlay