Techniques of Differentiation (Edexcel IGCSE Further Pure Maths)

Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

Product Rule

What is the product rule?

  • The product rule states that if  y equals straight f open parentheses x close parentheses straight g open parentheses x close parentheses  is the product of two functions straight f left parenthesis x right parenthesis and straight g left parenthesis x right parenthesis then

    •  fraction numerator straight d y over denominator straight d x end fraction equals straight f apostrophe open parentheses x close parentheses straight g open parentheses x close parentheses plus straight f open parentheses x close parentheses straight g apostrophe open parentheses x close parentheses

  • This is not given on the exam formula sheet, so you need to remember it

  • This is sometimes written as  y equals u v  where u and v are both functions of x

    • Then y apostrophe equals u apostrophe v plus u v apostrophe

      • where  y apostrophe equals fraction numerator straight d y over denominator straight d x end fraction,  u apostrophe equals fraction numerator straight d u over denominator straight d x end fraction  and  v apostrophe equals fraction numerator straight d v over denominator straight d x end fraction

  • 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)

      • space sin left parenthesis cos space x right parenthesis is a function of a function, “sin of cos of x

      •  sin space x cos space x is a product of two functions, “sin x times cos x

How do I use the product rule?

  • To differentiate  y equals straight f open parentheses x close parentheses straight g open parentheses x close parentheses

    • you'll need to make it clear what straight f open parentheses x close parentheses comma space straight g open parentheses x close parentheses comma space straight f to the power of apostrophe open parentheses x close parentheses and straight g to the power of apostrophe open parentheses x close parentheses are

      • Arranging them in a square can help

  • STEP 1
    Identify the two functions, straight f open parentheses x close parentheses and straight g open parentheses x close parentheses

    • Then differentiate each one with respect tospace x to find straight f to the power of apostrophe open parentheses x close parentheses and straight g to the power of apostrophe open parentheses x close parentheses

  • STEP 2
    Obtain fraction numerator straight d y over denominator straight d x end fraction by applying the product rule formula

    • If  y equals straight f open parentheses x close parentheses straight g open parentheses x close parentheses,  then  fraction numerator straight d y over denominator straight d x end fraction equals straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis straight g stretchy left parenthesis x stretchy right parenthesis plus straight f stretchy left parenthesis x stretchy right parenthesis straight g to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis

    • 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 apostrophe and v apostrophe

Examiner Tips and Tricks

  • Using u comma space v comma space u apostrophe and v apostrophe 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 apostrophe and v apostrophe

Worked Example

Find the derivative of space y equals 5 x squared cos space 3 x.

We'll use  y equals u v  form

Identify the functions u and v

u equals 5 x squared space space space space space space space space space space space v equals cos 3 x


Differentiate those to find u apostrophe and v apostrophe

u apostrophe equals 10 x space space space space space space space space space space space v apostrophe equals negative 3 sin 3 x


Put the pieces together using  y apostrophe equals u apostrophe v plus u v apostrophe

fraction numerator straight d y over denominator straight d x end fraction equals open parentheses 10 x close parentheses open parentheses cos 3 x close parentheses plus open parentheses 5 x squared close parentheses open parentheses negative 3 sin 3 x close parentheses


Expand the brackets


fraction numerator bold d bold y over denominator bold d bold x end fraction bold equals bold 10 bold italic x bold cos bold 3 bold italic x bold minus bold 15 bold italic x to the power of bold 2 bold sin bold 3 bold italic x

Quotient Rule

What is the quotient rule?

  • The quotient rule states if  y equals fraction numerator straight f open parentheses x close parentheses over denominator straight g open parentheses x close parentheses end fraction  is the quotient of two functions straight f open parentheses x close parentheses and straight g open parentheses x close parentheses then

    • fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight f to the power of apostrophe open parentheses x close parentheses straight g open parentheses x close parentheses minus straight f open parentheses x close parentheses straight g to the power of apostrophe open parentheses x close parentheses over denominator open square brackets straight g open parentheses x close parentheses close square brackets squared end fraction

      • This is given on the exam formula sheet, so you don't need to remember it

  • This is sometimes written as  y equals u over v  where u and v are both functions of x

    • Then  y apostrophe equals fraction numerator u apostrophe v minus u v apostrophe over denominator v squared end fraction

      • where  y apostrophe equals fraction numerator straight d y over denominator straight d x end fraction,  u apostrophe equals fraction numerator straight d u over denominator straight d x end fraction  and  v apostrophe equals fraction numerator straight d v over denominator straight d x end fraction

  • 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  fraction numerator k over denominator straight g open parentheses x close parentheses end fraction equals k open parentheses straight g open parentheses x close parentheses close parentheses to the power of negative 1 end exponent

      • The chain rule can be used here

      • Note that open parentheses straight g open parentheses x close parentheses close parentheses to the power of negative 1 end exponent is different from the inverse function straight g to the power of negative 1 end exponent open parentheses x close parentheses

    • If the denominator is a constant, treat it as a factor of the expression

      • fraction numerator straight f open parentheses x close parentheses over denominator k end fraction equals 1 over k straight f open parentheses x close parentheses

    • 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 equals fraction numerator straight f open parentheses x close parentheses over denominator straight g open parentheses x close parentheses end fraction

    • you'll need to make it clear what straight f open parentheses x close parentheses comma space straight g open parentheses x close parentheses comma space straight f to the power of apostrophe open parentheses x close parentheses and straight g to the power of apostrophe open parentheses x close parentheses are

      • Arranging them in a square can help

  • STEP 1
    Identify the two functions, straight f open parentheses x close parentheses and straight g open parentheses x close parentheses

    • Then differentiate each one with respect tospace x to find straight f to the power of apostrophe open parentheses x close parentheses and straight g to the power of apostrophe open parentheses x close parentheses

  • STEP 2
    Obtain fraction numerator straight d y over denominator straight d x end fraction by applying the quotient rule formula

    • If  y equals fraction numerator straight f open parentheses x close parentheses over denominator straight g open parentheses x close parentheses end fraction,  then  fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis straight g stretchy left parenthesis x stretchy right parenthesis minus straight f stretchy left parenthesis x stretchy right parenthesis straight g to the power of apostrophe open parentheses x close parentheses over denominator open square brackets straight g open parentheses x close parentheses close square brackets squared end fraction

    • 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 apostrophe and v apostrophe

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 apostrophe and v apostrophe 

  • Look out for functions of the form space y equals straight f left parenthesis x right parenthesis left parenthesis straight g left parenthesis x right parenthesis right parenthesis to the power of negative 1 end exponent

    • 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 equals fraction numerator straight f open parentheses x close parentheses over denominator straight g open parentheses x close parentheses end fraction

      • and then use the quotient rule

Worked Example

Given the function straight f stretchy left parenthesis x stretchy right parenthesis equals fraction numerator cos space 2 x over denominator 3 x plus 2 end fraction,  find straight f to the power of apostrophe open parentheses x close parentheses.

We'll use y equals fraction numerator straight f open parentheses x close parentheses over denominator straight g open parentheses x close parentheses end fraction form

Identify the functions straight f open parentheses x close parentheses and straight g open parentheses x close parentheses

straight f open parentheses x close parentheses equals cos 2 x space space space space space space space space space space space straight g open parentheses x close parentheses equals 3 x plus 2


Differentiate those to find straight f apostrophe open parentheses x close parentheses and straight g to the power of apostrophe open parentheses x close parentheses

straight f to the power of apostrophe open parentheses x close parentheses equals negative 2 sin 2 x space space space space space space space space space space space straight g to the power of apostrophe open parentheses x close parentheses equals 3

Put the pieces together using  fraction numerator straight d over denominator straight d x end fraction open parentheses fraction numerator straight f open parentheses x close parentheses over denominator straight g open parentheses x close parentheses end fraction close parentheses equals fraction numerator straight f to the power of apostrophe open parentheses x close parentheses straight g open parentheses x close parentheses minus straight f open parentheses x close parentheses straight g to the power of apostrophe open parentheses x close parentheses over denominator open square brackets straight g open parentheses x close parentheses close square brackets squared end fraction from the exam formula sheet

straight f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses negative 2 sin 2 x close parentheses open parentheses 3 x plus 2 close parentheses minus open parentheses cos 2 x close parentheses open parentheses 3 close parentheses over denominator open parentheses 3 x plus 2 close parentheses squared end fraction


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


bold f to the power of bold apostrophe stretchy left parenthesis x stretchy right parenthesis bold equals fraction numerator bold minus bold 2 stretchy left parenthesis 3 x plus 2 stretchy right parenthesis bold sin bold 2 bold x bold minus bold 3 bold cos bold 2 bold x over denominator stretchy left parenthesis 3 x plus 2 stretchy right parenthesis to the power of bold 2 end fraction

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

    • fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight d y over denominator straight d u end fraction cross times fraction numerator straight d u over denominator straight d x end fraction

      • 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 equals straight f left parenthesis straight g left parenthesis x right parenthesis right parenthesis,  then  fraction numerator straight d y over denominator straight d x end fraction equals straight f apostrophe left parenthesis straight g left parenthesis x right parenthesis right parenthesis straight g apostrophe left parenthesis x right parenthesis

  • The chain rule makes it possible to differentiate a function of a function

    • For example  y equals open parentheses 5 x to the power of 4 minus 2 x close parentheses to the power of 7

      • y equals u to the power of 7

      • u equals 5 x to the power of 4 minus 2 x

    • Or  y equals cos open parentheses 3 x squared minus 1 close parentheses

      • y equals cos u

      • u equals 3 x squared minus 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 equals open parentheses x squared minus 5 x plus 7 close parentheses to the power of 7

      • The other option would require trying to expand open parentheses x squared minus 5 x plus 7 close parentheses to the power of 7 first!

    • You may need to use laws of indices

      • e.g. finding the derivative of  y equals fraction numerator 1 over denominator square root of 2 x minus 3 end root end fraction

      • Rewrite first as y equals open parentheses 2 x minus 3 close parentheses to the power of negative 1 half end exponent

  • A power of a function can be differentiated using this special case of the chain rule:

    • If y equals open square brackets straight f open parentheses x close parentheses close square brackets to the power of n

      • i.e. if y is the function straight f open parentheses x close parentheses raised to the power n

    • then  space fraction numerator straight d y over denominator straight d x end fraction equals n straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis stretchy left square bracket straight f open parentheses x close parentheses stretchy right square bracket to the power of n minus 1 end exponent

      • i.e. n times the derivative of straight f open parentheses x close parentheses, times straight f open parentheses x close parentheses to the power of n minus 1

      • compare fraction numerator straight d over denominator straight d x end fraction open parentheses x to the power of n close parentheses equals n x to the power of n minus 1 end exponent

    • 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 equals cos open parentheses 3 x squared minus 1 close parentheses

      • y equals cos u

      • u equals 3 x squared minus 1

  • STEP 2
    Find  fraction numerator straight d y over denominator straight d u end fraction and fraction numerator straight d u over denominator straight d x end fraction

    • fraction numerator straight d y over denominator straight d u end fraction equals negative sin u

    • fraction numerator straight d u over denominator straight d x end fraction equals 6 x

  • STEP 3
    Substitute into  fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight d y over denominator straight d u end fraction cross times fraction numerator straight d u over denominator straight d x end fraction

    • fraction numerator straight d y over denominator straight d x end fraction equals open parentheses negative sin u close parentheses cross times open parentheses 6 x close parentheses

  • STEP 4
    Substitute the expression for u back in

    • This gets fraction numerator straight d y over denominator straight d x end fraction entirely in terms of x

      • fraction numerator straight d y over denominator straight d x end fraction equals open parentheses negative sin open parentheses 3 x squared minus 1 close parentheses close parentheses cross times open parentheses 6 x close parentheses equals negative 6 x sin open parentheses 3 x squared minus 1 close parentheses

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 ofspace y equals left parenthesis x squared minus 5 x plus 7 right parenthesis to the power of 7.

This is y equals open square brackets straight f open parentheses x close parentheses close square brackets to the power of n with n equals 7 and straight f open parentheses x close parentheses equals x squared minus 5 x plus 7

Start by finding the derivative of straight f open parentheses x close parentheses

fraction numerator straight d over denominator straight d x end fraction open parentheses x squared minus 5 x plus 7 close parentheses equals 2 x minus 5


Now use  space fraction numerator straight d y over denominator straight d x end fraction equals n straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis stretchy left square bracket straight f open parentheses x close parentheses stretchy right square bracket to the power of n minus 1 end exponent


fraction numerator straight d y over denominator straight d x end fraction equals 7 open parentheses 2 x minus 5 close parentheses open parentheses x squared minus 5 x plus 7 close parentheses to the power of 7 minus 1 end exponent


fraction numerator bold d bold y over denominator bold d bold x end fraction bold equals bold 7 stretchy left parenthesis 2 x minus 5 stretchy right parenthesis stretchy left parenthesis x squared minus 5 x plus 7 stretchy right parenthesis to the power of bold 6

 

(b) Find the derivative of  y equals square root of sin x end root.

First use laws of indices to write as a power


y equals open parentheses sin x close parentheses to the power of 1 half end exponent


This is y equals open square brackets straight f open parentheses x close parentheses close square brackets to the power of n with n equals 1 half and straight f open parentheses x close parentheses equals sin x

Find the derivative of straight f open parentheses x close parentheses


fraction numerator straight d over denominator straight d x end fraction open parentheses sin x close parentheses equals cos x


Now use  space fraction numerator straight d y over denominator straight d x end fraction equals n straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis stretchy left square bracket straight f open parentheses x close parentheses stretchy right square bracket to the power of n minus 1 end exponent

table row cell fraction numerator straight d y over denominator straight d x end fraction end cell equals cell open parentheses 1 half close parentheses open parentheses cos x close parentheses open parentheses sin x close parentheses to the power of 1 half minus 1 end exponent end cell row blank equals cell fraction numerator cos x over denominator 2 end fraction open parentheses sin x close parentheses to the power of negative 1 half end exponent end cell end table


And by laws of indices  open parentheses sin x close parentheses to the power of negative 1 half end exponent equals 1 over open parentheses sin x close parentheses to the power of 1 half end exponent equals fraction numerator 1 over denominator square root of sin x end root end fraction


fraction numerator bold d bold y over denominator bold d bold x end fraction bold equals fraction numerator bold cos bold x over denominator bold 2 square root of bold sin bold x end root end fraction

(c) Find the derivative of y equals straight e to the power of cos x end exponent.

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 equals straight e to the power of u

u equals cos x


Differentiate


fraction numerator straight d y over denominator straight d u end fraction equals straight e to the power of u

fraction numerator straight d u over denominator straight d x end fraction equals negative sin x


Substitute into  fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight d y over denominator straight d u end fraction cross times fraction numerator straight d u over denominator straight d x end fraction

fraction numerator straight d y over denominator straight d x end fraction equals open parentheses straight e to the power of u close parentheses cross times open parentheses negative sin x close parentheses


Substitute the expression for u back in

fraction numerator straight d y over denominator straight d x end fraction equals open parentheses straight e to the power of cos x end exponent close parentheses cross times open parentheses negative sin x close parentheses


fraction numerator bold d bold y over denominator bold d bold x end fraction bold equals bold minus stretchy left parenthesis sin x stretchy right parenthesis bold e to the power of bold cos bold x end exponent

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Paul

Author: Paul

Expertise: Maths Content Creator (Previous)

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.