Chain Rule (Cambridge O Level Additional Maths)

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Chain Rule

What is the chain rule?

  •  The chain rule states ifbold space bold italic y is a function ofbold space bold italic u andbold space bold italic u is a function ofbold space bold italic x then

space y equals straight f left parenthesis u left parenthesis x right parenthesis right parenthesis

space fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals fraction numerator bold d bold italic y over denominator bold d bold italic u end fraction bold cross times fraction numerator bold d bold italic u over denominator bold d bold italic x end fraction

  • In function notation this could be written

space y equals straight f left parenthesis straight g left parenthesis x right parenthesis right parenthesis

space 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

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 (usuallyspace x) does not ‘appear alone’
      • space sin space x – not a composite function, x ‘appears alone’
      • sin left parenthesis 3 x plus 2 right parenthesis 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 ofspace u; space y equals straight f left parenthesis u right parenthesis
 Write u as a function ofspace xspace u equals straight g left parenthesis x right parenthesis

 STEP 2
Differentiate y with respect to u to getspace fraction numerator straight d y over denominator straight d u end fraction
Differentiate u with respect to x to getspace fraction numerator straight d u over denominator straight d x end fraction

 STEP 3
Obtain fraction numerator straight d y over denominator straight d x end fraction by applying the formulaspace 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 and substitutespace u back in forspace straight g left parenthesis x right parenthesis
 
  • 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
      • fraction numerator d u over denominator d x end fraction equals a and fraction numerator d y over denominator d u end fraction equals n u to the power of n minus 1 end exponent
    • Put both parts into the chain rule
      • fraction numerator straight d y over denominator straight d x end fraction equals blank fraction numerator straight d y over denominator straight d u end fraction blank cross times blank fraction numerator straight d u over denominator straight d x end fraction equals a blank cross times blank n u to the power of n minus 1 end exponent blank equals blank a n u to the power of n minus 1 end exponent blank
    • Substitute u = ax + b back into your answer
      • fraction numerator straight d y over denominator straight d x end fraction equals a n left parenthesis a x plus blank b right parenthesis to the power of n minus 1 end exponent

How do I differentiate √(ax+b)?

  • The chain rule allows us to use substitution to differentiate any function in the form y equals square root of a x plus b end root
  • Rewrite square root of a x plus b end root equals left parenthesis a x plus b right parenthesis to the power of 1 half end exponent 
    • Let u = ax + b, then y = u½
    • Differentiate both parts separately
      • fraction numerator d u over denominator d x end fraction equals a and fraction numerator d y over denominator d u end fraction equals 1 half u to the power of negative 1 half end exponent
    • Put both parts into the chain rule
      • fraction numerator straight d y over denominator straight d x end fraction equals blank fraction numerator straight d y over denominator straight d u end fraction blank cross times blank fraction numerator straight d u over denominator straight d x end fraction equals a blank cross times blank 1 half u to the power of negative 1 half end exponent blank equals blank a over 2 u to the power of negative 1 half end exponent blank
    • Substitute u = ax + b back into your answer
      • fraction numerator straight d y over denominator straight d x end fraction equals blank a over 2 open parentheses a x plus b close parentheses to the power of negative 1 half end exponent blank equals fraction numerator a over denominator 2 square root of a x plus b end root end fraction
  • 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
    • Ifsize 16px space size 16px y size 16px equals begin mathsize 16px style stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis end style to the power of size 16px n thenspace 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 straight f stretchy left parenthesis x stretchy right parenthesis to the power of n minus 1 end exponent
    • IfAlternative text not available thenAlternative text not available
    • Ifspace y equals ln stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis thenspace fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight f to the power of straight apostrophe stretchy left parenthesis x stretchy right parenthesis over denominator straight f stretchy left parenthesis x stretchy right parenthesis end fraction
    • Ifspace y equals sin stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis thenspace fraction numerator straight d y over denominator straight d x end fraction equals straight f to the power of straight apostrophe stretchy left parenthesis x stretchy right parenthesis cos stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis
    • Ifspace y equals cos stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis thenspace fraction numerator straight d y over denominator straight d x end fraction equals negative straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis sin stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis
    • If size 16px y size 16px equals size 16px tan begin mathsize 16px style stretchy left parenthesis straight f open parentheses x close parentheses stretchy right parenthesis end style then Alternative text not available

Examiner Tip

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


STEP 1   Identify the two functions and rewrite

y equals u to the power of 7
           u equals x squared minus 5 x plus 7
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 7 u to the power of 6         fraction numerator straight d u over denominator straight d x end fraction equals 2 x minus 5

STEP 3   Apply the chain rule, 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 7 u to the power of 6 open parentheses 2 x minus 5 close parentheses space

Substitute u in terms of x back in
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 7 open parentheses x squared minus 5 x plus 7 close parentheses to the power of 6 open parentheses 2 x minus 5 close parentheses

b)
Find the derivative ofspace y equals sin left parenthesis straight e to the power of 2 x end exponent right parenthesis.


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

"... differentiate sin space box enclose blank end enclose, ignore straight e to the power of 2 x end exponent"

fraction numerator straight d y over denominator straight d x end fraction equals cos open parentheses straight e to the power of 2 x end exponent close parentheses cross times box enclose space space space space space space space space space end enclose

"... multiply by the derivative of straight e to the power of 2 x end exponent": d
ifferentiate straight e to the power of 2 x end exponent using the result "if y equals straight e to the power of straight f stretchy left parenthesis x stretchy right parenthesis end exponent, then "

fraction numerator straight d y over denominator straight d x end fraction equals cos open parentheses straight e to the power of 2 x end exponent close parentheses cross times 2 straight e to the power of 2 x end exponent

 

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Amber

Author: Amber

Expertise: Maths

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.