Chain Rule (CIE AS Maths: Pure 1)

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Chain Rule

What is the chain rule? 

  • If y is a function of u, and u is a function of x, then the chain rule tells us that

begin mathsize 22px style 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 end style

  • The chain rule allows us to differentiate more complicated expressions and composite functions
  • You will often see and use the chain rule with different variables
    • This is particularly useful for connected rates of change

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

How do I differentiate (f(x))n?

  • This method can be used for any linear or non – linear expression
    • Let u = f(x) and follow the method above
    • In general if space y equals left parenthesis straight f left parenthesis x right parenthesis right parenthesis to the power of n then fraction numerator straight d y over denominator straight d x end fraction equals n straight f to the power of straight apostrophe open parentheses x close parentheses open parentheses straight f open parentheses x close parentheses close parentheses to the power of n minus 1 end exponent  
  • With practice you will be able to carry out this method without the need for u
    • This is essential for learning the reverse chain rule later in the course

Worked example

5-1-4-chain-rule-we-solution-part-1

5-1-4-chain-rule-we-solution-part-2

Examiner Tip

If using as a substitution don't forget to substitute x back into your final answer.

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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.