Differentiating Inverse Functions (DP IB Analysis & Approaches (AA)): Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

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Differentiating inverse functions

What is meant by an inverse function?

  • Some functions are easier to handle with x (rather than y) as the subject

    • i.e.  in the form x equals f left parenthesis y right parenthesis

    • This can be particularly true when dealing with inverse functions

  • If y equals f left parenthesis x right parenthesis the inverse would be written as y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis

    • Finding f to the power of negative 1 end exponent left parenthesis x right parenthesis can be awkward

    • But a function and its inverse 'cancel' each other

      • This means space y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis space rightwards double arrow space f open parentheses y close parentheses equals f open parentheses f to the power of negative 1 end exponent left parenthesis x right parenthesis close parentheses space rightwards double arrow space f open parentheses y close parentheses equals x

    • So you can write x equals f left parenthesis y right parenthesis instead

      • This expresses the same relationship between x and y as y equals f to the power of negative 1 end exponent open parentheses x close parentheses does

How do I differentiate inverse functions?

  • With x equals f left parenthesis y right parenthesis it is easier to differentiate “x with respect to y” rather than “y with respect to x

    • I.e. to find fraction numerator straight d x over denominator straight d y end fraction rather than fraction numerator straight d y over denominator straight d x end fraction

    • Note that fraction numerator straight d x over denominator straight d y end fraction will be in terms of y

  • STEP 1
    For the function y equals f left parenthesis x right parenthesis, the inverse will be y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis

    Rewrite y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis as x equals f left parenthesis y right parenthesis

    • E.g. Find an expression in terms of y for the derivative fraction numerator straight d y over denominator straight d x end fraction of y equals f to the power of negative 1 end exponent open parentheses x close parentheses, where f open parentheses x close parentheses equals x cubed minus 1

y equals f to the power of negative 1 end exponent open parentheses x close parentheses space space rightwards double arrow space space x equals f open parentheses y close parentheses space space rightwards double arrow space space x equals y cubed minus 1

  • STEP 2
    From x equals f left parenthesis y right parenthesis find fraction numerator straight d x over denominator straight d y end fraction

fraction numerator straight d x over denominator straight d y end fraction equals 3 y squared

  • STEP 3
    Find fraction numerator straight d y over denominator straight d x end fraction using fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator 1 over denominator fraction numerator straight d x over denominator straight d y end fraction end fraction  (this will usually be in terms of y)

fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator 1 over denominator fraction numerator straight d x over denominator straight d y end fraction end fraction equals fraction numerator 1 over denominator 3 y squared end fraction

  • This can be used to find the gradient of the curve y equals f to the power of negative 1 end exponent open parentheses x close parentheses at a given point

    • Substitute the bold italic y-coordinate of the point into your expression for fraction numerator straight d y over denominator straight d x end fraction

      • If the y-coordinate is not given, you should be able to work it out from the original function and x-coordinate

    • E.g. for the function used above, find the gradient of y equals f to the power of negative 1 end exponent open parentheses x close parentheses at the point where x equals 7

      • You need to know the value of y equals f to the power of negative 1 end exponent open parentheses 7 close parentheses

        • y equals f to the power of negative 1 end exponent open parentheses 7 close parentheses space space rightwards double arrow space space f open parentheses y close parentheses equals 7 space space rightwards double arrow space space y cubed minus 1 equals 7 space space rightwards double arrow space space y equals 2

      • Substitute y equals f to the power of negative 1 end exponent open parentheses 7 close parentheses equals 2 into the expression for fraction numerator straight d y over denominator straight d x end fraction

fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator 1 over denominator 3 open parentheses 2 close parentheses squared end fraction equals 1 over 12

Examiner Tips and Tricks

With x's and y's everywhere this can soon get confusing! Be clear about the key information and steps, and set your working out accordingly:

  • The original function,  y equals f left parenthesis x right parenthesis

  • Its inverse,  y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis

  • Rewriting the inverse,  x equals f left parenthesis y right parenthesis

  • Finding fraction numerator straight d x over denominator straight d y end fraction first, then finding its reciprocal for fraction numerator straight d y over denominator straight d x end fraction

Your GDC can help when numerical derivatives (gradients) are required.

Worked Example

a)  Let f left parenthesis x right parenthesis equals square root of left parenthesis 5 x plus 1 right parenthesis cubed end root. Find the gradient of the curve y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis at the point where y equals 3.

5-8-1-ib-hl-aa-only-we2a-soltn

b)       By considering y equals straight e to the power of x, show that the derivative of y equals ln space x is 1 over x.

5-8-1-ib-hl-aa-only-we2b-soltn-

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Paul

Author: Paul

Expertise: Maths Content Creator

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.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject 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.