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

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Differentiating Inverse Functions

What is meant by an inverse function?

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

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

  • This is particularly true when dealing with inverse functions

    • e.g.  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

      • so write x equals f left parenthesis y right parenthesis instead

How do I differentiate inverse functions?

  • Since 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.  find fraction numerator d x over denominator d y end fraction rather than fraction numerator d y over denominator d x end fraction

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

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 this as x equals f left parenthesis y right parenthesis

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

STEP 3
Find fraction numerator d y over denominator d x end fraction using fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator begin display style fraction numerator d x over denominator d y end fraction end style end fraction - this will usually be in terms of y

  • If an algebraic solution in terms of x is required substitute f left parenthesis x right parenthesis for y in fraction numerator d y over denominator d x end fraction

  • If a numerical derivative (e.g. a gradient) is required then use the y-coordinate

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

Examiner Tips and Tricks

  • With x's and y's everywhere this can soon get confusing!

    • Be clear of the key information and steps - and set your wokring out accordingly

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

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

Worked Example

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

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

b)       Given that y equals 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.