AP®MathsCollege BoardCalculus ABRevision NotesUnit 3: Differentiation of Composite, Implicit & Inverse FunctionsDifferentiation of Composite & Inverse FunctionsDerivatives of Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions (College Board AP® Calculus AB)

Revision Note

Jamie Wood

Expertise

Maths

Derivatives of inverse trigonometric functions

How do I differentiate inverse trig functions?

  • The inverse trigonometric functions (sin to the power of negative 1 end exponent x etc) can be differentiated using:

    • The inverse function theorem, written as either

      • fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator open parentheses fraction numerator d x over denominator d y end fraction close parentheses end fraction

      • or g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction where g open parentheses x close parentheses equals f to the power of negative 1 end exponent open parentheses x close parentheses

    • The chain rule

    • Trigonometric identities

  • Note that you may also see inverse trig functions referred to with "arc" notation

    • E.g. sin to the power of negative 1 end exponent x equals arcsin space x

How do I differentiate inverse sine?

  • Using the inverse function theorem, g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction where g open parentheses x close parentheses equals f to the power of negative 1 end exponent open parentheses x close parentheses

  • Let g open parentheses x close parentheses equals sin to the power of negative 1 end exponent x and f open parentheses x close parentheses equals sin space x

    • So f to the power of apostrophe open parentheses x close parentheses equals cos space x

  • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction equals fraction numerator 1 over denominator cos open parentheses sin to the power of negative 1 end exponent x close parentheses end fraction

  • Recall the identity sin squared x space plus space cos squared x identical to 1

    • This rearranges to cos space x equals square root of 1 minus sin squared x end root

  • Use this identity for the denominator

    • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator square root of 1 minus sin squared open parentheses sin to the power of negative 1 end exponent x close parentheses end root end fraction

    • sin squared open parentheses sin to the power of negative 1 end exponent x close parentheses is the same as open parentheses sin open parentheses sin to the power of negative 1 end exponent x close parentheses close parentheses squared equals open parentheses x close parentheses squared

    • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction

  • fraction numerator d over denominator d x end fraction open parentheses sin to the power of negative 1 end exponent x close parentheses equals fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction

  • This result is only true for when sin space x has an inverse

    • For purposes of defining the 'official' inverse of sin x, the domain of sin x is restricted to negative straight pi over 2 less or equal than x less or equal than straight pi over 2

    • The range of sin x is open square brackets negative 1 comma space 1 close square brackets, so the domain for sin to the power of negative 1 end exponent x must be negative 1 less or equal than x less or equal than 1 or open vertical bar x close vertical bar less or equal than 1

    • However the derivative of sin to the power of negative 1 end exponent x is only defined for negative 1 less than x less than 1 or open vertical bar x close vertical bar less than 1

      • This can be seen by inspecting the denominator of g to the power of apostrophe open parentheses x close parentheses

      • The derivative becomes unbounded for x equals 1 or x equals negative 1

How do I differentiate inverse cosine?

  • Using the inverse function theorem, g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction where g open parentheses x close parentheses equals f to the power of negative 1 end exponent open parentheses x close parentheses

  • Let g open parentheses x close parentheses equals cos to the power of negative 1 end exponent x and f open parentheses x close parentheses equals cos space x

    • So f to the power of apostrophe open parentheses x close parentheses equals negative sin space x

  • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction equals fraction numerator 1 over denominator negative sin open parentheses cos to the power of negative 1 end exponent x close parentheses end fraction

  • Recall the identity sin squared x space plus space cos squared x identical to 1

    • This rearranges to sin space x equals square root of 1 minus cos squared x end root

  • Use this identity for the denominator

    • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator negative square root of 1 minus cos squared open parentheses cos to the power of negative 1 end exponent x close parentheses end root end fraction

    • cos squared open parentheses cos to the power of negative 1 end exponent open parentheses x close parentheses close parentheses is the same as open parentheses cos open parentheses cos to the power of negative 1 end exponent x close parentheses close parentheses squared equals open parentheses x close parentheses squared

    • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator negative square root of 1 minus x squared end root end fraction

  • fraction numerator d over denominator d x end fraction open parentheses cos to the power of negative 1 end exponent x close parentheses equals negative fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction

  • This result is only true for when cos space x has an inverse

    • For purposes of defining the 'official' inverse of cos x, the domain of cos x is restricted to 0 less or equal than x less or equal than straight pi

    • The range of cos x is open square brackets negative 1 comma space 1 close square brackets, so the domain for cos to the power of negative 1 end exponent x must be negative 1 less or equal than x less or equal than 1 or open vertical bar x close vertical bar less or equal than 1

    • However the derivative of cos to the power of negative 1 end exponent x is only defined for negative 1 less than x less than 1 or open vertical bar x close vertical bar less than 1

      • This can be seen by inspecting the denominator of g to the power of apostrophe open parentheses x close parentheses

      • The derivative becomes unbounded for x equals 1 or x equals negative 1

How do I differentiate inverse tangent?

  • Using the inverse function theorem, g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction where g open parentheses x close parentheses equals f to the power of negative 1 end exponent open parentheses x close parentheses

  • Let g open parentheses x close parentheses equals tan to the power of negative 1 end exponent x and f open parentheses x close parentheses equals tan space x

    • So f to the power of apostrophe open parentheses x close parentheses equals sec squared x

  • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses end fraction equals fraction numerator 1 over denominator sec squared open parentheses tan to the power of negative 1 end exponent x close parentheses end fraction

  • Recall the identity tan squared x space plus space 1 identical to sec squared x

    • This can be derived from sin squared x space plus space cos squared x identical to 1 by dividing by cos squared x

  • Use this identity for the denominator

    • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator tan squared open parentheses tan to the power of negative 1 end exponent x close parentheses space plus space 1 end fraction

    • tan squared open parentheses tan to the power of negative 1 end exponent open parentheses x close parentheses close parentheses is the same as open parentheses tan open parentheses tan to the power of negative 1 end exponent x close parentheses close parentheses squared equals open parentheses x close parentheses squared

    • g to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator x squared plus 1 end fraction

  • fraction numerator d over denominator d x end fraction open parentheses tan to the power of negative 1 end exponent x close parentheses equals fraction numerator 1 over denominator 1 plus x squared end fraction

  • This result is only true for when tan space x has an inverse

    • For purposes of defining the 'official' inverse of tan x, the domain of tan x is restricted to negative straight pi over 2 less than x less than straight pi over 2

    • The range of tan x is open parentheses negative infinity comma space infinity close parentheses, i.e. all real numbers, so the domain for tan to the power of negative 1 end exponent x is all real numbers

    • The derivative of tan x is also defined for all real numbers x

      • The derivative goes to zero as x goes to plus-or-minus infinity

Summary of derivatives of inverse trig functions

  • The methods above show how to find the derivatives of the three most common inverse trig functions

  • The derivatives of the inverses of the reciprocal trig functions can be found in a similar way

  • The table below summarizes the derivatives of all six inverse trig functions

Table of derivatives of inverse trig functions

f open parentheses x close parentheses

f to the power of apostrophe open parentheses x close parentheses

sin to the power of negative 1 end exponent x

fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction, space minus 1 less than x less than 1

cos to the power of negative 1 end exponent x

negative fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction, space minus 1 less than x less than 1

tan to the power of negative 1 end exponent x

fraction numerator 1 over denominator 1 plus x squared end fraction

csc to the power of negative 1 end exponent x

negative fraction numerator 1 over denominator open vertical bar x close vertical bar square root of x squared minus 1 end root end fraction, space x less than negative 1 or x greater than 1

sec to the power of negative 1 end exponent x

fraction numerator 1 over denominator open vertical bar x close vertical bar square root of x squared minus 1 end root end fraction, space x less than negative 1 or x greater than 1

cot to the power of negative 1 end exponent x

negative fraction numerator 1 over denominator 1 plus x squared end fraction

Worked Example

Find the derivative of f open parentheses x close parentheses equals arcsin open parentheses 2 x cubed plus e to the power of 2 x end exponent close parentheses.

Answer:

Recall that arcsin x is the same as sin to the power of negative 1 end exponent x, you can use whichever notation you prefer

Differentiating this function will require the chain rule, as it is a function within a function

y equals sin to the power of negative 1 end exponent open parentheses 2 x cubed plus e to the power of 2 x end exponent close parentheses

Let u equals 2 x cubed plus e to the power of 2 x end exponent, so that y equals sin to the power of negative 1 end exponent u

Differentiate both functions

fraction numerator d u over denominator d x end fraction equals 6 x squared plus 2 e to the power of 2 x end exponent

fraction numerator d y over denominator d u end fraction equals fraction numerator 1 over denominator square root of 1 minus u squared end root end fraction

Apply the chain rule

fraction numerator d y over denominator d x end fraction equals fraction numerator d y over denominator d u end fraction cross times fraction numerator d u over denominator d x end fraction equals fraction numerator 1 over denominator square root of 1 minus u squared end root end fraction times open parentheses 6 x squared plus 2 e to the power of 2 x end exponent close parentheses

Substitute u back in

fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator square root of 1 minus open parentheses 2 x cubed plus e to the power of 2 x end exponent close parentheses squared end root end fraction times open parentheses 6 x squared plus 2 e to the power of 2 x end exponent close parentheses

f to the power of apostrophe open parentheses x close parentheses equals fraction numerator 6 x squared plus 2 e to the power of 2 x end exponent over denominator square root of 1 minus open parentheses 2 x cubed plus e to the power of 2 x end exponent close parentheses squared end root end fraction

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    Jamie Wood

    Author: Jamie Wood

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.