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Derivatives of Sine and Cosine Functions (College Board AP® Calculus AB)

Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Derivatives of sine and cosine functions

How do I differentiate sin x and cos x?

  • If f open parentheses x close parentheses equals sin space x,

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

  • If g open parentheses x close parentheses equals cos space x,

    • then g to the power of apostrophe open parentheses x close parentheses equals negative sin space x

  • This can be continued...

  • If h open parentheses x close parentheses equals negative sin space x,

    • then h to the power of apostrophe open parentheses x close parentheses equals negative cos space x

  • If space j open parentheses x close parentheses equals negative cos space x,

    • then space j to the power of apostrophe open parentheses x close parentheses equals sin space x

  • The sequence then repeats

How do I differentiate sin kx and cos kx?

  • If f open parentheses x close parentheses equals sin space k x,

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

  • If g open parentheses x close parentheses equals cos space k x,

    • then g to the power of apostrophe open parentheses x close parentheses equals negative k sin space k x

  • These occur as a result of applying the chain rule

Worked Example

Differentiate the following functions.

(a) f open parentheses x close parentheses equals cos space x space minus space 4 sin space x

Answer:

cos space x differentiates to negative sin space x
sin space x differentiates to cos space x

f to the power of apostrophe open parentheses x close parentheses equals negative sin space x minus 4 open parentheses cos space x close parentheses

Simplify

f to the power of apostrophe open parentheses x close parentheses equals negative sin space x minus 4 cos space x

(b) g open parentheses x close parentheses equals 9 sin space open parentheses 2 over 3 x close parentheses minus 2 cos open parentheses 3 x close parentheses

Answer:

sin space k x differentiates to k cos space k x
cos space k x differentiates to negative k sin space k x

g to the power of apostrophe open parentheses x close parentheses equals 9 times 2 over 3 cos space open parentheses 2 over 3 x close parentheses space minus space 2 open parentheses negative 3 sin space 3 x close parentheses

Simplify

g to the power of apostrophe open parentheses x close parentheses equals 6 cos open parentheses 2 over 3 x close parentheses plus 6 sin space 3 x equals 6 open parentheses cos space 2 over 3 x space plus space sin space 3 x close parentheses

How do I use the definition of a derivative to differentiate sin x and cos x?

  • The definition of a derivative as a limit can be used to obtain the above results

    • f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of fraction numerator f open parentheses x plus h close parentheses minus f open parentheses x close parentheses over denominator h end fraction

  • You should know the following two trigonometric addition formulae:

    • sin open parentheses A plus B close parentheses equals sin A cos B plus cos A sin B

    • cos open parentheses A plus B close parentheses equals cos A cos B minus sin A sin B

  • You should know the following two trigonometric limit theorems:

    • limit as x rightwards arrow 0 of fraction numerator sin x over denominator x end fraction equals 1

    • limit as x rightwards arrow 0 of fraction numerator cos x minus 1 over denominator x end fraction equals 0

  • If f open parentheses x close parentheses equals sin space x, then using the definition of a derivative,

    • f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of fraction numerator sin open parentheses x plus h close parentheses minus sin open parentheses x close parentheses over denominator h end fraction

  • Using the addition formula sin open parentheses A plus B close parentheses equals sin A cos B plus cos A sin B,

    • f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of fraction numerator sin space x cos space h plus cos space x sin space h minus sin space x over denominator h end fraction

  • Factorizing so that the terms fraction numerator sin space h over denominator h end fraction and fraction numerator cos space h space minus 1 over denominator h end fraction are present

    • f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of open parentheses sin space x open parentheses fraction numerator cos space h space minus 1 over denominator h end fraction close parentheses plus cos space x open parentheses fraction numerator sin space h over denominator h end fraction close parentheses close parentheses

  • Applying the limits limit as x rightwards arrow 0 of fraction numerator sin x over denominator x end fraction equals 1 and limit as x rightwards arrow 0 of fraction numerator cos x minus 1 over denominator x end fraction equals 0

    • f to the power of apostrophe open parentheses x close parentheses equals open parentheses sin space x open parentheses 0 close parentheses plus cos space x open parentheses 1 close parentheses close parentheses

  • So it can be concluded that f to the power of apostrophe open parentheses x close parentheses equals cos space x

  • The method for cos space x is shown in the worked example below

Worked Example

Use the definition of a derivative as a limit to show that the derivative of f open parentheses x close parentheses equals cos space x is f to the power of apostrophe open parentheses x close parentheses equals negative sin space x.

Answer:

Write down the definition of a derivative as a limit, and apply it to f open parentheses x close parentheses equals cos space x

f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of fraction numerator f open parentheses x plus h close parentheses minus f open parentheses x close parentheses over denominator h end fraction

f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of fraction numerator cos open parentheses x plus h close parentheses minus cos open parentheses x close parentheses over denominator h end fraction

Use the addition formula cos open parentheses A plus B close parentheses equals cos A cos B minus sin A sin B

f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of fraction numerator cos space x space cos space h space minus space sin space x space sin space h minus cos space x over denominator h end fraction

Factorize so that the terms fraction numerator sin space h over denominator h end fraction and fraction numerator cos space h space minus 1 over denominator h end fraction are present

f to the power of apostrophe open parentheses x close parentheses equals limit as h rightwards arrow 0 of open parentheses cos space x open parentheses fraction numerator cos space h space minus 1 over denominator h end fraction close parentheses minus sin space x space open parentheses fraction numerator sin space h over denominator h end fraction close parentheses close parentheses

Apply the limits limit as x rightwards arrow 0 of fraction numerator sin x over denominator x end fraction equals 1 and limit as x rightwards arrow 0 of fraction numerator cos x minus 1 over denominator x end fraction equals 0

f to the power of apostrophe open parentheses x close parentheses equals cos space x open parentheses 0 close parentheses minus sin space x open parentheses 1 close parentheses

Simplify

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

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    Author
    Reviewer
    Jamie Wood

    Author: Jamie Wood

    Expertise: Maths

    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.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths 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.