AP®MathsCollege BoardCalculus ABStudy GuidesUnit 2: Differentiation Definition & Fundamental PropertiesFundamental Properties of DifferentiationDerivatives of Tangent and Reciprocal Trig Functions

Derivatives of Tangent and Reciprocal Trig Functions (College Board AP® Calculus AB)

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Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Derivative of the tangent function

What is the derivative of tan x?

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

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

  • This can be shown using the identity tan space x space identical to fraction numerator sin space x space over denominator cos space x end fraction and the quotient rule

    • The quotient rule states that if y equals u over v, then y to the power of apostrophe equals fraction numerator u to the power of apostrophe v space minus space u v to the power of apostrophe over denominator v squared end fraction

    • Let u equals sin space x and v equals cos space x

    • So u to the power of apostrophe equals cos space x and v to the power of apostrophe equals negative sin space x

    • Applying the quotient rule

      • y to the power of apostrophe equals fraction numerator cos space x times cos space x space minus space sin space x times open parentheses negative sin space x close parentheses over denominator open parentheses cos space x close parentheses squared end fraction

    • Simplifying

      • y to the power of apostrophe equals fraction numerator cos squared space x space plus space sin squared space x over denominator cos squared space x end fraction

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

      • y to the power of apostrophe equals fraction numerator 1 over denominator cos squared space x end fraction

    • This is the definition of sec squared space x, as sec x equals fraction numerator 1 over denominator cos x end fraction

      • y to the power of apostrophe equals sec squared space x

What is the derivative of tan kx?

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

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

    • This is a result of applying the chain rule

  • This can also be shown in the same way as above, using the quotient rule for fraction numerator sin space k x over denominator cos space k x end fraction

Worked Example

Differentiate the following functions.

(a) f open parentheses x close parentheses equals tan space 3 x

Answer:

tan space x differentiates to k space sec squared space k x

f to the power of apostrophe open parentheses x close parentheses equals 3 sec squared 3 x

(b) g open parentheses x close parentheses equals 3 x squared space tan space x

Answer:

This is a product of two terms, so use the product rule, y to the power of apostrophe equals u to the power of apostrophe v plus u v to the power of apostrophe

u equals 3 x squared and v equals tan space x

u to the power of apostrophe equals 6 x and v to the power of apostrophe equals sec squared space x

y to the power of apostrophe equals 6 x times tan space x space plus space 3 x squared times sec squared space x equals 3 x open parentheses 2 tan space x space plus thin space x sec squared space x close parentheses

This derivative is correct, but could also be written in other forms using trigonometric identities

E.g. using tan squared space x space plus space 1 space equals space sec squared space x it could be written all in terms of tan space x

y to the power of apostrophe equals 6 x space tan space x space plus thin space 3 x squared open parentheses tan squared space x space plus space 1 close parentheses equals 3 x open parentheses x tan squared space x space plus space 2 tan space x space plus space x close parentheses

Derivatives of reciprocal trig functions

What are the reciprocal trig functions?

  • The reciprocal trigonometric functions are:

    • fraction numerator 1 over denominator sin space x end fraction equals csc space x

    • fraction numerator 1 over denominator cos space x end fraction equals sec space x

    • fraction numerator 1 over denominator tan space x end fraction equals cot space x

  • Sometimes you may see csc written as cosec

    • In this form, you can remember the functions by looking at the third letter of each

      • cosec is the reciprocal of sin

      • sec is the reciprocal of cos

      • cot is the reciprocal of tan

What are the derivatives of the reciprocal trig functions?

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

    • then f to the power of apostrophe open parentheses x close parentheses equals negative cot space x space csc space x

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

    • then g to the power of apostrophe open parentheses x close parentheses equals tan space x space sec space x

  • If h open parentheses x close parentheses equals cot space x,

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

  • These results can be remembered, or they can be derived using the reciprocal trig function definitions and the quotient rule

How do I derive the derivative of csc x?

  • Recall that csc space x space equals fraction numerator space 1 over denominator sin space x end fraction

  • Apply the quotient rule, y to the power of apostrophe equals fraction numerator u to the power of apostrophe v minus u v to the power of apostrophe over denominator v squared end fraction

    • u equals 1 and v equals sin space x

    • u to the power of apostrophe equals 0 and v to the power of apostrophe equals cos space x

    • y to the power of apostrophe equals fraction numerator 0 minus cos space x over denominator open parentheses sin space x close parentheses squared end fraction

  • Simplify using the identities fraction numerator cos space x over denominator sin space x end fraction equals cot space x and fraction numerator 1 over denominator sin space x end fraction equals csc space x

    • y to the power of apostrophe equals fraction numerator negative cos space x over denominator sin squared space x end fraction equals negative fraction numerator cos space x over denominator sin space x end fraction times fraction numerator 1 over denominator sin space x end fraction equals negative cot space x space csc space x

How do I derive the derivative of sec x?

  • Recall that sec space x space equals fraction numerator space 1 over denominator cos space x end fraction

  • Apply the quotient rule, y to the power of apostrophe equals fraction numerator u to the power of apostrophe v minus u v to the power of apostrophe over denominator v squared end fraction

    • u equals 1 and v equals cos space x

    • u to the power of apostrophe equals 0 and v to the power of apostrophe equals negative sin space x

    • y to the power of apostrophe equals fraction numerator 0 minus open parentheses negative sin space x close parentheses over denominator open parentheses cos space x close parentheses squared end fraction

  • Simplify using the identities fraction numerator sin space x over denominator cos space x end fraction equals tan space x and fraction numerator 1 over denominator cos space x end fraction equals sec space x

    • y to the power of apostrophe equals fraction numerator sin space x over denominator cos squared space x end fraction equals fraction numerator sin space x over denominator cos space x end fraction times fraction numerator 1 over denominator cos space x end fraction equals tan space x space sec space x

How do I derive the derivative of cot x?

  • Recall that cot space x space equals fraction numerator space 1 over denominator tan space x end fraction equals fraction numerator cos space x over denominator sin space x end fraction

  • Apply the quotient rule, y to the power of apostrophe equals fraction numerator u to the power of apostrophe v minus u v to the power of apostrophe over denominator v squared end fraction

    • u equals cos space x and v equals sin space x

    • u to the power of apostrophe equals negative sin space x and v to the power of apostrophe equals cos space x

    • y to the power of apostrophe equals fraction numerator negative sin space x times sin space x minus cos space x times cos space x over denominator open parentheses sin space x close parentheses squared end fraction

  • Simplify using the identities sin squared space x space plus thin space cos squared space x equals 1 and fraction numerator 1 over denominator sin space x end fraction equals csc space x

    • y to the power of apostrophe equals fraction numerator negative sin squared space x minus cos squared x over denominator sin squared space x end fraction equals fraction numerator negative open parentheses sin squared space x plus cos squared space x close parentheses over denominator sin squared space x end fraction equals negative fraction numerator 1 over denominator sin squared space x end fraction equals negative csc squared space x

Worked Example

Show that the derivative of f open parentheses x close parentheses equals csc space x space sec space x is f blank to the power of apostrophe open parentheses x close parentheses equals sec squared space x minus csc squared space x.

Answer:

This is a product of two functions, so use the product rule, y to the power of apostrophe equals u to the power of apostrophe v plus u v to the power of apostrophe

u equals csc space x and v equals sec space x

Differentiate using the known results

u to the power of apostrophe equals negative cot space x space csc space x and v to the power of apostrophe equals tan space x space sec space x

Apply the product rule

y to the power of apostrophe equals negative cot space x space csc space x times sec space x space plus space csc space x times tan space x space sec space x

We now need to use trigonometric identities to rearrange to sec squared space x minus csc squared space x

Swap the reciprocal functions for their 'regular' counterparts
I.e. sec space x space equals fraction numerator 1 over denominator cos space x end fraction, csc space x space equals fraction numerator 1 over denominator sin space x end fraction and cot space x space equals fraction numerator cos space x over denominator sin space x end fraction
This can often make rearranging and simplifying easier

y to the power of apostrophe equals open parentheses fraction numerator negative cos space x over denominator sin space x end fraction times fraction numerator 1 over denominator sin space x end fraction times fraction numerator 1 over denominator cos space x end fraction close parentheses plus open parentheses fraction numerator 1 over denominator sin space x end fraction times fraction numerator sin space x over denominator cos space x end fraction times fraction numerator 1 over denominator cos space x end fraction close parentheses y to the power of apostrophe equals fraction numerator negative cos space x over denominator sin squared space x space cos space x end fraction plus fraction numerator sin space x over denominator sin space x space cos squared space x end fraction y to the power of apostrophe equals fraction numerator negative 1 over denominator sin squared space x end fraction plus fraction numerator 1 over denominator cos squared space x end fraction y to the power of apostrophe equals negative open parentheses fraction numerator 1 over denominator sin space x end fraction close parentheses squared plus open parentheses fraction numerator 1 over denominator cos space x end fraction close parentheses squared

f to the power of apostrophe open parentheses x close parentheses equals sec squared space x space minus space csc squared 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.