AP®MathsCollege BoardCalculus BCStudy GuidesUnit 2: Differentiation Definition & Fundamental PropertiesFundamental Properties of DifferentiationDerivatives of Exponentials and Logarithms

Derivatives of Exponentials and Logarithms (College Board AP® Calculus BC): Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

College Board Calculus BCCollege Board Calculus AB

Derivative of the exponential function

How do I differentiate the exponential function?

  • It can be shown that:

    • limit as h rightwards arrow 0 of fraction numerator e to the power of x plus h end exponent minus e to the power of x over denominator h end fraction equals e to the power of x for all real x

    • limit as x rightwards arrow a of fraction numerator e to the power of x minus e to the power of a over denominator x minus a end fraction equals e to the power of a for all real a

Examiner Tips and Tricks

You do not need to learn how to derive this result as it is beyond the scope of this course.

  • This means that f open parentheses x close parentheses equals e to the power of x is an important function because its rate of change is equal to itself

    • f to the power of apostrophe open parentheses x close parentheses equals e to the power of x

  • For the function g open parentheses x close parentheses equals e to the power of k x end exponent, its rate of change is proportional to itself

    • g to the power of apostrophe open parentheses x close parentheses equals k e to the power of k x end exponent

    • This occurs as a result of applying the chain rule

  • If there is a constant multiple of the exponential, the same approach used for powers of x can be applied

    • If h open parentheses x close parentheses equals a e to the power of k x end exponent then h to the power of apostrophe open parentheses x close parentheses equals a k e to the power of k x end exponent

Worked Example

Given that f open parentheses x close parentheses equals e to the power of x plus e to the power of 2 x end exponent minus 4 e to the power of 7 x end exponent, find f apostrophe open parentheses x close parentheses.

Answer:

e to the power of x differentiates to itself
e to the power of k x end exponent differentiates to k e to the power of k x end exponent
a e to the power of k x end exponent differentiates to a k e to the power of k x end exponent

f to the power of apostrophe open parentheses x close parentheses equals e to the power of x plus 2 e to the power of 2 x end exponent minus 4 times 7 e to the power of 7 x end exponent

Simplify

f apostrophe open parentheses x close parentheses equals e to the power of x plus 2 e to the power of 2 x end exponent minus 28 e to the power of 7 x end exponent

How do I differentiate a number raised to the power of x?

  • For a positive constant raised to the power of x,

    • If f open parentheses x close parentheses equals a to the power of x where a greater than 0 then f to the power of apostrophe open parentheses x close parentheses equals a to the power of x space ln space a

  • This can be shown by using the identity a to the power of x equals e to the power of x ln a end exponent

    • f apostrophe open parentheses x close parentheses equals e to the power of x ln a end exponent times ln a equals a to the power of x ln a

  • If the power is a multiple of x,

    • g open parentheses x close parentheses equals a to the power of k x end exponent

    • g to the power of apostrophe open parentheses x close parentheses equals a to the power of k x end exponent space k space ln space a

    • This occurs as a result of applying the chain rule

Worked Example

Given that g open parentheses x close parentheses equals 3 to the power of x plus 3 to the power of 2 x end exponent, find g apostrophe open parentheses x close parentheses.

Answer:

a to the power of x differentiates to a to the power of x space ln space a
a to the power of k x end exponent differentiates to a to the power of k x end exponent space k space ln space a

g to the power of apostrophe open parentheses x close parentheses equals 3 to the power of x space ln space 3 space plus space 3 to the power of 2 x end exponent times 2 times ln space 3

g to the power of apostrophe open parentheses x close parentheses equals 3 to the power of x space ln space 3 space plus space 2 times 3 to the power of 2 x end exponent ln 3

Derivative of the natural logarithmic function

How do I differentiate a natural logarithm?

  • For a natural logarithm, it can be shown using inverse functions that

    • If f open parentheses x close parentheses equals ln space x for x greater than 0 then f to the power of apostrophe open parentheses x close parentheses equals 1 over x

Examiner Tips and Tricks

In Unit 3, you learn how to differentiate inverse functions using the rule:

fraction numerator d over denominator d x end fraction open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses equals fraction numerator 1 over denominator f apostrophe open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses end fraction

ln x is the inverse of e to the power of x. And the derivative of e to the power of x is e to the power of x. Therefore,

fraction numerator d over denominator d x end fraction open parentheses ln x close parentheses equals 1 over e to the power of ln x end exponent equals 1 over x

  • This means that:

    • limit as h rightwards arrow 0 of fraction numerator ln open parentheses x plus h close parentheses minus ln x over denominator h end fraction equals 1 over x for all x greater than 0

    • limit as x rightwards arrow a of fraction numerator ln x minus ln a over denominator x minus a end fraction equals 1 over a for all a greater than 0

  • If there is a constant multiple of the logarithm, the same approach used for powers of x can be applied

    • If h open parentheses x close parentheses equals a space ln space x then h to the power of apostrophe open parentheses x close parentheses equals a space open parentheses 1 over x close parentheses equals a over x

  • If there is a constant multiple of x inside the logarithm,

    • g open parentheses x close parentheses equals ln space k x

    • This can be rewritten using the laws of logarithms

      • g open parentheses x close parentheses equals space ln space k space plus space ln space x

    • ln space k is a constant, which means it has a derivative of zero

    • Therefore g to the power of apostrophe open parentheses x close parentheses equals 1 over x

Examiner Tips and Tricks

Don't forget that the derivative of space ln k x space is 1 over x

  • I.e. it is exactly the same as the derivative for ln x

  • Differentiating space ln k x space as k over x is a common mistake on the exam!

Worked Example

Find the derivative of the following function

f open parentheses x close parentheses equals ln open parentheses 2 x to the power of 5 close parentheses plus ln open parentheses 2 x close parentheses

Answer:

Rewrite both logarithms using the laws of logarithms

f open parentheses x close parentheses equals ln space 2 space plus space ln space x to the power of 5 space plus space ln 2 space plus space ln space x f open parentheses x close parentheses equals ln space 2 space plus space 5 ln space x space plus space ln 2 space plus space ln space x

Simplify

f open parentheses x close parentheses equals 2 ln space 2 plus 6 ln space x

2 ln space 2 is a constant so differentiates to zero
ln space x differentiates to 1 over x

f to the power of apostrophe open parentheses x close parentheses equals 6 times 1 over x

f to the power of apostrophe open parentheses x close parentheses equals 6 over x

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Author
Reviewer
Jamie Wood

Author: Jamie Wood

Expertise: Maths Content Creator

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

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.