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

Derivatives of Exponentials and Logarithms (College Board AP® Calculus AB)

Revision Note

Jamie Wood

Expertise

Maths

Derivative of the exponential function

How do I differentiate the exponential function?

  • 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

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

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

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

  • For a constant raised to the power of x,

    • f open parentheses x close parentheses equals a to the power of x

    • f to the power of apostrophe open parentheses x close parentheses equals a to the power of x space ln space 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

  • This result can also be shown using logarithms and the result for differentiating e to the power of k x end exponent

    • y equals a to the power of k x end exponent can be rewritten as y equals e to the power of ln space open parentheses a to the power of k x end exponent close parentheses end exponent equals e to the power of k x ln space a end exponent

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

    • So fraction numerator d y over denominator d x end fraction equals k ln space a space times e to the power of k x ln space a end exponent

    • This simplifies to fraction numerator d y over denominator d x end fraction equals k ln space a space times space a to the power of k x end exponent

Worked Example

Differentiate the following functions.

(a) 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

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

(b) g open parentheses x close parentheses equals 3 to the power of x plus 3 to the power of 2 x end exponent

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

Simplify

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 open parentheses 3 to the power of 2 x end exponent close parentheses 2 ln 3

This answer could also be factorised

g to the power of apostrophe open parentheses x close parentheses equals 3 to the power of x ln 3 open parentheses 1 plus 2 times 3 to the power of x close parentheses

Derivative of the natural logarithmic function

How do I differentiate a natural logarithm?

  • For a natural logarithm,

    • f open parentheses x close parentheses equals ln space x

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

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

    • h open parentheses x close parentheses equals a space ln space x

    • 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

Exam Tip

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

Differentiate 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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    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.