AP®MathsCollege BoardCalculus ABRevision NotesUnit 3: Differentiation of Composite, Implicit & Inverse FunctionsSummary of DifferentiationSelecting Procedures for Calculating Derivatives

Selecting Procedures for Calculating Derivatives (College Board AP® Calculus AB)

Revision Note

Jamie Wood

Expertise

Maths

Selecting procedures for calculating derivatives

  • You should be familiar with all the different methods for differentiating functions

  • This way you can choose the most appropriate method to use for an exam question

  • You should also know that the derivative of f open parentheses x close parentheses is defined as

    • 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 generally only need to use this if asked to do so

Estimating derivatives

  • You can estimate a derivative at a point using a table or graph

    • Remember that the derivative at a point is equal to the slope of the tangent at that point

    • To approximate the slope of the tangent to the graph of f open parentheses x close parentheses at x equals a:

      • Find the slope of line segments joining nearby coordinates that lie on the graph

    • This is useful when you only have a graph or table

Finding the derivative as an expression

Basic differentiation

  • If you know the function or the equation of the graph, you can differentiate it using several methods

  • Powers of x are differentiated according to the following formula:

    • Ifspace f left parenthesis x right parenthesis equals x to the power of n thenspace f to the power of apostrophe left parenthesis x right parenthesis equals n x to the power of n minus 1 end exponent 

  • When differentiating sums or differences of powers of x,

    • the derivative is simply the sum (or difference) of the derivatives of the terms

  • Constant multiples of powers of x are differentiated according to the following formula:

    • Ifspace f left parenthesis x right parenthesis equals a x to the power of n thenspace f to the power of apostrophe open parentheses x close parentheses equals a n x to the power of n minus 1 end exponent 

  • Remember the two special cases

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

    • If g open parentheses x close parentheses equals a then g to the power of apostrophe open parentheses x close parentheses equals 0

  • You may need to expand brackets or simplify expressions before differentiating

    • This may involve using laws of exponents, e.g. rewriting square root of x as x to the power of 1 half end exponent

Differentiating exponentials and logarithms

  • The table below summarises the key results for derivatives of exponentials and logarithms

f open parentheses x close parentheses

f to the power of apostrophe open parentheses x close parentheses

e to the power of k x end exponent

k e to the power of k x end exponent

a to the power of k x end exponent

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

ln space k x

1 over x

Differentiating trigonometric functions

  • The table below summarises the key results for derivatives of

    • trigonometric functions,

    • reciprocal trigonometric functions,

    • and inverse trigonometric functions

f open parentheses x close parentheses

f to the power of apostrophe open parentheses x close parentheses

sin space k x

k space cos space k x

cos space k x

negative k space sin space k x

tan space k x

k space sec squared k x

csc space k x

negative k space cot space k x space csc space k x

sec space k x

k space tan space k x space sec space k x

cot space k x

negative k space csc squared k x

arcsin space x

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

arccos space x

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

arctan space x

fraction numerator 1 over denominator 1 plus x squared end fraction

  • The most important results to remember are for sin and cos

  • The results for tan, csc, sec, and cot can all be derived using the quotient rule and trigonometric identities

  • The results for the inverse trigonometric functions can be derived using either the inverse function theorem or implicit differentiation

Product rule and quotient rule

  • The product rule is used when two functions are multiplied together, it states that

    • If f open parentheses x close parentheses equals g open parentheses x close parentheses times h open parentheses x close parentheses,

    • then f to the power of apostrophe open parentheses x close parentheses equals g to the power of apostrophe open parentheses x close parentheses times h open parentheses x close parentheses space plus space g open parentheses x close parentheses times h to the power of apostrophe open parentheses x close parentheses

  • It can also be written as,

    • if y equals u times v,

    • then fraction numerator d y over denominator d x end fraction equals fraction numerator d u over denominator d x end fraction times v space plus space u times fraction numerator d v over denominator d x end fraction

    • Or in a more concise form: y to the power of apostrophe equals u to the power of apostrophe v space plus space u v to the power of apostrophe

  • The quotient rule is used when one function is divided by another, it states that

    • If f open parentheses x close parentheses equals fraction numerator g open parentheses x close parentheses over denominator h open parentheses x close parentheses end fraction,

    • then f to the power of apostrophe open parentheses x close parentheses equals fraction numerator g to the power of apostrophe open parentheses x close parentheses times h open parentheses x close parentheses space minus space g open parentheses x close parentheses times h to the power of apostrophe open parentheses x close parentheses over denominator open parentheses h open parentheses x close parentheses close parentheses squared end fraction

  • It can also be written as,

    • If y equals u over v,

    • then fraction numerator d y over denominator d x end fraction equals fraction numerator fraction numerator d u over denominator d x end fraction times v space minus space u times fraction numerator d v over denominator d x end fraction over denominator v squared end fraction

    • Or in a more concise form: 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

The chain rule

  • The chain rule is used for composite functions (a function within a function); it states that

    • If y equals f open parentheses u close parentheses and u equals g open parentheses x close parentheses (i.e. y is a function of u, and u is a function of x),

    • then 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

  • Or in function notation, if h open parentheses x close parentheses equals f open parentheses g open parentheses x close parentheses close parentheses

    • h to the power of apostrophe open parentheses x close parentheses equals f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses times g to the power of apostrophe open parentheses x close parentheses

The inverse function theorem

  • The inverse function theorem can be used to find the derivative of the inverse of a function, it states that

    • For a function f, the derivative of its inverse will be given by:

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

  • You may also see this written as:

    • g to the power of apostrophe open parentheses a close parentheses equals fraction numerator 1 over denominator f to the power of apostrophe open parentheses g open parentheses a close parentheses close parentheses end fraction

    • Where g open parentheses a close parentheses equals f to the power of negative 1 end exponent open parentheses a close parentheses

  • Or if y equals f to the power of negative 1 end exponent open parentheses x close parentheses so that x equals f open parentheses y close parentheses,

    • then the inverse function theorem can be written as

    • 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

Implicit differentiation

  • Implicit differentiation is used for functions written implicitly

    • E.g. 3 x squared minus 7 x y squared equals 3 or x squared plus y squared equals 25

  • Every term in the equation is differentiated

  • For terms that are only in terms of x, this is straightforward

  • For terms that involve y, we apply the chain rule

    • fraction numerator d over denominator d x end fraction f open parentheses y close parentheses equals f to the power of apostrophe open parentheses y close parentheses times y to the power of apostrophe equals f to the power of apostrophe open parentheses y close parentheses times fraction numerator d y over denominator d x end fraction

  • In short, this means:

    • Differentiate the function that is in terms of y, with respect to bold italic y,

    • and then multiply it by the term fraction numerator d y over denominator d x end fraction

  • Once each term has been differentiated with respect to x, rearrange to make fraction numerator d y over denominator d x end fraction the subject

Exam Tip

Exam questions will often involve a combination of the above skills.

  • Break the question down into smaller pieces

    • If one term of an expression is hard to differentiate, work on it separately off to one side

  • Be purposeful with your notation

    • If you have already used u and v to represent expressions, choose another variable like w for the next one

Exam Tip

Being comfortable with both fraction numerator d y over denominator d x end fraction notation and f to the power of apostrophe open parentheses x close parentheses notation can be helpful for different scenarios, for example:

  • The chain rule may be easier to remember as 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 than ash to the power of apostrophe open parentheses x close parentheses equals f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses times g to the power of apostrophe open parentheses x close parentheses

  • The inverse function theorem in the form open parentheses f to the power of negative 1 end exponent close parentheses apostrophe open parentheses a close parentheses equals fraction numerator 1 over denominator f apostrophe open parentheses f to the power of negative 1 end exponent open parentheses a close parentheses close parentheses end fraction is useful for finding the derivative of the inverse at a point x equals a,

    • whereas the form 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 is more useful for finding an expression in terms of x for the derivative of the inverse

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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.