Implicit Differentiation (College Board AP® Calculus AB): Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

Derivatives of implicit functions

What is an implicit function?

  • An equation in the form y equals f open parentheses x close parentheses or x equals f open parentheses y close parenthesesis said to be written explicitly

    • E.g. y equals 3 x squared plus 2 x minus 3

  • Equations in terms of both x and y which cannot be written as y in terms of x are referred to as implicit functions

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

    • For such an equation

      • We cannot write the relationship between x and y explicitly by expressing y as a function of x

      • But values of y satisfying the equation still depend implicitly on the values of x

What is implicit differentiation?

  • The method by which implicit functions are differentiated is known as implicit differentiation

  • To differentiate an implicit function with respect to x, each term is differentiated with respect to x

  • For terms only involving x, this is straightforward

  • However, to differentiate an expression that is in terms of y, but differentiate it with respect to x, 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

  • This could also be written as

    • fraction numerator d u over denominator d x end fraction equals fraction numerator d u over denominator d y end fraction times fraction numerator d y over denominator d x end fraction where u is a function in terms of y

  • 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

    • you may have to factorize out fraction numerator d y over denominator d x end fraction to do this

  • The final expression for fraction numerator d y over denominator d x end fraction will often be in terms of both x and y

    • To then find the value of the derivative at a point, you would need the coordinates of a point open parentheses x comma space y close parentheses to substitute in

      • And substitute in both the x and y values

    • This is only valid if the point open parentheses x comma space y close parentheses lies on the curve in question

      • To check this, substitute the x and y values into the original equation of the curve, and check that the equation is satisfied

        • If it is then the point lies on the curve

        • If it is not, then the point does not lie on the curve

How do I use implicit differentiation?

  • For example, to differentiate x squared plus y squared equals 4 x

  • The derivative would be fraction numerator d over denominator d x end fraction open parentheses x squared close parentheses plus fraction numerator d over denominator d x end fraction open parentheses y squared close parentheses equals fraction numerator d over denominator d x end fraction open parentheses 4 x close parentheses

  • The terms which are only in terms of x are straight forward

    • 2 x plus fraction numerator d over denominator d x end fraction open parentheses y squared close parentheses equals 4

  • Then apply the result 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 fraction numerator d y over denominator d x end fraction to the remaining term, which is in terms of y

    • 2 x plus open parentheses 2 y times fraction numerator d y over denominator d x end fraction close parentheses equals 4

  • Rearrange to find an expression for fraction numerator d y over denominator d x end fraction

    • fraction numerator d y over denominator d x end fraction equals fraction numerator 4 minus 2 x over denominator 2 y end fraction equals fraction numerator 2 minus x over denominator y end fraction

Worked Example

A curve is defined by the equation

4 x cubed plus 2 x plus 2 y cubed plus 6 y equals 44

(a) Confirm that the point open parentheses 2 comma space 1 close parentheses lies on the curve.

Answer:

Substitute x equals 2 and y equals 1 into the equation

table row cell 4 open parentheses 2 close parentheses cubed plus 2 open parentheses 2 close parentheses plus 2 open parentheses 1 close parentheses cubed plus 6 open parentheses 1 close parentheses end cell equals 44 row cell 32 plus 4 plus 2 plus 6 end cell equals 44 row 44 equals 44 end table

The equation is satisfied, so the point lies on the curve

open parentheses 2 comma space 1 close parentheses lies on the curve

(b) Find the value of the derivative fraction numerator d y over denominator d x end fractionat the point open parentheses 2 comma space 1 close parentheses.

Answer:

Differentiate each term individually with respect to x

Use the result 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 fraction numerator d y over denominator d x end fraction for the y-terms

Remember that the constant 44 on the right-hand side of the equation will differentiate to zero

12 x squared plus 2 plus 6 y squared times fraction numerator d y over denominator d x end fraction plus 6 times fraction numerator d y over denominator d x end fraction equals 0

Factorize out the fraction numerator d y over denominator d x end fraction term

12 x squared plus 2 plus fraction numerator d y over denominator d x end fraction open parentheses 6 y squared plus 6 close parentheses equals 0

Rearrange to make fraction numerator d y over denominator d x end fraction the subject and simplify

table row cell fraction numerator d y over denominator d x end fraction end cell equals cell fraction numerator negative 12 x squared minus 2 over denominator 6 y squared plus 6 end fraction end cell row cell fraction numerator d y over denominator d x end fraction end cell equals cell fraction numerator negative 6 x squared minus 1 over denominator 3 y squared plus 3 end fraction end cell end table

To find the value of the derivative at open parentheses 2 comma space 1 close parentheses substitute in x equals 2 and y equals 1

fraction numerator d y over denominator d x end fraction equals fraction numerator negative 6 open parentheses 2 squared close parentheses minus 1 over denominator 3 open parentheses 1 squared close parentheses plus 3 end fraction equals fraction numerator negative 24 minus 1 over denominator 6 end fraction equals negative 25 over 6

The value of the derivative at open parentheses 2 comma space 1 close parentheses is negative 25 over 6 space

How might implicit differentiation questions be made harder?

  • Implicit differentiation may be combined with other skills including

    • Chain rule

    • Product rule

    • Quotient rule

    • Derivatives of exponentials, logarithms, and trigonometric functions

  • The following result can be useful when using the product rule

    • fraction numerator d over denominator d x end fraction open parentheses x y close parentheses equals y plus x fraction numerator d y over denominator d x end fraction

Worked Example

Given that x to the power of 4 plus y cubed equals 12 x y, find an expression for fraction numerator d y over denominator d x end fraction.

Answer:

Differentiate each term individually with respect to x

Use the result 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 fraction numerator d y over denominator d x end fraction for the terms with y

4 x cubed plus 3 y squared times fraction numerator d y over denominator d x end fraction equals fraction numerator d over denominator d x end fraction open parentheses 12 x y close parentheses

To differentiate 12 x y with respect to x, use the product rule

u equals 12 x and v equals y

u to the power of apostrophe equals 12 and v to the power of apostrophe equals 1 times fraction numerator d y over denominator d x end fraction

u to the power of apostrophe v plus u v to the power of apostrophe equals 12 y plus 12 x times fraction numerator d y over denominator d x end fraction

So the differentiated expression is

4 x cubed plus 3 y squared times fraction numerator d y over denominator d x end fraction equals 12 y plus 12 x times fraction numerator d y over denominator d x end fraction

Rearrange to make fraction numerator d y over denominator d x end fraction the subject

table row cell 3 y squared times fraction numerator d y over denominator d x end fraction minus 12 x times fraction numerator d y over denominator d x end fraction end cell equals cell 12 y minus 4 x cubed end cell row cell fraction numerator d y over denominator d x end fraction open parentheses 3 y squared minus 12 x close parentheses end cell equals cell 12 y minus 4 x cubed end cell row cell fraction numerator d y over denominator d x end fraction end cell equals cell fraction numerator 12 y minus 4 x cubed over denominator 3 y squared minus 12 x end fraction end cell end table

fraction numerator d y over denominator d x end fraction equals fraction numerator 12 y minus 4 x cubed over denominator 3 y squared minus 12 x end fraction

Equivalent answers would also be correct, for example:

fraction numerator d y over denominator d x end fraction equals fraction numerator 4 x cubed minus 12 y over denominator 12 x minus 3 y squared end fraction

Worked Example

Find fraction numerator d y over denominator d x end fraction given that cos open parentheses x plus y close parentheses equals 2 x squared.

Answer:

Differentiate both sides with respect to x

Use the chain rule for the term cos open parentheses x plus y close parentheses

The chain rule states that f open parentheses g open parentheses x close parentheses close parentheses differentiates to 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

When differentiating y, remember to multiply by fraction numerator d y over denominator d x end fraction

table row cell negative sin open parentheses x plus y close parentheses times open parentheses fraction numerator d over denominator d x end fraction open parentheses x plus y close parentheses close parentheses end cell equals cell 4 x end cell row blank blank blank row cell negative sin open parentheses x plus y close parentheses times open parentheses 1 plus 1 times fraction numerator d y over denominator d x end fraction close parentheses end cell equals cell 4 x end cell end table

Rearrange to make fraction numerator d y over denominator d x end fraction the subject

1 plus fraction numerator d y over denominator d x end fraction equals fraction numerator 4 x over denominator negative sin open parentheses x plus y close parentheses end fraction
fraction numerator d y over denominator d x end fraction equals negative 1 minus fraction numerator 4 x over denominator sin open parentheses x plus y close parentheses end fraction

fraction numerator d y over denominator d x end fraction equals negative open parentheses 1 plus fraction numerator 4 x over denominator sin open parentheses x plus y close parentheses end fraction close parentheses

Derivatives of inverse functions using implicit differentiation

How can I differentiate inverse functions using implicit differentiation?

  • Implicit differentiation provides an alternative method for finding the derivative of inverse functions

  • Consider differentiating y equals sin to the power of negative 1 end exponent x

    • This can be rewritten as x equals sin space y

  • Differentiate using implicit differentiation

    • fraction numerator d over denominator d x end fraction open parentheses x close parentheses equals fraction numerator d over denominator d x end fraction open parentheses sin space y close parentheses

    • 1 equals cos space y times fraction numerator d y over denominator d x end fraction

  • Rearrange to make fraction numerator d y over denominator d x end fraction the subject

    • fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator cos space y end fraction

  • Recall that y equals sin to the power of negative 1 end exponent x

    • fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator cos open parentheses sin to the power of negative 1 end exponent x close parentheses end fraction

  • Use the identity sin squared x plus cos squared x equals 1 rearranged to cos space x equals square root of 1 minus sin squared x end root

    • fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator square root of 1 minus sin squared open parentheses sin to the power of negative 1 end exponent x close parentheses end root end fraction

    • fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction

  • A similar method can be used to show why the derivative of a to the power of x is a to the power of x ln space a

    • If y equals a to the power of x

      • then ln space y equals x space ln space a

    • Differentiate using implicit differentiation

      • Remember that fraction numerator d over denominator d x end fraction open parentheses ln space x close parentheses equals 1 over x and that ln space a is a constant

      • 1 over y times fraction numerator d y over denominator d x end fraction equals ln space a

    • Rearrange

      • fraction numerator d y over denominator d x end fraction equals y space ln space a

    • Recall that y equals a to the power of x

      • fraction numerator d y over denominator d x end fraction equals a to the power of x space ln space a

Worked Example

Given that y equals arctan space x, use implicit differentiation to show that fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator x squared plus 1 end fraction.

Answer:

Rewrite the equation in terms of tan rather than arctan

tan space y equals x

Differentiate using implicit differentiation

tan space y differentiates to sec squared y, and remember to multiply by fraction numerator d y over denominator d x end fraction as we are differentiating with respect to x

table row cell fraction numerator d over denominator d x end fraction open parentheses tan space y close parentheses end cell equals cell fraction numerator d over denominator d x end fraction open parentheses x close parentheses end cell row cell sec squared y times fraction numerator d y over denominator d x end fraction end cell equals 1 end table

Make fraction numerator d y over denominator d x end fraction the subject

fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator sec squared y end fraction

Recall that y equals arctan x

fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator sec squared open parentheses arctan x close parentheses end fraction

Use the identity tan squared x plus 1 equals sec squared x

fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator tan squared open parentheses arctan x close parentheses plus 1 end fraction equals fraction numerator 1 over denominator open parentheses tan open parentheses arctan x close parentheses close parentheses squared plus 1 end fraction

Simplify

fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator x squared plus 1 end fraction

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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.

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