Implicit Differentiation (DP IB Analysis & Approaches (AA)): Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

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Implicit differentiation

What is implicit differentiation?

  • An equation connecting x and y is not always easy to write explicitly in the form y equals f left parenthesis x right parenthesis or x equals f left parenthesis y right parenthesis

    • In such cases the equation is written implicitly

      • as a function of x and y

      • e.g. in the form f left parenthesis x comma y right parenthesis equals 0

  • Such equations can be differentiated implicitly using the chain rule

fraction numerator straight d over denominator d x end fraction open square brackets f left parenthesis y right parenthesis close square brackets equals f to the power of apostrophe left parenthesis y right parenthesis fraction numerator straight d y over denominator straight d x end fraction

  • A shortcut way of thinking about this is that ‘y is a function of x

    • And when differentiating a function of y chain rule says “differentiate with respect to y, then multiply by the derivative of y” (which is fraction numerator straight d y over denominator straight d x end fraction)

How do I find a derivative using implicit differentiation?

  • In order to find fraction numerator straight d y over denominator straight d x end fraction for an implicit relation between x and y, follow these steps

  • STEP 1
    Differentiate both sides of the equation implicitly with respect to x

    • E.g.  2 x squared plus y to the power of 4 equals 48

table row cell fraction numerator straight d over denominator straight d x end fraction open parentheses 2 x squared plus y to the power of 4 close parentheses end cell equals cell fraction numerator straight d over denominator straight d x end fraction open parentheses 48 close parentheses end cell row cell 4 x plus 4 y cubed fraction numerator straight d y over denominator straight d x end fraction end cell equals 0 end table
 

  • STEP 2
    Rearrange the new equation to make fraction numerator straight d y over denominator straight d x end fraction the subject

table row cell fraction numerator straight d y over denominator straight d x end fraction end cell equals cell negative x over y cubed end cell end table

Examiner Tips and Tricks

In more complicated implicit differentiation questions, you may need to combine other differentiation techniques (e.g. chain rule, product rule, quotient rule) with implicit differentiation.

Applications of implicit differentiation

What type of problems could involve implicit differentiation?

  • Broadly speaking there are three types of problem that could involve implicit differentiation

    • Algebraic problems involving graphs, derivatives, tangents, normals, etc.,

      where it is not practical to write y explicitly in terms of x

      • Usually in such cases, fraction numerator straight d y over denominator straight d x end fraction will be in terms of x and y

    • Optimisation problems that involve time derivatives

      • More than one variable may be involved too

        e.g.  Volume of a cylinder, V equals pi r squared h

        e.g.  The side length and (therefore) area of a square increase over time

    • Any problem that involves differentiating with respect to an extraneous variable

      • e.g. y equals f left parenthesis x right parenthesis but the derivative fraction numerator straight d y over denominator straight d theta end fraction is required (rather than fraction numerator straight d y over denominator straight d x end fraction)

How do I apply implicit differentiation to algebraic problems?

  • Algebraic problems revolve around values of the derivative (gradient) open parentheses fraction numerator straight d y over denominator straight d x end fraction close parentheses

  • Particular problems focus on special case tangent values

    • Horizontal tangents

      • Also referred to as tangents parallel to the x-axis

      • This is when fraction numerator straight d y over denominator straight d x end fraction equals 0

    • Vertical tangents

      • Also referred to as tangents parallel to the y-axis

      • This is when fraction numerator straight d x over denominator straight d y end fraction equals 0

    • Recall the relation fraction numerator straight d x over denominator straight d y end fraction equals fraction numerator 1 over denominator fraction numerator straight d y over denominator straight d x end fraction end fraction

  • Other problems may involve finding equations of (other) tangents and/or normals

  • For problems that involve finding the coordinates of points on a curve with a specified gradient the method below can be used
     

  • STEP 1
    Differentiate the equation of the curve implicitly
     

  • STEP 2
    Substitute the given or implied value of fraction numerator straight d y over denominator straight d x end fraction to create an equation linking x and y
     

  • STEP 3
    There are now two equations

    • the original equation

    • the linking equation

    Solve them simultaneously to find the x and y coordinates as required

Examiner Tips and Tricks

After rearranging following implicit differentiation, fraction numerator straight d y over denominator straight d x end fraction will often be in terms of both x and y. Unless specifically asked for in the question, there is usually no need to try to write fraction numerator straight d y over denominator straight d x end fraction in terms of x (or y) only.

If evaluating derivatives, you'll need both x and y coordinates, so one may have to be found from the other using the original equation.

Worked Example

The curve C has equation x squared plus 2 y squared equals 16.

a) Find the exact coordinates of the points where the normal to curve C has gradient 2.

5-8-2-ib-hl-aa-only-we3a-soltn

b) Find the equations of the tangents to the curve that are

(i)       parallel to the x-axis

(ii)      parallel to the y-axis.

5-8-2-ib-hl-aa-only-we3b-soltn

How do I apply implicit differentiation to optimisation problems?

  • For a single variable use chain rule to differentiate implicitly

    • e.g. A square with side length changing over time, A equals x squared

      • Differentiating both sides with respect to t gives

fraction numerator straight d A over denominator straight d t end fraction equals 2 x fraction numerator straight d x over denominator straight d t end fraction

  • For more than one variable use product rule (and chain rule) to differentiate implicitly

    • e.g. A square-based pyramid with base length x and height h changing over time, V equals 1 third x squared h

      • Differentiating both sides with respect to t gives

fraction numerator straight d V over denominator straight d t end fraction equals 1 third open square brackets x squared fraction numerator straight d h over denominator straight d t end fraction plus 2 x fraction numerator straight d x over denominator straight d t end fraction h close square brackets equals 1 third x open parentheses x fraction numerator straight d h over denominator straight d t end fraction plus 2 h fraction numerator straight d x over denominator straight d t end fraction close parentheses

  • After differentiating implicitly the rest of the question should be similar to any other optimisation problem

    • Be aware of phrasing

      • “the rate of change of the height of the pyramid” (over time) is fraction numerator straight d h over denominator straight d t end fraction

    • When finding the location of minimum and maximum problems

      • The solution is not necessarily at a turning point

      • The minimum or maximum could be at the start or end of a given or appropriate interval

Examiner Tips and Tricks

If you are struggling to tell which derivative is needed for a question, writing all possibilities down may help you. You don't need to work them out at this stage but if you consider them it may nudge you to the next stage of the solution.

For example, for V equals pi r to the power of italic 2 h, possible derivatives are fraction numerator straight d V over denominator straight d r end fraction comma space fraction numerator straight d V over denominator straight d h end fraction and fraction numerator straight d V over denominator straight d t end fraction.

Worked Example

The radius, r cm, and height, h cm, of a cylinder are increasing with time.  The volume, V cm3, of the cylinder at time t seconds is given by V equals pi r squared h.

a)       Find an expression for fraction numerator straight d V over denominator straight d t end fraction.

5-8-2-ib-hl-aa-only-we2a-soltn

b) At time T seconds, the radius of the cylinder is 4 cm, expanding at a rate of 2 cm s-1. At the same time, the height of the cylinder is 10 cm, expanding at a rate of 3 cm s-1.

Find the rate at which the volume is expanding at time T seconds.

5-8-2-ib-hl-aa-only-we2b-soltn

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Paul

Author: Paul

Expertise: Maths Content Creator

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams.

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.