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ImplicitDifferentiation

mc-TY-implicit-2009-1 Sometimes functions are given not in the formy=f(x)but in a more complicated form in which it is difficult or impossible to expressyexplicitly in terms ofx. Such functions are called implicit functions. In this unit we explain how these can be differentiated using implicit differentiation. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to:

•differentiate functions defined implicitly

Contents

1.Introduction2

2.Revision of the chain rule2

3.Implicit differentiation4

www.mathcentre.ac.uk 1c?mathcentre 2009

1. IntroductionIn this unit we look at how we might differentiate functions ofywith respect tox.

Consider an expression such as

x

2+y2-4x+ 5y-8 = 0

It would be quite difficult to re-arrange this soywas given explicitly as a function ofx. We could perhaps, given values ofx, use the expression to work out the values ofyand thereby draw a graph. In general even if this is possible, it will be difficult. A function given in this way is said to be definedimplicitly. In this unit we study how to differentiate a function given in this form. It will be necessary to use a rule known as thethe chain ruleor the rule for differentiating a function of a function. In this unit we will refer to it as the chain rule. There is a separate unit which covers this particular rule thoroughly, although we will revise it briefly here.

2. Revision of the chain rule

We revise the chain rule by means of an example.

Example

Suppose we wish to differentiatey= (5 + 2x)10in order to calculatedy dx. We make a substitution and letu= 5 + 2xso thaty=u10.

The chain rule states

dy dx=dydu×dudx Now ify=u10thendy du= 10u9 and ifu= 5 + 2xthendu dx= 2 hence dy dx=dydu×dudx = 10u9×2 = 20u9 = 20(5 + 2x)9 So we have used the chain rule in order to differentiate the functiony= (5 + 2x)10. www.mathcentre.ac.uk 2c?mathcentre 2009 In quoting the chain rule in the formdydx=dydu×dudxnote that we writeyin terms ofu, andu in terms ofx. i.e. y=y(u)andu=u(x) We will need to work with different variables. Suppose we havezin terms ofy, andyin terms ofx, i.e. z=z(y)andy=y(x)

The chain rule would then state:dz

dx=dzdy×dydx

Example

Supposez=y2. It follows thatdz

dy= 2y. Then using the chain rule dz dx=dzdy×dydx = 2y×dy dx = 2ydy dx Notice what we have just done. In order to differentiatey2with respect toxwe have differentiated y

2with respect toy, and then multiplied bydy

dx, i.e. d dx?y2?=ddy?y2?×dydx

We can generalise this as follows:

to differentiate a function ofywith respect tox, we differentiate with respect toyand then multiply by dy dx.

Key Point

d dx(f(y)) =ddy(f(y))×dydx We are now ready to do some implicit differentiation. Remember, every time we want to differ- entiate a function ofywith respect tox, we differentiate with respect toyand then multiply by dy dx. www.mathcentre.ac.uk 3c?mathcentre 2009

3. Implicit differentiationExampleSuppose we want to differentiate the implicit function

y

2+x3-y3+ 6 = 3y

with respectx.

We differentiate each term with respect tox:

d dx?y2?+ddx?x3?-ddx?y3?+ddx(6) =ddx(3y) Differentiating functions ofxwith respect toxis straightforward. But when differentiating a function ofywith respect toxwe must remember the rule given in the previous keypoint. We findd dy?y2?×dydx+ 3x2-ddy?y3?×dydx+ 0 =ddy(3y)×dydx that is 2ydy dx+ 3x2-3y2dydx= 3dydx

We rearrange this to collect all terms involving

dy dxtogether.

3x2= 3dy

dx-2ydydx+ 3y2dydx then

3x2=?3-2y+ 3y2?dy

dx so that, finally, dy dx=3x23-2y+ 3y2

This is our expression for

dy dx.

Example

Suppose we want to differentiate, with respect tox, the implicit function siny+x2y3-cosx= 2y As before, we differentiate each term with respect tox. d dx(siny) +ddx?x2y3?-ddx(cosx) =ddx(2y) Recognise that the second term is a product and we will need the product rule. We will also use the chain rule to differentiate the functions ofy. We find d dy(siny)×dydx+? x

2ddx?y3?+y3ddx?x2??

+ sinx=ddy(2y)×dydx so that cosydy dx+? x

2.ddy?y3?dydx+y3.2x?

+ sinx= 2dydx www.mathcentre.ac.uk 4c?mathcentre 2009

Tidying this up gives

cosydy dx+x23y2dydx+ 2xy3+ sinx= 2dydx

We now start to collect together terms involving

dy dx.

2xy3+ sinx= 2dy

dx-cosydydx-3x2y2dydx

2xy3+ sinx= (2-cosy-3x2y2)dy

dx so that, finally dy dx=2xy3+ sinx2-cosy-3x2y2 We have deliberately included plenty of detail in this calculation. With practice you will be able to omit many of the intermediate stages. The following two examples show how you should aim to condense the solution.

Example

Suppose we want to differentiatey2+x3-xy+ cosy= 0to finddy dx. The condensed solution may take the form: 2ydy dx+ 3x2-ddx(xy)-sinydydx= 0 (2y-siny)dy dx+ 3x2-? xdydx+y.1? = 0 (2y-siny-x)dy dx+ 3x2-y= 0 (2y-siny-x)dy dx=y-3x2 so that dy dx=y-3x22y-siny-x

Example

Suppose we want to differentiate

y

3-xsiny+y2

x= 8

The solution is as follows:

3y2dy dx-? xcosydydx+ siny.1? +x2ydy dx-y2.1 x2= 0 www.mathcentre.ac.uk 5c?mathcentre 2009

Multiplying through byx2gives:

3x2y2dy

dx-x3cosydydx-x2siny+ 2xydydx-y2= 0 dy dx?3x2y2-x3cosy+ 2xy?=x2siny+y2 so that dy dx=x2siny+y23x2y2-x3cosy+ 2xy

Exercises

1. Find the derivative, with respect tox, of each of the following functions (in each casey

depends onx). a)yb)y2c)sinyd) e2ye)x+y f)xyg)ysinxh)ysinyi)cos(y2+ 1)j)cos(y2+x)

2. Differentiate each of the following with respect toxand finddy

dx. a)siny+x2+ 4y= cosx. b)3xy2+ cosy2= 2x3+ 5. c)5x2-x3siny+ 5xy= 10. d)x-cosx2+y2 x+ 3x5= 4x3. e)tan5y-ysinx+ 3xy2= 9.

Answers to Exercises on Implicit Differentiation

1. a)dy dxb)2ydydxc)cosydydx d)2e2ydy dxe)1 +dydxf)xdydx+y g)ycosx+ sinxdy dxh)(siny+ycosy)dydxi)-2ysin(y2+ 1)dydx j)-? 2ydy dx+ 1? sin(y2+x) 2. a)dy dx=-sinx-2x4 + cosyb)dydx=6x2-3y26xy-2ysiny2 c) dy dx=10x-3x2siny+ 5yx3cosy-5xd)dydx=12x4-15x6+y2-2x3sinx2-x22xy e)dy dx=ycosx-3y25sec25y-sinx+ 6xy www.mathcentre.ac.uk 6c?mathcentre 2009quotesdbs_dbs21.pdfusesText_27

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