2.5 Implicit Differentiation. After this lesson, you should be able to: Distinguish between...

2.5 Implicit Differentiation

After this lesson, you should be able to:

Distinguish between functions written in implicit form and explicit form.Use implicit differentiation to find the derivative of a function.

To this point we’ve done quite a few derivatives, but they have all been derivatives of functions of the form y = f (x).

Ex. y = f (x) = 5x6 – x2 + xsin4x

Unfortunately not all the functions that we’re going to look at will fall into this form. Let’s take a look at an example of this.

Example 1 Find y’ for xy = l

Implicit Differentiation

SolutionThere are actually two solution methods for this problem. Solution 1 :This is the simple way of doing the problem. Just solve for y to get the function in the form that we’re used to dealing with.


So, that’s easy enough to do. However, there are some functions for which this can’t be done. That’s where the second solution technique comes into play.

2

x

ly

x

ly


Solution 2 :In this case we’re going to leave the function in the form that we were given. We do want to remember however that we are thinking of y as a function of x. In other words, y = y(x). Let’s rewrite the equation to note this.lxxy )(Now, we will differentiate both sides with respect to x.

The right side is easy. It’s just the derivative of a constant. The left side is also easy, but we’ve got to recognize that we’ve actually got a product here. So to do the derivative of the left side we’ll need to do the product rule.

)()( ldx

dxxy

dx

d


Note that we dropped the (x) on the y as it’s no longer really needed. We just wanted it in the equation to recognize the product rule. So, just what were we after here? We were after y’ and notice that there is now a y’ in the equation. So, to get the derivative all that we need to do is solve the equation for y’.

This is not what we got from the first solution. Or at least it doesn’t look like the same derivative. Recall however, that we do know what y is in terms of x and if we plug that in we will get,

0' xyyx

yy '

2'

x

ly


The process that we used in the second solution to the previous example is called Implicit Differentiation. In the previous example we were able to just solve for y and avoid implicit differentiation. However, that won’t always be the case.

SolutionThis is just a circle and while can solve for y this would give,

922 yxExample 2 Find y’ for the equation

This would cause us problems in getting the derivative and so will do us no good. We want a single function for the derivative and using this would give us two.

29 xy


The process that we used in the second solution to the previous example is called Implicit Differentiation. In the previous example we were able to just solve for y and avoid implicit differentiation. However, that won’t always be the case.Guideline for Implicit Differentiation

1. Differentiate both sides of the equation with respect to x.

2. Collect all terms involving y’ on the left side of the equation and move all other terms to the right side of the equation.

3. Factor y’ out of the left side of the equation.

4. Solve for y’.

SolutionThis is just a circle and while can solve for y this would give,

922 yxExample 2 Find y’ for the equation

This would cause us problems in getting the derivative and so will do us no good. We want a single function for the derivative and using this would give us two.


In this example we really are going to need to do implicit differentiation. We’ll do the same thing we did in the first example and remind ourselves that y is really a function of x and write y as y(x).

29 xy


Unlike the Example 1, we can’t just plug in for y since we wouldn’t know which of the two roots to use. Most answers from implicit differentiation will involve both x and y so don’t get excited about that when it happens.

)9()( 22

dx

dxyx

dx

d 0)(')(22 xyxyx

0'22 yyx y

xy '


SolutionFirst note that unlike all the other tangent line problems we’ve done in previous sections we need to be given both the x and the y values of the point. Notice as well that this point does lie on the graph of the circle Recall that all we need is the slope of the tangent line and this is nothing more than the derivative evaluated at the point. We’ve got the derivative from the previous example so,

922 yx

Example 3 Find the equation of the tangent line to

at the point ).5 ,2(


The tangent line is then

5)2(5

2 xy

5

2'

5 ,2

yxym

922 yx

Example 3 Find the equation of the tangent line to

at the point ).5 ,2(

Example 3 – Continued

y

xy '

5

9

5

2 xy


Example 4 Find y’ for each of the following.

Solution(a) First differentiate both sides with respect to x and notice that the first time on left side will be a product rule.

(a)

(c)

(b)

18353 yxyx

xxyyx 2sectan 102

2332 )ln( xxye yx

'243 '53 24352 yyyyxyx

33'5 '24 52432 yxyyxyy


)1(3')524( 52232 yxyyxy

)524(

)1(3'

232

52

yxy

yxy

(b) So, we’ve got two product rules to deal with this time.

2tansecsec'10')(sectan2 10922 xxyxyyyyxyx

xxyyx 2sectan 102

yxxxyyxyyx tan2tansec2')sec10sec( 10922

xyyx

yxxxyy

sec10sec

tan2tansec2'

922

10

Implicit Differentiation(c) We’re going to need to be careful with this problem. We’ve got a couple chain rules that we’re going to need to deal with.

xxy

yxyyye yx 2

'3)'32(

3

2332

xxy

yxy

xy

yeye yxyx 2

'3'32

3

2

3

33232

xy

y

xeye yxyx 2

'31'32 3232

xexy

ye yxyx 1

22')1

(3 3232

)(3

22'

132

132

ye

xexy

yx

yx


Example 5 Determine the slope of the graph of

Solution

xyyx 100)(3 222 at the point of ).1 ,3(

xydx

dyx

dx

d100)(3 222

'100)'22)()(2(3 22 xyyyyxyx '25)')((3 22 xyyyyxyx

)(325'25')(3 2222 yxxyxyyyxy

)(325'25)(3 2222 yxxyyxyxy

xyxy

yxxyy

25)(3

)(325'

22

22


Example 5 Determine the slope of the graph of

Solution

xyyx 100)(3 222 at the point of ).1 ,3(

Example 5 -- Continued

At the point (3, 1), the slope of the graph is

9

13

7530

9025

)3(25)13)(1(3

)13)(3(3)1(25'

22

22

1 ,3

yxy

xyxy

yxxyy

25)(3

)(325'

22

22


Example 6 Given the equation

Solution

10022 yx

find

022 dx

dyyx

2

2

dx

yd

xdx

dyy

y

x

dx

dyy '

Differentiating a second time with respect to x produces

22

2 ')1(

y

xyy

dx

yd

2

)/(

y

yxxy 3

22

y

xy

)100(

1002xy


Example 7 Find implicitly for the equation

Solution

xy sinThen find the largest interval of the form –a < y < a on which y is a differentiable function of x.

][][sin xdx

dyy

dx

dy

dx

dy

1cos dx

dyy

ydx

dy

cos

1

The largest interval symmetry to origin for which y is a differentiable function of x is –/2 < y < /2 .


Example 7 Find implicitly for the equation

Solution

xy sinThen find the largest interval of the form –a < y < a on which y is a differentiable function of x.

dx

dy

The given equation is that x is a function of y. If we restrict y value to the interval –/2 < y < /2, we could write the cos y, or equivalently, dy/dx explicitly.

,1sin1cos 22 xyy 22

y

Then

21

1

xdx

dy

or

21

1][arcsin

xx

dx

d

11 x

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2.5 Implicit Differentiation. After this lesson, you should be able to: Distinguish between...

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