tDon'You can do it!!!

2.5 Implicit Differentiation

How would you find the derivative in the equationx2 – 2y3 + 4y = 2

where it is very difficult to express y as a functionof x?

To do this, we use a procedure called implicitdifferentiation.

This means that when we differentiate termsinvolving x alone, we can differentiate as usual. But when we differentiate terms involving y, wemust apply the Chain Rule.

Watch the examples very carefully!!!

Differentiate the following with respect to x.

3x2

2y3

x + 3y

xy2

6x

6y2 y’

1 + 3y’

Product rule

x(2y)y’ + y2(1) = 2xyy’ + y2

Find dy/dx given that y3 + y2 – 5y – x2 = -4

02523 2 =−−+ xdxdy

dxdy

ydxdy

y Isolate dy/dx’s

xdx

dy

dx

dyy

dx

dyy 2523 2 =−+ Factor out dy/dx

( ) xyydx

dy2523 2 =−+

( )523

22 −+

=yy

x

dx

dy

What are the slopes at the following points?

(2,0)

(1,-3)

x = 0

(1,1)

5

4−=m

8

1=m

0=m

undefined

Determine the slope of the tangent line to the graph of x2 + 4y2 = 4 at the point .( )21,2 −

-2 -1 1 2

( )21,2 −

082 =+dxdy

yx

xdx

dyy 28 −=

y

x

dx

dy

8

2−=

y

x

dx

dy

4

−=

⎟⎠

⎞⎜⎝

⎛−−

==

2

14

2

dx

dym

4

2

1

2 −⋅

−=

2

1

4

2==m

Differentiate sin y = x

1cos =dxdy

yydx

dy

cos

1=

Differentiate x sin y = y cos x Product Rule

x cos y (y’) + sin y (1) = y (-sin x) + cos x (1)(y’)

x cos y (y’) - cos x (y’) = -sin y - y sin x

y’(x cos y - cos x) = -sin y - y sin x

xyx

xyyy

coscos

sinsin'

−−−

=

Given x2 + y2 = 25, find y”

0'22 =+ yyx

y

x

y

xy −=

−=

22

'

2

')1("

y

xyyy

−−=

Now replace y’ withy

x−

2"

y

yx

xy

y⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

Multiply top and bottom by y

⎟⎟⎠

⎞⎜⎜⎝

⎛y

y3

22

y

yx +−=

What can we substitute in for x2 + y2?

3

25

y−=