1 Implicit Differentiation. 2 Introduction Consider an equation involving both x and y: This...

1

Implicit Differentiation

2

Introduction

• Consider an equation involving both x and y:

• This equation implicitly defines a function in x

• It could be defined explicitly

2 2 49x y

2 49 ( 7)y x where x

3

Differentiate

• Differentiate both sides of the equation– each term– one at a time– use the chain rule for terms containing y

• For we get

• Now solve for dy/dx

2 2 49x y

2 2 0dy

x ydx

4

Differentiate• Then gives us

• We can replace the y in the results with the explicit value of y as needed

• This gives usthe slope on the curve for any legal value of x

2 2 0dy

x ydx

2

2

dy x x

dx y y

2 49

dy x

dx x

5

Guidelines for Implicit Differentiation

6

Slope of a Tangent Line

• Given x3 + y3 = y + 21find the slope of the tangent at (3,-2)

• 3x2 +3y2y’ = y’

• Solve for y’2

2

3'

1 3

xy

y

Substitute x = 3, y = -2 27

11slope

7

Second Derivative

• Given x2 –y2 = 49

• y’ =??

• y’’ =

'x

yy

2

2 2

2 2

2 2

2 2 3 3

'

49

d y y x y

dx y

x xy y

y xy yy y y y

Substitute

8

Find the derivative with respect to x.

22 2 2 2dy dy dx

y x y ydx dx

dy dx dy dxx y x x y

dx dx xdx dx d

2 2 23. x y y xy xy 22 2x y xyy xy

2 22 2 2dy dy dy dy

x xy y x y xy ydx dx dx dx

2 22 2 2dy dy dy dy

x y x xy y xy ydx dx dx dx

2 22 2 2dy

x y x xy y xy ydx

2

2

2

2 2

dy y xy y

dx x y x xy

9

1. Find the equation of the tangent line to at the point (2,1).

2 4 8x y

2 4 2 0dy

x y xdx

2

2

4

dy xy

dx x

42,1

8

dy

dx

1

2

11 2

2y x

12

2y x

10

2. Find the second derivative with respect to x.

2 2 16x y

2 2 0dy

x ydx

dy x

dx y

2

2

d y

dx

2

1d

x

y

ydx

y

2

xy

y x

y

2

2

xy

yy

y

y

2 2

3

y x

y

3

16

y

2 2 16x y

Look for a substitution of the original.

11

3. Find the points at which the graph has horizontal and vertical tangents.

2 225 16 200 160 400 0x y x y

0slope undefinedslope

50 32 200 160 0dy dy

x ydx dx

50 200

32 160

dy x

dx y

Horizontal: Vertical:

50 200x 32 160y

0 0

4x 5y

2 225 4 16 200 4 160 400 0y y 2225 16 5 200 160 5 400 0x x 216 160 0y y 16 10 0y y

0,10y

and 4,0 4,10

225 200 0x x 25 8 0x x

0, 8x

and0,5 8,5

12

We need the slope. Since we can’t solve for y, we use

implicit differentiation to solve for .dy

dx

Find the equations of the lines tangent and normal to the

curve at .2 2 7x xy y ( 1, 2)

2 2 7x xy y

2 2 0dydy

x yx ydxdx

Note product rule.

2 2 0dy dy

x x y ydx dx

22dy

y xy xdx

2

2

dy y x

dx y x

2 2 1

2 2 1m

2 2

4 1

4

5

13

Find the equations of the lines tangent and normal to the

curve at .2 2 7x xy y ( 1, 2)

4

5m tangent:

42 1

5y x

4 42

5 5y x

4 14

5 5y x

normal:

52 1

4y x

5 52

4 4y x

5 3

4 4y x

14

Higher Order Derivatives

Find if .2

2

d y

dx3 22 3 7x y

3 22 3 7x y

26 6 0x y y

26 6y y x

26

6

xy

y

2xy

y

2

2

2y x x yy

y

2

2

2x xy y

y y

2 2

2

2x xy

y

x

yy

4

3

2x xy

y y

Substitute back into the equation.

y