EXERCISE 4-5 185

5. y - 5x2 + 3 = 0; (1, 2) Using implicit differentiation:

!

d

dx(y) -

!

d

dx(5x2) +

!

d

dx(3) =

!

d

dx(0)

y' - 10x = 0 y' = 10x

y'

(1,2) = 10(1) = 10 = 10

7. x2 - y3 - 3 = 0; (2, 1)

!

d

dx(x2) -

!

d

dx(y3) -

!

d

dx(3) =

!

d

dx(0)

2x - 3y2y' = 0

3y2y' = 2x

y' =

2x

3y2

y'

(2,1) =

4

3

9. y2 + 2y + 3x = 0; (-1, 1)

d

dx(y2) +

d

dx(2y) +

d

dx(3x) =

d

dx(0)

2yy' + 2y' + 3 = 0

2y'(y + 1) = -3

y' = -

!

3

2(y + 1)

y'

(!1,1) =

!

"3

2(2) = -

3

4

11. xy - 6 = 0

d

dxxy -

d

dx6 =

d

dx(0)

xy' + y - 0 = 0 xy' = -y

y' = -

y

x

y' at (2, 3) = -32

13. 2xy + y + 2 = 0

2

d

dxxy +

d

dxy +

d

dx2 =

d

dx(0)

2xy' + 2y + y' + 0 = 0 y'(2x + 1) = -2y

y' =

!2y

2x + 1

y' at (-1, 2) =

!

"2(2)

2("1) + 1 = 4

15. x2y - 3x2 - 4 = 0

d

dxx2y -

d

dx3x2 -

d

dx4 =

d

dx(0)

x2y' + y

d

dx(x2) - 6x - 0 = 0

x2y' + y2x - 6x = 0 x2y' = 6x - 2yx

y' =

6x ! 2yx

x2 or

6 ! 2y

x

y' (2,4)

=

6 ! 2 " 2 ! 4 ! 2

22=12 " 16

4 = -1

186 CHAPTER 5 ADDITIONAL DERIVATIVE TOPICS

17. ey = x2 + y2

d

dxey =

d

dxx2 +

d

dxy2

eyy' = 2x + 2yy' y'(ey - 2y) = 2x

y' =

2x

ey ! 2y

y' (1,0)

=

2 ! 1

e0 " 2 ! 0=

2

1 = 2

19. x3 - y = ln y

d

dxx3 -

d

dxy =

d

dxln y

3x2 - y' =

y'

y

3x2 =

!

1 +1

y

"

# $

%

& ' y'

3x2 =

y + 1

yy'

y' =

3x2y

y + 1

y' (1,1)

=

3 ! 12 ! 1

1 + 1 =

3

2

21. x ln y + 2y = 2x3

d

dx[x ln y] +

d

dx2y =

d

dx2x3

ln y·

d

dxx + x

d

dxln y + 2y' = 6x2

ln y·1 + x ·

y'

y + 2y' = 6x2

y'

!

x

y+ 2

"

# $

%

& ' = 6x2 - ln y

y' =

6x2y ! y ln y

x + 2y

y' (1,1)

=

6 ! 12 ! 1 " 1 ! ln 1

1 + 2 ! 1 =

6

3 = 2

23. x2 - t2x + t3 + 11 = 0

d

dtx2 -

d

dt(t2x) +

d

dtt3 +

d

dt11 =

d

dt0

2xx' - [t2x' + x(2t)] + 3t2 + 0 = 0

2xx' - t2x' - 2tx + 3t2 = 0

x'(2x - t2) = 2tx - 3t2

x' =

2tx ! 3t2

2x ! t2

x' (!2,1)

=

!

2("2)(1) " 3("2)2

2(1) " ("2)2

=

!4 ! 12

2 ! 4=

!16

!2 = 8

EXERCISE 4-5 187

25. (x - 1)2 + (y - 1)2 = 1. Differentiating implicitly, we have:

d

dx(x - 1)2 +

d

dx(y - 1)2 =

d

dx(1)

2(x - 1) + 2(y - 1)y' = 0 y' = -

(x ! 1)

(y ! 1)

To find the points on the graph where x = 1.6, we solve the given equation for y:

(y - 1)2 = 1 - (x - 1)2

y - 1 = ±

!

1 " (x " 1)2

y = 1 ±

!

1 " (x " 1)2

Now, when x = 1.6, y = 1 +

!

1 " 0.36 = 1 +

!

0.64 = 1.8 and y = 1 -

!

0.64 = 0.2. Thus, the points are (1.6, 1.8) and (1.6, 0.2). These values can be verified on the graph.

y' (1.6,1.8)

= -

(1.6 ! 1)

(1.8 ! 1) = -

0.6

0.8 = -

3

4

y' (1.6,0.2)

= -

(1.6 ! 1)

(0.2 ! 1) = -

0.6

(!0.8) =

3

4

27. xy - x - 4 = 0 When x = 2, 2y - 2 - 4 = 0, so y = 3. Thus, we want to find the equation of the tangent line at (2, 3).

First, find y'.

d

dxxy -

d

dxx -

d

dx4 =

d

dx0

xy' + y - 1 - 0 = 0 xy' = 1 - y y' =

1 ! y

x

y' (2,3)

=

1 ! 3

2 = -1

Thus, the slope of the tangent line at (2, 3) is m = -1. The equation of the line through (2, 3) with slope m = -1 is:

(y - 3) = -1(x - 2) y - 3 = -x + 2 y = -x + 5

188 CHAPTER 5 ADDITIONAL DERIVATIVE TOPICS

29. y2 - xy - 6 = 0 When x = 1, y2 - y - 6 = 0 (y - 3)(y + 2) = 0 y = 3 or -2. Thus, we want to find the equations of the tangent lines at (1, 3) and (1, -2). First, find y'.

d

dxy2 -

d

dxxy -

d

dx6 =

d

dx0

2yy' - xy' - y - 0 = 0 y'(2y - x) = y y' =

y

2y ! x

y' (1,3)

=

!

3

2(3) " 1=3

5 [Slope at (1, 3)]

The equation of the tangent line at (1, 3) with m =

3

5 is:

(y - 3) =

3

5(x - 1)

y - 3 =

3

5x -

3

5

y =

3

5x +

12

5

y' (1,!2)

=

!

"2

2("2) " 1 =

2

5 [Slope at (1, -2)]

Thus, the equation of the tangent line at (1, -2) with m =

2

5 is:

(y + 2) =

2

5(x - 1)

y + 2 =

2

5x -

2

5

y =

2

5x -

12

5

31. xey = 1 Implicit differentiation: x ·

d

dxey + ey

d

dxx =

d

dx 1

xeyy' + ey = 0

y' = -

ey

xey = -

1

x

Solve for y: ey =

1

x

y = ln

!

1

x

"

# $

%

& ' = -ln x (see Section 2-3)

y' = -

1

x

In this case, solving for y first and then differentiating is a little easier than differentiating implicitly.

EXERCISE 4-5 189

33. (1 + y)3 + y = x + 7

d

dx(1 + y)3 +

d

dxy =

d

dxx +

d

dx7

3(1 + y)2y' + y' = 1 y'[3(1 + y)2 + 1] = 1

y' =

!

1

3(1 + y)2 + 1

y' (2,1)

=

!

1

3(1 + 1)2 + 1=

1

13

35. (x - 2y)3 = 2y2 - 3

d

dx(x - 2y)3 =

d

dx(2y2) -

d

dx(3)

3(x - 2y)2(1 - 2y') = 4yy' - 0 [Note: The chain rule is applied to the left-hand side.] 3(x - 2y)2 - 6(x - 2y)2y' = 4yy'

-6(x - 2y)2y' - 4yy' = -3(x - 2y)2

-y'[6(x - 2y)2 + 4y] = -3(x - 2y)2

y' =

3(x ! 2y)2

6(x ! 2y)2 + 4y

y' (1,1)

=

3(1 ! 2 " 1)2

6(1 ! 2)2 + 4=

3

10

37.

!

7 + y2 - x3 + 4 = 0 or (7 + y2)1/2 - x3 + 4 = 0

d

dx(7 + y2)1/2 -

d

dxx3 +

d

dx4 =

d

dx0

1

2(7 + y2)-1/2

d

dx(7 + y2) - 3x2 + 0 = 0

1

2(7 + y2)-1/22yy' - 3x2 = 0

yy'

(7 + y2)1 2 = 3x2

y' =

3x2(7 + y2)1 2

y

y' (2,3)

=

3 ! 22(7 + 32)1 2

3=12(16)1 2

3 = 16

190 CHAPTER 5 ADDITIONAL DERIVATIVE TOPICS

39. ln(xy) = y2 - 1

d

dx[ln(xy)] =

d

dxy2 -

d

dx1

1

xy ·

d

dx(xy) = 2yy'

1

xy(x · y' + y) = 2yy'

1

y · y' - 2yy' +

1

x = 0

xy' - 2xy2y' + y = 0 y'(x - 2xy2) = -y y' =

!y

x ! 2xy2=

y

2xy2 ! x

y' (1,1)

=

1

2 ! 1 ! 12 " 1 = 1

41. First find point(s) on the graph of the equation with abscissa x = 1: Setting x = 1, we have y3 - y - 1 = 2 or y3 - y - 3 = 0 Graphing this equation on a graphing utility, we get y ≈ 1.67.

Now, differentiate implicitly to find the slope of the tangent line at the point (1, 1.67):

d

dxy3 + x

d

dxy + y

d

dxx -

d

dxx3 =

d

dx2

3y2y' - xy' - y - 3x2 = 0 (3y2 - x)y' = 3x2 + y

y' =

3x2 + y

3y2 ! x;

y' (1,1.67)

=

!

3 + 1.67

3(1.67)2 " 1=4.67

7.37 ≈ 0.63

Tangent line: y - 1.67 = 0.63(x - 1) or y = 0.63x + 1.04

43. x = p2 - 2p + 1000

d(x)

dx =

d(p2)

dx!d(2p)

dx+d(1000)

dx

1 = 2p

dp

dx - 2

dp

dx + 0

1 = (2p - 2)

dp

dx

Thus,

dp

dx = p' =

1

2p ! 2.