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EXERCISE 4-5 185
5. y - 5x2 + 3 = 0; (1, 2) Using implicit differentiation:
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