!
5. y - 5x2 + 3 = 0; (1, 2)
Using implicit differentiation:
d
d
d
d
(y) (5x2) +
(3) =
(0)
dx
dx
dx
dx
y' - 10x = 0
!
y' = 10x
! y'
!
!
(1,2) = 10(1) = 10
7. x2 - y3 - 3 = 0; (2, 1)
d
d
d
d
(x2) (y3) (3) =
(0)
dx
dx
dx
dx
2x - 3y2y' = 0
3y2y' = 2x
!
2x
= 10 y'=
3y 2
!
y' (2,1)=
9. y2 + 2y + 3x = 0; (-1, 1)
d
d
d
d
(y2) +
(2y) +
(3x) =
(0)
dx
dx
dx
dx
2yy' + 2y' + 3 = 0
2y'(y + 1) = -3
y' = y' (!1,1) =
3
2(y + 1)
3
"3
= 4
2(2)
!
4
3
11. xy - 6 = 0
d
d
d
xy 6 =
(0)
dx
dx
dx
xy' + y - 0 = 0
xy' = -y
y
y' = x
3
y' at (2, 3) = - 2
!
13. 2xy + y + 2 = 0
d
d
d
d
2
xy +
y +
2=
(0)
!
dx
dx
dx
dx
2xy' + 2y + y' + 0 = 0
y'(2x + 1) = -2y
!2y
y' =
2x + 1
"2(2)
y' at (-1, 2) =
= 4
2("1) + 1
15. x2y - 3x2 - 4 = 0
d 2
d
d
x y 3!
x2 4=
dx
dx
dx
d
x2y' + y
(x2) - 6x - 0 =
dx
x2y' + y2x - 6x =
x2y' =
d
(0)
dx
0
0
6x - 2yx
6x ! 2yx
6 ! 2y
y' =
or
2
x
x
6! 2" 2! 4! 2
12 " 16
=
y'
=
= -1
2
(2,4)
2
4
EXERCISE 4-5
185
17. ey = x2 + y2
19. x3 - y = ln y
d y
d 2
d 2
d 3
d
e =
x +
y
x y=
dx
dx
dx
dx
dx
eyy' = 2x + 2yy'
3x2 - y' =
y'(ey - 2y) = 2x
2x
3x2 =
y' = y
e ! 2y
2! 1
2
3x2 =
=
y'
= 0
= 2
(1,0)
e " 2!0
1
d
ln y
dx
y'
y
"
1%
$1 + ' y'
y&
#
y + 1
y'
y
2
! y' = 3x y
y + 1
3 ! 12 ! 1
3
y'
=
=
(1,1)
1+ 1
2
21. x ln y + 2y = 2x3
d
d
[x ln y] +
2y
dx
dx
d
d
ln y·
x + x
ln y + 2y'
dx
dx
y'
ln y·1 + x ·
+ 2y'
y
"x
%
y' $ + 2'
#y
&
!
d
2x3
dx
= 6x2
=
= 6x2
= 6x2 - ln y
6x 2y ! y ln y
y' =
x + 2y
6 ! 12 ! 1 " 1 ! ln 1
6
y'
=
=
= 2
(1,1)
1+ 2!1
3
23. x2 - t2x + t3 + 11 = 0
d 2
d
d 3
d
d
x (t2x) +
t +
11 =
0
dt
dt
dt
dt
dt
2xx' - [t2x' + x(2t)] + 3t2 + 0 = 0
2xx' - t2x' - 2tx + 3t2 = 0
x'(2x - t2) = 2tx - 3t2
x' =
x'
(!2,1)
=
=
!
186
CHAPTER 5
ADDITIONAL DERIVATIVE TOPICS
2tx ! 3t2
2x ! t2
2("2)(1) " 3("2)2
2(1) " ("2)2
!4 ! 12
!16
= 8
=
2 ! 4
!2
25. (x - 1)2 + (y - 1)2 = 1.
Differentiating implicitly, we have:
d
d
d
(x - 1)2 +
(y - 1)2 =
(1)
dx
dx
dx
2(x - 1) + 2(y - 1)y' = 0
(x ! 1)
y' = (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).
! be verified on the graph.
These values can
!
!
(1.6
(1.6,1.8)
(1.8
(1.6
y'
= (1.6,0.2)
(0.2
y'
= -
!
!
!
!
!3
1)
0.6
= = 1)
0.8
4
1)
0.6
3
= =
1)
(!0.8)
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
d
d
xy x 4=
dx
dx
dx
xy' + y - 1 - 0 =
xy' =
d
0
dx
0
1 - y
1 ! y
y' =
x
1 ! 3
y'
=
= -1
(2,3)
2
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
EXERCISE 4-5
187
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 2
d
d
d
y xy 6 =
0
dx
dx
dx
dx
2yy' - xy' - y - 0 = 0
y'(2y - x) = y
y
y' =
2y ! x
3
3
=
y'
=
[Slope at (1, 3)]
(1,3)
2(3) " 1
5
3
The equation of the tangent line at (1, 3) with m =
is:
5
3
(y - 3) !
=
(x - 1)
5
3
3
y - 3= x 5
5
3
12
y= x +
5
5
2
"2
y'
=
=
[Slope at (1, -2)]
(1,! 2)
5
2("2) " 1
2
Thus, the equation of the tangent line at (1, -2) with m =
is:
5
2
(y + 2) =
(x - 1)
5
!
2
2
y + 2= x 5
5
2
12
y= x 5
5
31. xe y = 1
d y
d
d
e + ey
x =
1
dx
dx
dx
xe y y' + e y = 0
ey
1
y' = - y = xe
x
Implicit differentiation:
Solve for y: e y =
1
x
x·
"1%
y = ln $ ' = -ln x (see Section 2-3)
#x &
1
y' = x
In this case, solving for y first and then differentiating is a
!
little easier than differentiating implicitly.
188
CHAPTER 5
ADDITIONAL DERIVATIVE TOPICS
33. (1 + y)3 + y = x + 7
d
d
d
d
(1 + y)3 +
y=
x +
7
dx
dx
dx
dx
3(1 + y)2y' + y' = 1
y'[3(1 + y)2 + 1] = 1
1
y' =
3(1 + y)2 + 1
y'
(2,1)
=
1
2
3(1 + 1) + 1
=
1
13
!
35. (x - 2y)3 = 2y2 - 3
!d
d
d
(x - 2y)3 =
(2y2) (3)
dx
dx
dx
[Note: The chain rule is applied to the
left-hand side.]
3(x - 2y)2(1 - 2y') = 4yy' - 0
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
3(1 ! 2 " 1)2
3
=
y'
=
2
(1,1)
6(1 ! 2) + 4
10
37.
!
7 + y 2 - x3 + 4 = 0
or
(7 + y2)1/2 - x3 + 4 = 0
d
d 3
d
(7 + y2)1/2 x +
4 =
dx
dx
dx
1
d
(7 + y2)-1/2
(7 + y2) - 3x2 + 0 =
2
dx
1
(7 + y2)-1/22yy' - 3x2 =
2
yy '
=
(7 + y 2)1 2
y' =
y'
(2,3)
=
d
0
dx
0
0
3x2
3x 2(7 + y 2)1 2
y
3 ! 22(7 + 32)1 2
12(16)1 2
= 16
=
3
3
EXERCISE 4-5
189
39. ln(xy) = y2 - 1
d
[ln(xy)] =
dx
1
d
·
(xy) =
xy
dx
1
(x · y' + y) =
xy
1
1
· y' - 2yy' +
=
y
x
xy' - 2xy2y' + y =
y'(x - 2xy2) =
d 2
d
y 1
dx
dx
2yy'
2yy'
0
0
-y
!y
y
=
2
2
x ! 2xy
2xy ! x
1
y'
=
= 1
(1,1)
2 ! 1 ! 12 " 1
y' =
41. First find point(s) on the graph of the equation with abscissa x = 1:
Setting x = 1, we have
y 3 - y - 1 = 2 or y 3 - 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
d 3
d
d
d 3
d
at the point (1, 1.67):
y + x
y + y
x x =
2
dx
dx
dx
dx
dx
3y 2 y' - xy' - y - 3x 2 = 0
(3y 2 - x)y' = 3x 2 + y
3x 2 + y
y' =
;
3y 2 ! x
3 + 1.67
4.67
y'
=
=
≈ 0.63
2
(1,1.67)
7.37
3(1.67) " 1
Tangent line: y - 1.67 = 0.63(x - 1) or y = 0.63x + 1.04
43. x = p2 - 2p + 1000
!
d(x) d(p 2) d(2p) d(1000)
=
!
+
dx
dx
dx
dx
dp
dp
1 = 2p
- 2
+ 0
dx
dx
dp
1 = (2p - 2)
dx
1
dp
Thus,
= p' =
.
2p ! 2
dx
190
CHAPTER 5
ADDITIONAL DERIVATIVE TOPICS