Section 2.5
156. f(x) = x Inx, f(l) = 0
Implicit Differentiation
145
3
f'(x) = I + Inx, f(l) = I
1
r(x) = -,
x r(1) = I
-,~.
-1
= f(l) + f'(l)(X - 1) = x - 1, PI(l) = 0
Pl(X)
P2(x)= f(1) + f'(l)(x - 1) + ~r(l)(X - 1)2
1
= (x- 1)+ 2(x - 1)2, P2(l) = 0
PI'(x) = 1, PI'(l) = 1
P/(x) = 1 + (x - I) = x, P/(l) = 1
P/'(x) = x, P2"(1)= 1
The values off, PI, P2' and their first derivatives agree at x
offandP2 agree atx = 1.
157. False. Ify = (1 - X)I/2,theny' =!(l
- X)-1/2(_I).
= 1.The values of the second derivatives
= sin22x, thenf'(x) = 2(sin 2x)(2 cos 2x).
158. False. Iff(x)
159. True
Section 2.5
1. r
+
Implicit Differentiation
T = 16
2.
2x + 2yy' = 0
-x
y'=y
3.
y2
=
16
X
y'= y
Xl/2 + yl/2 = 9
4.
x3+y3=8
3r + 3y2y' = 0
r
y'= _X-l/2 = -
y-l/2
-V;Ii
x3-xy+y2=4
y'
ry + y2x = - 2
ry'
(2y- x)y' = y - 3r
y'=-
= - y2
6.
3r - xy'- y + 2yy'= 0
7.
-
2x - 2yy' = 0
I
1
-X-l/2
+ _y-l/2y'
= 0
2
2
5.
X2
y - 3r
2y- x
xeY- lOx+ 3y = 0
xeY-dy + eY - 10 + 3 -dy = 0
dx
dx
dy (xeY+ 3) = 10 - eY
dx
dy_lO-eY
dx-xeY+3
T + 2yxy' = 0
(r + 2xy)y'= _(y2+ 2xy)
+ 2xy +
, -y(y + 2x)
y = x(x + 2y)
8.
eXY+r-y2=
(x:
+ y)eXY+ 2x - 2y:
10
=0
dy
dx (xeXY- 2y) = -yeXY- 2x
dy=
dx
yeXY+2x
xeXY- 2y
146
Chapter 2
Differentiation
x3 - 2x2y+ 3xy2= 38
9.
10.
3x2 - 2x2y' - 4xy + 6xyy' + 3y2 = 0
2x(3y
-
2 sinxcosy
2[sin x( - sin y)y' + COgy(cos x)] = 0
x)y' = 4xy - 3x2 - 3y2
, cosxcosy
y = sinxsiny
y' = 4xy - 3X2 -3y2
2x(3y - x)
11.
sin x + 2cos 2y
cosx - 4(sin2y)y'
=I
=0
=I
==cot x coty
12.
(sin TTX + COg 7Ty)2
2(sin 7TX + COS7TY)[7T COg TTX
cosx
y'=4sin2y
7T COg TTX
-
7T(sin7Ty)y1
= 2
= 0
- 7T(sin7TY)y'= 0
COg TTX
y' = sin TTY
14.
13. sinx = x(l + tany)
COgx = X(sec2y)y' + (l + tany)(l)
,
y
coty=x-y
(-csc2y)y' = I - y'
I
y'=I-csc2y
cosx-tany-I
=
xsec2y
=~
-cot2 Y = -tan2Y
15.
y
= sin(xy)
I
16. x=,secy
y' = [xy' + y] cos(xy)
y'
I
I = --sec-tany2
y
y' - xcos(xy)y' = ycos(xy)
, - y cos(xy)
y - I - x cos(xy)
17.
x3y3_y_X=0
I
Y
y' = sec(l/~~~(l/y)
18.
= _y2 cos(~)cot(~)
(xy)1/2- X + 2y
=0
I
~xy)-1/2(xy'
+ y) - I + 2y' = 0
2
3x3yZy'+ 3x2l - y' - I = 0
(3x3y2- l)y' = I - 3x2y3
, - I - 3x2l
y-3x3y2-I
-2-y'
2Fxy
+ -L- - I + 2y' = 0
2Fxy
xy' + Y - 2Fxy + 4Fxyy' = 0
, 2Fxy - y
y =
4Fxy + x
19.
20.
x2 - 3 In y + y2 = 10
3dy
2x---+2y-=0
ydx
dy
dx
2x =
In(xy)+ 5x = 30
Inx + Iny + 5x = 30
:
(~ - 2Y)
dy2x
- 2xy
dx - (3/y) - 2y - 3 - 2y2
-1 + --I dy +
x
ydx
5 = 0
I dy
--=---5
ydx
I
x
dy = _lx - 5y = dx
y + 5XY
(
x
)
Section 2.5
21.
xy
=4
22.
xy' + y(I) = 0
xl-y3=O
2x - 3y2y' = 0
xy' = -y
y'=
2x
y' = 3y2
-y
X
At(-4,-1):
2
At (1,1): y' = 3"
1
-4'
y'=
xl - 9
y2 = xl + 9
23.
Implicit Differentiation
24.
(x + y)3 = x3 + y3
x3 + 3x2y + 3xy2 + y3
2
' -
yy -
+ 9)(2x) - (xl - 9)2x
(X2
= x3 +
y3
3x2y + 3xy2 = 0
(xl + 9)2
,
18x
y = (X2+ 9)2y
x2y + xy2 = 0
x2y' + 2xy + 2.xyy' + y2 = 0
At (3, 0): y' is undefined.
(xl + 2xy)y' = _(y2 + 2xy)
,
y(y + 2x)
y = - x(x + 2y)
At(-l,
25.
y2/3
=5
2
+ ::;y-1/3y'
3
=0
X2/3
2
-X-I/3
3
+
26.
- -
y-I/3 -
(3y2- 2x)y' = 2y - 3xl
~~
3
Y
, 2y - 3xl
y = 3y2- 2x
1
At (8,1): y' = -2:"
27.
x3+y3_2xy=0
3xl + 3y2y' - 2.xy'- 2y = 0
-X-I/3
y' = -
1): y'=-1.
At (1, 1): y' = -1.
28.
tan(x + y) = x
xcosy = 1
x[-y' sin y] + cosy
(I + y~ sec2(x+ y) = 1
, 1 - sec2(x+ y)
Y = sec2(X+ y)
cosy
y'= xsiny
- -tan2(x + y)
- tan2(x+ y) + 1
xl
--xl+1
1
=-cotyx
At (2,~):
y' = 2~'
At (0, 0): y' = O.
29.
2eX}'
- x = 0,
(2,0)
30.
y2 = Inx,
2eX}'xy'= I - 2yeX)'
, - 1 - 2yex)'
y 2xeX)'
I
(e,l)
I
2eX}'[xy'+ y] - 1 = 0
At (2,0): y'=4'
=0
2yy' =
~
1
y' = 2xy
I
At(e,1):
y'=2e'
- cot Y
x
147
148
Chapter 2
31.
Differentiation
Jx+Jy=3
2-- x-I
Y - X2 + 1
32.
1
1
-x-1/2
+ -y-1/2y'
=0
2
2
,
2yy
(x2 + 1)(1) - (x - I)(2x)
=
(1/2)X-1/2
Jy
(1/2)y-1/2 = - Jx
y' =
x2+1-2r+2x
(x2 + 1)2
1
At (4, 1): y' = -2"
1+2x-x2
y' = 2Y(X2+ 1)2
Tangent line:
y - 1=
(x2+ 1)2
y"S
1
--(x
2
,
1+4-4
-~
: y = [(2y"S)/S](4+ 1)2- 1Oy"S.
( )
At 2, 5
- 4)
Tangent line:
1
y = --x2 + 3
y"S
y - -
S
1
1Oy"S
= -(x
- 2)
x + 2y - 6 = 0
lOy"Sy - 10 = x - 2
5
x - lOy"Sy + 8 = 0
~
-2~7
1
~
-1
-11I~=r
-1
33.
(X2
(X2
+ 4)y = 8
(4 - X)y2 = X3
34.
+ 4)y' + y(2x) = 0
(4 - x)(2yy~ + y2(_I)
= 3x2
3X2+y2
y' = 2y(4 - x)
'- -2xy
y-x2+4
At (2,2): y'= 2.
- _2x[8/(X2 + 4)]
x2+4
-16x
= (x2 + 4)2
At (2,1): y' =
-32
1
M
= -2.
(or, you could just solve for y: y = X2~ 4)
35.
(x2 + y2)2 = 4x2y
2(x2 + y2)(2x + 2yy~ = 4x2y' + y(8x)
36.
X3+y3_6xy=O
3X2+ 3y2y' - 6xy' - 6y = 0
4X3+ 4x2yy' + 4xy2 + 4yV = 4x2y' + 8xy
y'(3y2 - 6x) = 6y - 3x2
4x2yy' + 4yV - 4x2y' = 8xy - 4x3 - 4xy2
4y'(x2y + y3 -
X2)
= 4(2xy- x3 - xy2)
,_2xy-x3-xy2
Y - x2y+y3-x2
At (1,1): y' = O.
y'-
- (16/9) 32 4
(3' 3): y , = (16/3)
(64/9) - (8/3) = 40 = S.
4 8
At
6y -3x2
2y -x2
- 3y2 - 6x - y2 - 2x
Section 2.5
Implicit Differentiation
(c) Explicitly:
37. (a) x2 + y2 = 16
y2 = 16 - r
dy
dx
y = :t~16 - r
= :tl.(16
2
-
(b)
- r)-1/2(-2x)
+=x
-
- x
- ~16 - X2- :t~16 -
r
= -x
y
(d) Implicitly:
2x+2yy'=O
-6
x
-;;
y'=
38. (a) (X2- 4x + 4) + (y2 + 6y + 9) = -9 + 4 + 9
(x - 2)2 + (y + 3)2 = 4 (Circle)
(y + 3)2= 4
-
(x - 2)2
(c) Explicitly:
dy
dx
1
= :t¥4 - (x - 2)2]-1/2(-2)(x - 2)
+=(x- 2)
(~4 - (x - 2)2
Y = -3:t--/4 - (x - 2)2
(b)
-(x - 2)
:t~4 - (x - 2)2
-1
-(x-2)
-3 :t ~4 - (x - 2)2 + 3
-3
= - (x - 2)
y+3
(d) Implicitly:
2x + 2yy' - 4 + 6y' = 0
(2y + 6)y' = -2(x - 2)
y
(c) Explicitly:
39. (a) 16y2= 144 - 9X2
1
y2 = -(144
16
3
y = :t4~16
9
- 9X2)= -(16
16
- X2)
dy
-dx
3
= :t-(16
- r)-1/2(-2x)
8
3x
- X2
=
+=4~16
-3x
-
r
= 4(4/3)y =
(d) Implicitly:
(b)
18x + 32yy' = 0
-9x
-6
,_-(x-2)
y:t3
y' = 16y
-9x
16y
149
150
Chapter 2
40. (a) y2
Differentiation
= 1 + X24 = r 4+ 4
(c) Explicitly:
1
Y = -z-..}r
+4
2
r
dy = +.!.(
dx
-4
+ 4) -1/2(2x)
-zx
= 2../r + 4
(b)
-zx
_x
- 4(I/2)..}r + 4 - 4y
(d) Implicitly:
1
2yy' - -x
2 = 0
X
y' = 4y
41.
r+xy=5
2x + xy' + Y = 0
-(2x+y)
y'=~
2 + xy" + y' + y' = 0
xy"= -2(1 +y1
"
- 2[1 - (2x + y)/x]
=
Y
"
x
2(x + y)
r
y=
(Note: Youcouldwritey
x
2(x + y)
=~
10
.:;=x3
= (5 -
r)/x and calculate y' and y" directly.)
42.
ry2-2x=3
2x2yy' + 2xy2- 2 = 0
ryy' + xy2 - 1 = 0
y'-
l-xy2
-~
2xyy' +r(y12 + x2yy"+ 2xyy'+ y2 = 0
4xyy' + r(Y12 + x2yy"+ y2 = 0
4
- -4xy2 +
X
4xy2
-
(I
-
xy2)2
x2y2
4ry4 + 1 - 2xy2 + ry4
+ x2yy"+ y2 = 0
+ x4y3y" +
H =0
x4y3y" = 2x2y4- 2xy2 - 1
y" =
2x2,.4- 2xy2 - 1
.f
x4y3
Section 2.5
43.
X2
- y2 = 16
44.1-xy=x-y
y-xy=x-l
2x - 2yy' = 0
X
y
y'=
x-I
y=-=-1
1- x
y' = 0
x - yy' = 0
y"=
1 - yy" - (y ~2 = 0
1
,,
-yY-y
Implicit Differentiation
0
X 2
()
=0
y2 - y3y" = X2
y" = y2 - xl = -16
y3
y3
45.
46.
y2=x3
2yy' = 4
2yy' = 3X2
,
3X2 3x2 xy
3y
y ----.---.--- 2y - 2y xy - 2x
, 2
y =Y
x3
3y
y2 - 2x
"
2x(3y~ - 3y(2)
y =
4xl
-
2x[3
y2 = 4x
. (3y/2x)]
-
y" = -2y-2y' =
[ ] . ~ --2
2
Y
Y
- -4
y3
- 6y
4X2
-ll.
- 3x
- 4X2 - 4y
47. X2
+ y2 =
25
-x
Y'=y
6
At (4, 3):
-4
Tangent line: y - 3 = 3(x
- 4) ~
Normal line: y - 3 = ~(x - 4) ~
~
4x + 3y - 25 = 0
3x - 4y = O.
-9lEZJ
6
At (- 3,4):
Tangent line: y - 4 = ~(x + 3) ~
line:
y - 4 = 3(x
~
3x - 4y + 25 = 0
-4
Normal
9
-6
-9
+ 3) ~
4x
+
3y = O.
[SJ2]-6
9
151
152
Chapter 2
48. x2
+ y2 =
Differentiation
9
-x
4
Y'=y
EZELJ
At (0, 3):
Tangent line: y = 3
-6[TJ6
Normal line: x = O.
-4
4
At (2, ,)5):
-2
Tangentline: y - .,/5 = .,/5 (x - 2) => 2x + .,/5y - 9 = 0
Normalline: y - .,/5 =
49.
~
(x - 2) => .,/5x - 2y = O.
UE3
-6
LB2J
6
-4
x2 + y2 = r2
2x + 2yy' = 0
-x
y' = - = slopeof tangentline
y
lx
= slope of normal line
Let (xo'y~ be a point on the circle. If Xo= 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes
through the origin. If Xo*- 0, then the equation of the normal line is
y - y
0
= Yo(x Xo
y=-x
x )
0
Yo
Xo
which passes through the origin.
50.
y2 = 4x
2yy' = 4
y'=-=
2
y
lat(I,2)
Equation of normal at (I, 2) is y - 2 = - I(x - I), Y = 3 - x. The centers of the circles must be on the normal and at a
distance of 4 units from (1,2). Therefore,
(x - 1)2+ [(3 - x) - 2]2= 16
2(x - 1)2= 16
x = I ::f:2J2.
Centersof the circles: (I + 2J2, 2 - 2J2) and(I - 2J2, 2 + 2J2)
Equations:(x - I - 2J2)2 + (y - 2 + 2J2)2 = 16
(x - 1 + 2J2)2 + (y - 2 - 2J2)2 = 16
Section 2.5
51. 25x2 + 16y2+ 200x
-
16Oy+ 400
5Ox + 32yy' + 200
-
Implicit Differentiation
=0
16Oy'
=0
, 200 + 5Ox
y = 160 - 32y
(-8.5)
= -4:
Horizontal tangents occur when x
-10 -8 -6 -4
-2
25(16) + 16y2+ 200(-4) - 16Oy+ 400 = 0
y(y - 10)= 0 ~ y = 0, 10
Horizontaltangents: (- 4,0); (- 4, 10).
Vertical tangents occur when y
25r + 400 + 200x
-
= 5:
800 + 400
=0
25x(x + 8) = O~x
= 0,-8
Verticaltangents: (0,5), (- 8, 5).
52. 4r + y2 - 8x + 4y + 4
=0
8x + 2yy' - 8 + 4y' = 0
8-8x
-1
4-4x
Horizontal tangents occur when x
4(1)2 + y2 - 8(1) + 4y + 4
-3
= I:
=0
= y(y + 4) = 0 ~
(2.-2)
(0,-2)l
y' = 2y + 4 = y + 2
y2 + 4y
(1.0)
I"""
I
-1
-4
-s
y
1(1,:'4)
= 0, -4
Horizontal tangents: (1,0), (1, -4).
Vertical tangents occur when y
= - 2:
4r + (- 2)2- 8x + 4(- 2) + 4 = 0
4r - 8x= 4x(x- 2) = 0 ~ x = 0,2
Verticaltangents: (0, -2), (2, -2).
53.
y = xR-=t
54.
I
Iny = Inx + ZIn(r - 1)
Hz)
=~+
~-
X2
~I
2x2-1
= ~(x - l)(x - 2)(x - 3)
I
Iny
= Z[In(x -
1) + In(x - 2) + In(x - 3)]
~(z) = Ux ~ 1 + x ~ 2 + x ~ 3]
_2x2-1
dx - Y x(r - 1) - R-=t
[
y
]
1
3X2- 12x + 11
I)(x - 2)(x - 3)
= Z[(x -
]
dy - 3r - 12x+ 11
dx 2y
3r - 12x+ 11
2~(x - l)(x - 2)(x - 3)
153
154
Differentiation
Chapter 2
X2§-=2
55.
y
=
y=
Y(
)
2
dx
=Y
3
2
= ~ + 2(3x - 2) - x - I
3x2 - 15x + 8
dy
[2x(3x
2)(x - I)
-
H2) =
]
-
x(x - 1)3/2
v"X+l
3
I
In y = In x + 2ln(x - I) - 2ln(x + I)
()
I
=I
2
59.
H2)=x~l+x~2-x~l-x~2
dy
]
( )= ~(I
xx
X2
X2
~(2) = (x -
- In x)
(2)
dy
- 6(X2 - 2)
= (x = (x +
dx=y~+lnx
[
I
Iny = -In(1
+ x)
x
-dx = y[ x-2'--
( )=1. (~ )+ In(1 + x)(-1- )
= ~
dx xx+1
[ - In(xx+ I) ]
=
x
[~x+1 - In(xx+ I)]
1. dy
= (x + I)C ~ 2) + In(x - 2)
= (x -
y = (I + X)l/x
I)ln(x - 2)
X+ I
ydx
+ In(x - 2)
]
[ x-2
xl+x
dy
X+ I
2)X+ 1 -
]
= xx-2 (x - I + xlnx)
62.
2)X+1
1)(;) + Inx
X- I
dy
+ In(x - 2)
]
y..
(I + X)l/x
]
(x + I)(x - I)(x + 2)(x - 2)
y =XX-1
dy
dx = 2y
X2 (I - In x) = 2x(2/x)-2(I - In x)
~
- I)(r - 4)
In y = (x - l)(In x)
( ) ()
Iny
-6X2+ 12
6(X2- 2)
(x - 1)2(x - 2)2
60.
1. dy = ~ 1. + Inx -~
y
] [(r
- 4 =Y
- (x - I)(x - 2)
2
Iny=-Inx
x
61.
-4
r
+ I)(x + 2) .
- (x
+ 4x - 2 = (2x2 + 2x - I)~
x(x2 - I)
(x + 1)3/2
y = X2/x
ydx
-2
[
dx = Y X2 - I +
x~ I]
4X2
[
Iny = In(x + I) + In(x + 2) - In(x- I) - In(x- 2)
( ) ( )
1-
+ 1)2/3(X2 - 1)4/3
(x + I)(x + 2)
y = (x - I)(x - 2)
58.
I
I
-2x+1
2=~[~+ x~
x ~ I - x ~ I]
-4x
- 3(r
Y=
I dY
I
3
ydx
=~+2x-1
~I-
HX2
dy
-4xy I) - - 3(x44x I) -virr -+ I
dx - 3(x4-
3x3 - 15r + 8x
- 2(x - 1)3~3x - 2
57.
+I
x2-1
I
Iny = 3[In(x2 + I) -In(x + I) -In(x - I)]
I
In y = 2 In x + 21n(3x - 2) - 2ln(x - I)
I dY
dx
~
3-
56.
(x - 1)2
X2
Section 2.5
63. Find the points of intersection by letting y2
2,x2+ 4x = 6
and
= 4x in
the equation 2,x2 + y2
Implicit Differentiation
= 6.
(x + 3)(x - I) = 0
The curves intersect at (1, :t2).
Ellipse:
Parabola:
4x + 2yy' = 0
2x
y'= -Y
2yy' = 4
2
y' =y
-6
At (1, 2), the slopes are:
y' = 1.
y' = -1
At (1, -2), the slopes are:
y' = -1.
y' = 1
Tangents are perpendicular.
64. Find the points of intersection by letting y2
2x2 + 3X3 =
5
and
= x3 in the
equation 2,x2 + 3y2
= 5.
3x3+2x2_5=0
Intersect when x = 1.
Points of intersection: (1,:t I)
y2 = X3:
2x2 + 3y2 = 5:
2yy' = 3X2
4x + 6yy' = 0
'- 3X2
Y -2y
2x
y' = - 3y
i=X3
At (1, 1), the slopes are:
2
y' = -'3'
3
y' =- 2
At (1, -1), the slopes are:
2
y' = '3'
3
y'=
-2
Tangents are perpendicular.
65. y
= -x
and x = siny
Point of intersection: (0, 0)
y = -x:
x=siny:
y' = -1
1 = y'cosy
y'=secy
At (0, 0), the slopes are:
y' = -1
y' = 1.
Tangents are perpendicular.
-6
6
155
156
Differentiation
Chapter 2
66. Rewriting each equation and differentiating,
x(3y - 29) = 3
x3 = 3(y - 1)
x3
y=-+3
y' =
I
y = ~(~ + 29)
I
X2
- X2'
y' =
-3
For each value of x, the derivatives are negative reciprocals of each other, Thus, the tangent lines are orthogonal at both points
of intersection,
67.
xy'
2
x2_y2=K
xy=C
+ Y = 0
2
2x - 2yy' = 0
-,~"c:s=:rn,
~
X
y'= _lX
y'=
Y
bl:ITJ
-2
-2
At any point of intersection (x, y) the product of the
= -1. The curves are orthogonal.
slopes is (-y/x)(x/y)
68.
2x +
2
y = Kx
x2+y2=C2
2
y'= K
2yy' = 0
X
y'= -y
.tm3.
-2
.~,
-2
At the point of intersection (x,y) the product of the
slopes is (- x/y)(K) = (- x/Kx)(K) = -1. The curvesareorthogonal.
69. 2y2 - 3x4 = 0
dy
dx
(b) 4y- dt - 12x3-dt
(a) 4yy' - 12x3= 0
=0
4yy'= 12x3
Y
,
ydy
dt
12x3 3x3
- 4y - y
----
= 3X3dx
dt
70. x2 - 3xy2 + y3 = 10
dx
dx
dy
dy
(b) 2x -dt - 3y2-dt - 6xy -dt + 3y2-dt = 0
(a) 2x - 3y2 - 6xyy' + 3y2y' = 0
(-6xy + 3y2)y' = 3y2 - 2x
.(2x - 3y2)dx=
dt (6xy - 3y2)dy
dt
3y 2 - 2x
3y2 - 6xy
.
y'=
71. cos TTY- 3 sin TTX= I
(a)
- 7T sin (77)')y' - 37T CDSTTX =
(b) -7Tsin(7Ty)dy
dt - 37Tcos(7Tx)dx
dt = 0
0
dy
-sin(7Tytdt
,
- 3 cos TTX
Y =
sin TTY
dx
= 3 COS(7TX7t
72. 4 sin x cos y = I
(a) 4 sin x(-sin y)y' + 4cosxcosy
= 0
,
cos x cos y
y = sin x sin y
(b) 4sinx(-siny)!
+ 4COSX~ cosy = 0
dx
.
. dy
cos x cos Y dt = sm x sm y dt
Section 2.5
73. (a) .0 = 4(4x2 - y2)
10
4y2 = 16x2 - .0
1
y2
y
(b)
= 4X2 - -.0
4
=
Ij:::
y
=3~
-"~,,
-10
- ~.0
1
= 4X2 - -.0
4
9
=
36
16x2 - .0
.0 - 16x2+ 36 = 0
X2 = 16 I ~256 - 144 = 8 + "28
2
-~£O
Note thatx2 = 8 I h8
= 8 I 2fi = (I Ifi)2.
Hence, there are four values of x:
- 1-
fi,
1-
fi,
-1+
fi,
1+
fi
3
' - x(8 - X2)
2 ' - 8
fi d h 1
To III t e s ope, yy - x - x ~ y 2(3) .
For x = -1 - fi, y' = Hfi
+ 7), andthe lineis
Yl = Hfi + 7)(x+ 1 + fi) + 3 = t[(fi + 7)x + 8fi + 23].
Forx = 1 - fi, y' = Hfi
yz = Hfi
- 7), andthe lineis
- 7)(x - 1 + fi)
Forx = -1 + fi,y'=
Y3= -Hfi
= -Hfi
- 7)x + 23 - 8fi].
- 7),andthelineis
- 7)(x + 1 - fi)
Forx = 1 + fi,y'
Y4= -Hfi
-Hfi
+ 3 = H(fi
+ 3 = -t[( fi - 7)x - (23 - 8fi)].
+ 7), andthe lineis
+ 7)(x - 1 - fi)
+ 3 = -t[( fi
+ 7)x - (8fi
+ 23)].
(c) Equating Y3 and Y4'
-1-(fi
3
- 7)(x + 1 - fi)
+ 3 = -1-(fi
3
+ 7)(x - 1 - fi)
+3
(fi - 7)(x+ 1 - fi) = (fi + 7)(x- 1 - fi)
fix + fi - 7 - 7x - 7 + 7fi = fix - fi - 7 + 7x - 7 - 7fi
16fi = 14x
8fi
x=7
If x = 8f,
theny = 5 andthe linesintersectat (8f,
5).
Implicit Differentiation
157
158
Chapter 2
74.
tany=x
Differentiation
y'sec2y = 1
,
1
~
~
y = -- < y < sec2y' 2
2
sec2y = 1 + tan2y = 1 + X2
1
1 +x2
y'=
75. Letf(x) = x" = xp/q, where p and q are nonzero integers and q > O.First consider the case where p =
f(x) = Xl/q is given by
~[Xl/q]
dx
= ~-->o
lim f(x
+ /lx)
/lx - f(x)
= lim
'-->xf(t)t --
f(x)
x
where t = x + /lx. Observe that
f(t)
- f(x) - t1/q - Xl/q t1/q - Xl/q
t- x
t - x - (tl/q)q - (Xl/q)q
t1/q - Xl/q
(tl/q - Xl/q)(tl-(l/q) + tl-(2/q)Xl/q +
. . . + tl/qxl-(2/q)+ Xl-(l/q»)
1
- tl-(l/q) + tl-(2/q)Xl/q + ... + tl/qxl-(2/q) + X1-(I/q)'
As t~x,
the denominator approaches qxl-(l/q). That is,
d
1
1
_[Xl/q] =
= -X(l/q)-l.
dx
qxl-(l/q)
q
Now consider f(x) = :xf'/q= (:xf')1/q.
From the Chain Rule,
f'(x) = .!.(:xf')(l/q)-l~:xf']
q
dx
= .!.(:xf')(l/q)-lp:xf'-l
q
76.
= !!..;,<p/q)-p]+(p-I) = !!..X(p/q)-bnx"-I
q
q
(n = q).
!!..
~+Jy=Jc
~+~dy=O
2~
2Jy dx
dy__Jy
dx- ~
Tangent line at (xo' Yo):
y
-
Yo = -
~
yXo
(x - xo)
x-intercept:(xo + ~-So,
0)
y-intercept:(0, Yo+ ~-So)
Sum of intercepts:
(xo + ~-50)
+ (Yo+ ~-50)
= Xo+ 2~-50
+ Yo= (~
+ -50)2 = (Jc)2 = C.