Section 2.6
Section
1.
Derivatives of Inverse Functions
Derivatives of Inverse Functions
2.6
f(x) = X3+ 2x - 1, f(1) = 2 = a
f'(x) = 3x2+ 2
1
1
1
1
(j-l)'(2) = f'(j-l(2)) = 1'(1) = 3(1)2+ 2 = "5
2.
f(x) = 2x5+x3 + 1, f(-I)=-2=a
f'(x) = lOx4+ 3X2
1
(j-l)'(-2)
3.
1
= f'(j-l(-2))
= 1'(-1)
1
1
= 10(-1)4 + 3(_1)2 =
13
f(x) = sinx, f(~) = ~ = a
f'(x) = cos X
-) '
(j
4.
1. -
1
)
~
-
) (2 - f'(j-l(1/2))
-
1
-
~
- f'(n/6) - cos(7T/6)- ~/2
- 2~
- 3
f(x) = cos 2x, f(O) = 1 = a
f'(x) = - 2 sin2x
(j-l)'(1) =
5.
1
f'(j-l(1))
4
f(x) = x3 - -,
x
=~
1'(0)
=
1
- 1
-2 sin 0 -
. .
0 WhICh
IS undefined.
f(2) = 6 = a
f'(x) = 3x2 + .!
X2
1
1
1
1
(j-l)'(6) = f'(j-l(6)) = f'(2) = 3(2)2+ (4/22) =
6.
f(x) =~,
f'(x) =
f(8) = 2 = a
1
2~
1
1
1
(j-l)'(2) = f'(j-l(2)) = 1'(8) = 1/(2~)
7.
13
f(x) = x3,
G,)
- 1 - 1/4 - 4
8.
f(x) = 3 - 4x,
(1,-1)
f'(x) = - 4
f'(x) = 3X2
1'(1) = -4
3-x
f-l(X) = ~'
f'G)=
r1(x) = 4'x,
1
(j-l)'(x) = 3W
(,)
1
(j-l)'(X) = -4
1
(j-l)'( -1) = -4
(j-l)'(i)
=
(-1,1)
159
160
Chapter 2
9.
Differentiation
(5,1)
f(x) = h"-=4,
10.
f(x) = 1
1
f'(x) =
2h"-=4
: x2'
(1,~)
-2x
f'(x) = (1 + x2)2
1'(5) =.!.2
1
1'(1) = -2
(1,5)
f-l(X) = x2 + 4,
~
U-l)'(X) = 2x
f-l(X)
1
(2' 1)
= -..J~'
U-l)/(1) = 2
lr--x
U-l)/(X)
=-
2x2-..J
~
U-l)'(4) = -2
11. f(x) = 2 arcsin(x- 1)
2
2
f'(x) = ""1 - (x - 1)2= ""2x - x2
x
13.
g(x)
x
15. f(x) = arctan-a
1
-1
-1
g/(x)= -J1=X )( 2.Jx) = 2.Jx""1 - x
(
f'(x) = C ~ JC~J
= 2.Jx(~ + x)
18. h(x) = x arctan x
17. g(x) = arcsin3x
x
x(3/~)
gx=
f'(x) = g=7
16. f(x) = arctan.Jx
l/a
- ~
f'(x) = 1 + (x2/a2)- a2 + x2
( )
2t
14. g(x) = arccos.Jx
= 3 arccos 2
-3(1/2)
-3
g/(x) = ""1 - (x2/4) = ""4 - x2
/
,
12. f(t) = arcsin t2
- arcsin3x
x2
X
h'(x) = 1 + x2 + arctan x
- 3x - ~
arcsin3x
x2""1- 9X2
20. f(x) = arcsec2x
19. h(x) = arccot 6x
-6
h'(x) = 1 + 36x2
21. h(t) = sin(arccost) =
f'(x) =
22. g(t)
1
h'(t) = -(1
2 - t2)-1/2(- 2t)
Ixl""4x2 - 1
= tan(arcsin t) = ""1 -
t2
~-t(-t/~)
g/(t) =
=~
~
f'(x) = 0
12x1""4x2- 1
1
=-
t
~
23. f(x) = arcsinx + arccosx =
2
-
1 - t2
1
- O-=-t2)3/2
f
7r
24. f(x)
= arcsec x
f'(x) = 0
+ arccsc x
= "2
Section 2.6
I I
25. Y = 2
x+ I
I + arctanx
(21nx -
f = ~C ~
I
)= 4'[ln(x + 1) -In(x
Derivatives of Inverse Functions
I
- I)] + 2 arctanx
I - x ~ I) + 112r = I ~ X4
26. Y = i(x~
+ arcsinx)
-x
dy =l x
dx 2[ (~
) + ~+
]= ~
I
~
27. Y = x arcsinx + ~
dy
1
(~
dx=x
.
x
)+arcsmx-
.
~=arcsmx
1
28. y = x arctan2x - 4'ln(I +
dy
dx
4X2)
2x
I
(
8x
)
= I + 4r + arctan(2x)- 4' I + 4r = arctan(2x)
29. f(x) = arccosx
f'(x) =
-I
.J3
= -2 whenx = :i:-.
2
~
Whenx = .J3/2, f( .J3/Z) = 7T/6.Whenx = - .J3/2,f( - .J3/Z) = 57T/6.
Tangent lines:
~ = -2(X- ~)
y -
y - 5; = -2(X + ~)
I
30. g(x)
= arctan x, g'(x)
Tangent line: y
y
~
= -2x + (~+ .J3)
~ y = -2x + e67T- .J3)
I
= I + r'
g'(1) =
2
- ~ = i(x - I)
I
31. f(x) = arcsin x,
I
7T
y=-x+--2
4
2
1
a =2
I.S
I
f'(x) = ~I -
1.0
r
x
r(x)
=
(I
-
p;
x2)3/2
' 1
iI
=J\2 + f 2)( x
I
'IT
2.J3
I
) ( - 2)= 6"+ 3 (x - 2)
I
I
1
I
1
z.J3
2.J3
I "1
,(1
)+Tx-2 ( )
P2(x)=f2( ) +f\2 )( x-2 ) +2f (2)(x-2 ) =6"+3x-2 (
p.(x)
2
'IT
2
161
162
Differentiation
Chapter 2
32. f(x) = arctan x,
a = 1
1
f'(x) = -1 + X2
-2x
f"(x) = (1 + x2)2
PI (x) = f(1) + f'(1)(x
'TT 1
- 1) = 4" + 2(x - 1)
,
1
'TT 1
1
P2(x) = f(1) + f (1)(x - 1) + -2 f"(1)(x - 1)2 = -4 + -(x
2 - 1) - -(x
4 - 1)2
h(t) = - 16t2 + 256
33. (a)
-16t2 + 256 = 0
(b) tan 0 = ~
500
when t = 4 sec.
= -16t2500+ 256
0 = arctan[51~ (- t2 + 16)]
dO - 8t/125
-1000t
dt - 1 + [(4/125)(-t2 + 16)]2 - 15,625 + 16(16 - t2)2
.~
500
When t = 1, dO/dt = -0.0520 rad/sec.
Whent=
2, dO/dt= -0.1l16rad/sec.
h
34. tan 0 = 300
300
dh = 5 ftlsec
dt
0
200
100
= arctan(3~)
dh
300
dO
1/300
dt = 1 + (h2/3002) dt = 3002 + h2(5)
()
100
200
300
1500
3
= 3002 + h2 = 200 rad/sec when h = 100.
35. (a) Let y = arctan u. Then
tany = u
2::J
2 dy = u'
see y dx
dy_~_~
dx - sec2y - 1 + U2'
(b) Let y = arcsec u. Then
secy=u
secytany:
= u'
dyu'
-~
dx - secytany - lulg=}'
~R
Note: The absolute value sign in the fonnula for the derivative of arcsec u is necessary
because the inverse secant function has a positive slope at every value in its domain.
-CONTINUED-
.
Section 2.6
Derivatives of Inverse Functions
35.-CONTINUED(c) Let Y = arccos u. Then
cosy = u
~~
-siny dy
dx = u'
- -~-
dy
-
dx
(d) Let y
u'
,Jf=U2'
sin y -
= arccotu. Then
coty
-CSC2Ydy
dx
=u
~,
= u'
dy-
u'
-
u'
u
dx - -csc2y - -I + u2"
(e) Let y
= arccsc u. Then
cscy=u
dy
,
-csc Ycotydx = u
dy=~
dx
-cscycoty
~,
u'
../u2-1
lul~'
Note: The
absolute value sign in the formula for the derivative of arccsc u is necessary
because the inverse cosecant function has a negative slope at every value in its domain.
36.
= kx +
f(x)
sin x
f'(x) = k + cos x ~ 0 for k ~ I
f'(x) = k + cos x :0;0 for k :0;-I
Therefore, f(x) = kx + sin x is strictly monotonic and has an inverse for k :0;- I or k ~ 1.
37. f(x)
g(x)
= sin x
38. f(x) = cosx
= arcsin(sin x)
(a) The range of y
-
= arcsin x is
7r/2 :0;Y :0; 7r/2.
(b) Maximum:7r/2
Minimum:-7r/2
g(x) = arccos(cosx)
(a) The range of y.= arccos x is
0:0; Y :0; 7r,
(b) Maximum:
Minimum:
7r
0
3
~3~~
-3
-2
f
-3
39. y
= 19.14-
(a)
15
0.18t- 37,10arccot(t)
(b) y' = -0.18 - 37.1O--=L
I + t2 = -0.18 + I37.10
+ t2
y'(20) = -0.087 millionlyr
,[5]0
y'(60) = -0.170 millionlyr
163