Unformatted text preview: Section 3.8
b f (x)
x→ g(x) (b) lim x2 lim a
lim x2 x→ lim x→ (c) lim
x→ f (x)
g(x) lim x→ x5 x dy
at
dx 1, which is 7. 2x 1 at the point (1, 2). The slope is x2
a2
x2 1
b
a x2 7. The slope of L is the reciprocal of the slope of its reflection 1 x2 8. 1
7 a2
x 1 y
gets reflected to become
x since a2 b
x
a x→ lim y
a2 x→ lim 6. The reflection of line L is the tangent line to the graph of a2 b
x
a x→ 111 Quick Review 3.8 2 a2
x2 1. Domain: [ 1, 1] 1 Range: s Section 3.8 Derivatives of Inverse
Trigonometric Functions (pp. 157–163)
Exploration 1 Finding a Derivative on an
Inverse Graph Geometrically
1. The graph is shown at the right. It appears to be a onetoone function At 1: , 22 2 2. Domain: [ 1, 1]
Range: [0, ]
At 1: 0
3. Domain: all reals
Range:
At 1: , 22 4 4. Domain: ( [ 4.7, 4.7] by [ 3.1, 3.1]
4 2. f (x) 5x
2. The fact that this function is always
positive enables us to conclude that f is everywhere
increasing, and hence onetoone. Range: 0, , 1] 2 2 At 1: 0 1 3. The graph of f is shown to the right, along with the
graph of f. The graph of f 1 is obtained from the graph of f
by reflecting it in the line y x. 5. Domain: all reals
Range: all reals
At 1: 1
6. f (x) y 3x y 8 3x x [ 4.7, 4.7] by [ 3.1, 3.1] 4. The line L is tangent to the graph of f 1 at the point (2, 1). y 8 8
3 Interchange x and y:
y
f 1 7. f (x) (x)
y [ 4.7, 4.7] by [ 3.1, 3.1] y 5. The reflection of line L is tangent to the graph of f at the
point (1, 2). 3 x x 8
3 x 8
3 3 x
y x 5
5 3 5 Interchange x and y:
y
f [ 4.7, 4.7] by [ 3.1, 3.1] 1 x3 5 (x) 3 5 x [1, )
, x
1
. It is .
y
7 112 Section 3.8 8. f (x) 8
x
8
y y
x 5. dy
ds 1 f
9. f (x) 2s 6.
2
x xy 3x 3)x 7. 2 x 3 1 4s 2s d
csc 1 (x 2
dx 3 1 y
(x 2 (x2
2x 1) x s
1 d
(5s)
1 ds (5s)2 5s s2 1 1 dy
dx x2 s 25s 2 1)
d
(x 2
1 dx 1)2 1)
2 4 2x 2 (x Note that the condition x x 2 x2 1) 2 0 is required in the last step. 2
3 x 8. x
arctan
3 y
x
,
3 tan y 4s d
sec 1 5s
ds 2 2 (x) 10. f (x) 1 1) (2) 2 1 2
y3 y 1) dy
ds Interchange x and y: 1 (2s 2 x f 1 d
(2s
ds
1 2 1 3x y (y 2s 8
x
8
x (x) 1) 1 Interchange x and y:
y d
sec 1 (2s
ds 3 tan y, 2 y 2 1 d
x
csc 1
dx
2 dx
1 dx 2 x2
2 x
2 2 y 2 dy
dx x2 x 4 2 Interchange x and y:
y
f 3 tan x,
1 (x) x 2 3 tan x, 2 9. 2 x dy
dt 1 1
d
sec 1
t
dt 1
t 2
1 Section 3.8 Exercises
1. dy
dx 1 d
cos 1 (x 2)
dx
1
x4 1 d2
(x )
dx
(x ) x4 1 1 1
t2 1 dy
dx 10. 1 1
d
cos 1
x
dx d
1 2 dx
x 1
1
1 1
x 2 1
x2 0 is required in the last step. dy
dt 1 d
3
sin 1 2
dt
t 1
x 1
1 1
x2 x t2 1 11. 1 dy
dt d
cot 1
dt 3
t2 6 6
t3 9
t4 d
3 2 dt
t2 1 2. 1 Note that the condition t 22 1
2x (2x) 12
t 1
t 1
t d1
dt t 2 t t4
1 t
1 ( 9
d
t)2 dt t 1 3. 4. dy
dt d
sin 1
dt dy
dt d
sin 1 (1
dt
1
2t t2 1 2t ( 2 t)2 1 1 t)
1 (1 d
(
dt 2 2 t) d
(1
2 dt
t) 1 t) 2 t (t 1) 2t 2 12. dy
dt d
cot 1
dt 1 1
t t 1 1
1 ( 1
12t 1 ) d
dt
1)
2 (t
1
2t t 1 t 1 1 113 Section 3.8 13. dy
ds d
(s
ds 1 (s)
2 d
(cos 1s)
ds s2) 1 ( 2s) 2 1 1 s (1 s) 1 14. dy
ds d
ds [sin 1 1 (2) 4x 2 1 (2x)]2 1 dy
dx 4x 2 sec 2 x, the slope at The tangent line is given by y
d
sec 1 s
ds 1 y 1 (2s) s2 (2x) 1 2 1 s2 1 1 sin 2 (2x)] s 19. (a) Since s2 2 s 2 1 dx 2 s2 2s 2d (2x)] 2 1
2 1 [sin [sin s 2 d
[sin 1(2x)] 1
dx 1 s2 1 2 dy
dx 1 s2)(1) 1 s s2
1 ( 18. 1 (b) Since s2 s 2x 1 2x 1
x2 1 , the slope at 1, s2 s
dy
15.
dx 1 d
(tan 1
dx x 1
1
1 1 x2 2 2 1) x2 x x 2 1 1 x2 x y 1 f x Note that the condition x dy
dx 1
d
cot 1
x
dx 5x 4 1 1) 4 1
.
2 , or 6x 2 1. Thus f (1) 3 and f (x) includes the point (1, 3) and (x) will include (3, 1) and the slope will be
1
. (We have
12 (x) is defined and differentiable at 5x 4 6x 2 1, which is never zero.) 21. (a) Note that f (x)
sin x 3, which is always between
2 and 4. Thus f is differentiable at every point on the
interval( , ) and f (x) is never zero on this interval,
so f has a differentiable inverse by Theorem 3. d
(tan 1 x)
dx
d1
dx x ) (3) 3. This is true by Theorem 3, because f (x) 1 is required in the last step. 1 1 and ( f 1 assumed that f 1 0 16. 1
1 + 12 . 1
. Thus, f 1(3)
12 x2 x is the slope of the graph is 12 at this point, the graph of 1 x x2 4 1 1 1 4 (b) Since the graph of y 1 1) (2x) x2 1
2 20. (a) Note that f (x)
f (1) 12. d
(csc 1 x)
dx d
(
1)2 dx x2 ( 1
x
2 y 1 2. 4 1, or 4 1
(x
2 The tangent line is given by y
ss , 1 is sec 2 1. 2 dy
dx 4 1 1
x2 1 ( 1
1
x2 1 ) 1
x2 1
x2 (b) f (0) cos 0 3(0)
f (0)
sin 0 3 x2 1 1;
3 (c) Since the graph of y 1 f (x) includes the point (0, 1) 1
x2 1 and the slope of the graph is 3 at this point, the graph
of y 1
1 0, x 1 x2 f 1 (x) will include (1, 0) and the slope will be 1
. Thus, f 1(1)
3 22. 0 The condition x 0 is required because the original
function was undefined when x 0.
17. dy
dx d
(x sin 1 x)
dx
1 (x) 1 sin 1 x x2 d
(
dx (sin 1
1 x 2) [ 2 , 2 ] by [ 4, 4] 1 x)(1)
2 1 x2 ( 2x) (a) All reals
(b) , 22 0 and (f 1 ) (1) 1
.
3 114 Section 3.8 22. continued (b) y (c) At the points x
(d) k , where k is an odd integer.
2 y 2 (c) None, since
30. (a) y 0. x2 x 1 0 (b) y 1 d
sec 1 x
dx 0 2 (c) None, since 1 d
csc 1 x
dx 0. x2 x 1 2π 1 x is
1. (b) None, since sin –2π 31. (a) None, since sin
undefined for x x 1 x is undefined for x (c) None, since –2 1 x is undefined for x (b) None, since cos 1 x is undefined for x 1 d
sin x
1 sin2 x dx
cos x which is (c) None, since 23. (a) v(t)
(b) a(t)
(c)
24. 1 which is always positive. 1
1 So cos t2
2t
which is always negative.
(1 t 2)2 x, x sin
1 sin (b) β 1 x x 2 . x α 2 d
cos 1 (x)
dx d
dx sin 2 1 x 1 tan
1 So tan x, x cot
cot (c) x2 1
d
cot 1 x
dx 1 d
sin 1 (x)
dx
1 0 25. 0. x cos
1 x2 1. α cos x is positive or negative.
dx
dt
dv
dt 1 1. β 1 1 depending on whether 1 d
cos 1 x
dx 33. (a) sin2 x 1 0. x2 1 32. (a) None, since cos d
sin 1 (sin x)
dx (e) f (x) 1 d
sin 1 x
dx 1. 1 1 x x 2 . β x
α d
tan 1 (x)
dx 2
d
0
tan 1 (x)
dx
1
1 x2 1
1 sec
So sec 1 x x, csc
csc 1 x 1 x
2 . 34.
d
26.
csc 1 (x)
dx d
dx 2 x (x) x 2 A
B 1 C 2 (b) y 2 (c) None, since
28. (a) y 1 d
sec 1 (x)
dx
1 0 27. (a) y sec d
tan 1 x
dx 1 0. x2 1 0 (b) y
(c) None, since
29. (a) y 2 d
cot 1 x
dx 1
1 x2 0. The “straight angle” with the arrows in it is the sum of the
three angles A, B, and C.
A is equal to tan 1 3 since the opposite side is 3 times as
long as the adjacent side.
B is equal to tan 1 2 since the side opposite it is 2 units
and the adjacent side is one unit.
C is equal to tan 1 1 since both the opposite and adjacent
sides are one unit long.
But the sum of these three angles is the “straight angle,”
which has measure radians. Section 3.9
35. s 6. Fold 1 log4 x 15 15 log4 x 12 12 log4 x log4 x ln Fold 2
s
2 Fold 3 s
s
s tan 1, so tan 1 tan 1 ln 19
ln 3 5t ln 5 1 1 tan 2 tan 1 t ln 5 s Section 3.9 Derivatives of Exponential
and Logarithmic Functions (pp. 163–171)
Exploration 1 Leaving Milk on the Counter 55 and solve;
72 30(0.98)t 4. ln (x 2
ln (x dy
dx d
(2e x)
dx 2e x dy
dx d 2x
(e )
dx e 2x dy
dx d
ex
dx dy
dx d
e 5x
dx e 5. dy
dx d 2x/3
e
dx e 2x/3 6. dy
dx d
e x/4
dx e dy
dx d
(xe 2)
dx dx
(e )
dx dy
dx d 2x
(x e )
dx d
(xe x)
dx (e x)(2x) tan x x 2e x
2 ln (x 2) ln x
x 4
2 ln (x 2)(x 2)
x2 2) 5. log2 (8x 5) log2 (23)x 5 log2 23x 15 2.71 Section 3.9 Exercises e x ln 7 4) ln 3
ln 2 ln 3 x d
(2x)
dx xd e dx 2e 2x ( x) 5x d dx x e ( 5x) d 2x
dx 3 5e 5x 2 2x/3
e
3 , ln 8
ln 5 ) ln 3 (x 2)(e x) ln (0.98) Quick Review 3.9 3. ln (e ln 2) 7. 17
30 0.343 degrees/minute. tan x x ln 2 4. 30 ln (0.98) (0.98)t. At t e ln 7 1) ln 3 x(ln 3 28.114 ln (0.98) ln 2. 7x ln 2x 3. 17
30 The milk reaches a temperature of 55 F after about
28 minutes. x 1 1.50 2x ln 3x
(x 1 8. ln 1. log5 8 ln (ln 5)
ln 5 2. 17
ln
30 t ln (0.98) dy
5.
dt
dy
dt ln 18 1. 55 17
30 (0.98)t t ln (ln 5) 3x 10. 2. The temperature of the room is 72 F, the limit to which y
tends as t increases.
3. The milk is warming up the fastest at t 0. The second
derivative y
30(ln(0.98))2(0.98)t is negative, so y (the
rate at which the milk is warming) is maximized at the
lowest value of t. ln 18 t 1. The temperature of the refrigerator is 42 F, the temperature
of the milk at time t 0. 4. We set y 18
ln 5
18
ln
ln 5 ln 5t 3. 2.68 18 5t
1 ln (12x 2) ln 19 From Exercise 34, we have
tan ln 3x ln 19 x
9. ln x 3 19 x ln 3 2. 0 ln (4x 4) 1 and 2, so s
2 (x 3)(12x 2)
3x ln 3x If s is the length of a side of the square, then
tan ln (12x 2) 3x 8. 5
,x
4 ln 3x 7. 3 ln x
s 15
12 3x 15 9.
10. x/4 d xe x 1 x/4
e
4 x
4 dx e2 ex [(x)(e x) (e x)(1)] ex dy
dx d
ex
dx e dy
dx d (x 2)
e
dx e (x ) xd dx
2 ( x) d2
(x )
dx e x 2x 2xe (x )
2 115 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.
 Spring '08
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