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35.1 Evaluate: (a) e2 ln x ; (b) ln e2 ; (c) e(ln u) 1 ; (d) 1x .
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(a) e2 ln x = eln x = x2 (b) ln e2 = 2 (c) e(ln u) 1 =
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[by Property (VIII)] [by Theorem 35.1] [by Theorem 35.1] [by Property (IV)] [by Theorem 35.1 and by Property(II)]
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(d) By de nition, ln 1x = x(ln 1) = x(0) = 0 = ln 1. Hence, 1x = 1.
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CHAP. 35]
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35.2 Find the derivatives of:
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(a) e
(b) 32x ; (c) xex ; (d) 3x
(a) Dx e x = e x Dx x
[by the chain rule]
1 e x = e x Dx x 1/2 = e x x 1/2 = 2 2 x
(b) Dx 32x = ln 3 32x Dx 2x
[by the chain rule and Theorem 35.4]
= ln 3 32x 2 = 2 ln 3 32x (c) Dx xex = xDx ex + Dx (x)ex [by the product rule] = xex + 1 ex = ex x + 1
(d) Dx 3x 2 = 3Dx x 2 = 3 2x 2 1 = 3 2x 2 1
2 2 2
35.3 Find the following antiderivatives (ex stands for e(x ) ):
(a) By Theorem 35.4, 10x dx = (b) Let u = x 2 . Then du = 2x dx, and xex dx =
10x dx; (b)
xex dx.
1 10x + C ln 10
eu du =
1 u 1 2 e + C = ex + C 2 2
35.4 Prove Properties (I) (VIII) of ax (Section 35.2).
By de nition, ln ax = x ln a. We shall use the fact that ln u = ln v implies u = v. (I) a0 = 1 ln 1 = 0 = 0 (ln a) = ln a0 . Hence, 1 = a0 . (II) a1 = a ln a = 1 ln a = ln a1 So, a = a1 . (III) au+v = au av ln au av = ln au + ln av = u ln a + v ln a = (u + v) ln a = ln au+v So, au+v = au av . au (IV) au v = v a By (III), au v av = a(u v)+v = av . Now divide by av . 1 (V) a v = v a Let u = 0 in (IV), and use (I). (VI) ab
= a x bx ln ax bx = ln ax + ln bx = x ln a + x ln b = x ln a + ln b = x ln ab = ln ab
So, ab
= a x bx .
EXPONENTIAL FUNCTIONS
[CHAP. 35
(VII)
ax a x = x b b By (VI),
x a a x x b = ax . Now divide by bx . b = b b
(VIII)
= auv ln au
= v ln au = v u ln a = uv ln a = ln auv
So, au
= auv .
35.5 Show that
g (x)eg(x) dx = eg(x) + C.
By Theorem 35.3 and the chain rule, d g(x) e = eg(x) g (x) dx Thus, eg(x) is a particular antiderivative of g (x)eg(x) , and so eg(x) + C is the general antiderivative.
dy 35.6 Use logarithmic differentiation to nd : dx
(a) ln y = ln x x = x ln x. Now differentiate,
(a) y =
xx ;
(b) y =
x + 1e5x 2
1 dy 1 = xDx ln x + Dx (x) ln x = x + 1 ln x = 1 + ln x y dx x dy = y 1 + ln x = x x 1 + ln x dx (b) ln y = ln x+1 e5x ln 2 x = ln x + 1 + ln e5x x ln 2
1/2
= ln x+1 Differentiate,
+ 5x ln 2
1 x = ln x+1 + 5x ln 2 x 1/2 2
1 1 ln 2 1/2 ln 2 1 1 dy = +5 x +5 = y dx 2 x+1 2 2(x + 1) 2 x dy x + 1 e5x ln 2 ln 2 1 1 +5 +5 =y = dx 2 x 2 x 2 x+1 2 x+1 2 x
35.7 Prove the following facts about ex : (a) ex > 0 (b) ex is increasing (c) ex > x (d)
x +
lim ex = +
x
lim ex = 0
(a) ax > 0, by de nition (Section 35.1). (b) Dx ex = ex > 0, and a function with positive derivative is increasing. More generally, the de nition of ax , together with the fact that ln y is increasing, implies that ax is increasing if a > 1. (c) We know (Problem 34.13(a)) that ln u < u. Hence, x = ln ex < ex . (d) This is a direct consequence of part (c). (e) Let u = x; then, by Property (V), 1 ex = e u = u e As x , u + , and the denominator on the right becomes arbitrarily large [by part (d)]. Hence, the fraction becomes arbitrarily small.
CHAP. 35]
EXPONENTIAL FUNCTIONS
35.8 Sketch the graph of y = ex , and show that it is the re ection in the diagonal line y = x of the graph of y = ln x.
From Problem 35.7 we know that ex is positive and increasing, and that it approaches + on the right and approaches 0 2 on the left. Moreover, since Dx ex = ex > 0, the graph will be concave upward for all x. The graph is shown in Fig. 35-1(a). By Theorem 35.1, y = ex is equivalent to x = ln y. So, a point (a, b) is on the graph of y = ex if and only if (b, a) is on the graph of y = ln x. But the points (a, b) and (b, a) are symmetric with respect to the line y = x, by Problem 6.4. In general, the graphs of any pair of inverse functions are mirror images of each other in the line y = x.
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