barcode in ssrs 2008 THE NATURAL LOGARITHM in .NET

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THE NATURAL LOGARITHM
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[CHAP. 34
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Fig. 34-2
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34.2 Prove: ln uv = ln u + ln v.
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In ln v =
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make the change of variable w = ut (u xed). Then dw = u dt, and the limits of integration, t = 1 and t = v, go over into w = u and w = uv, respectively, ln v = Then, by Theorem 30.4, ln u + ln v =
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u 1 1 uv u 1 u
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dw =
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uv 1 u
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dw =
uv 1 u
dt +
uv 1 u
dt =
uv 1 1
dt = ln uv
34.3 Prove: ln x r = r ln x for rational r.
By the chain rule, 1 1 1 Dx (ln x r ) = r Dx (x r ) = r (rx r 1 ) = r = rDx (ln x) = Dx (r ln x) x x x Then, by Corollary 29.2, ln x r = r ln x + C, for some constant C. Substituting x = 1, we nd that C = 0, and the proof is complete.
34.4 Evaluate: (a) Dx (ln(x 3 2x))
Use the chain rule. 1 1 3x 2 2 Dx (x 3 2x) = 3 (3x 2 2) = 3 (a) Dx(ln(x 3 2x)) = 3 x 2x x 2x x 2x cos x 1 1 Dx (sin x) = (cos x) = = cot x sin x sin x sin x 1 sin (ln x) (c) Dx (cos (ln x)) = sin (ln x)) Dx (ln x) = (sin (ln x)) = x x (b) Dx (ln (sin x)) =
(b) Dx (ln (sin x))
(c) Dx (cos (ln x))
CHAP. 34]
THE NATURAL LOGARITHM
34.5 Quick Formula II: Proof:
f (x) dx = ln f (x) + C. f (x) 1 f (x) f (x) [by the chain rule]
Dx (ln f (x) ) =
34.6 Find the following antiderivatives: (a) tan x dx (b) cot x dx (c) 1 dx 3x 1 (d) x dx x2 5
(a) Since Dx (cos x) = sin x, tan x dx = sin x dx = cos x = ln |cos x| + C = ln |cos x| 1 + C = ln |sec x| + C (b) Since Dx (sin x) = cos x, cot x dx = (c) (d) 1 1 dx = 3x 1 3 x 1 dx = 2 x2 5 cos x dx = ln |sin x| + C sin x [by Quick Formula II] sin x dx cos x [by Quick Formula II] [by Property 7] [since sec x = (cos x) 1 ]
3 1 dx = ln |3x 1| + C 3x 1 3 2x 1 dx = ln x 2 5 + C 2 x2 5
[by Quick Formula II] [by Quick Formula II]
34.7 Find
sec x dx.
The solution depends on a clever trick, sec x dx = sec x + tan x dx = sec x + tan x = ln |sec x + tan x| + C (sec x) sec2 x + sec x tan x dx sec x + tan x
Here we have applied Quick Formula II, using the fact that Dx (sec x + tan x) = sec x tan x + sec2 x.
34.8 (Logarithmic Differentiation)
Find the derivative of y= x 2 8x + 5 (2x 1)3
Instead of using the product and quotient rule for differentiation, it is easier to nd the logarithm of the absolute values,1 ln |y| = ln x 2 8x + 5 (2x 1)3 [by Property 6] [by Property 5] [by Property 8]
= ln (x 2 8x + 5) ln |2x 1|3 = ln (x 2 ) + ln (8x + 5)1/2 ln |2x 1|3 = 2 ln x + 1 ln (8x + 5) 3 ln |2x 1| 2
1 We take the absolute values to make sure that the logarithm is de ned. In practice, we shall omit the absolute values when it is clear that the functions are nonnegative.
THE NATURAL LOGARITHM
[CHAP. 34
and then to differentiate, 1 1 Dx y = 2 y x Therefore, Dx y = y 4 6 2 + x 8x + 5 2x 1 = + 2 4 6 1 1 1 8 3 2 = + 2 8x + 5 2x 1 x 8x + 5 2x 1 4 6 x 2 8x + 5 2 + 8x + 5 2x 1 (2x 1)3 x
This procedure of rst taking the logarithm and then differentiating is called logarithmic differentiation.
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