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The Natural Logarithm solved questions
The Natural Logarithm solved questions
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CHAPTER 23The Natural Logarithm
23.1 State the definition of In x, and show that D^(ln jc) =
_ !_
In* = dt for x > 0 Hence (Problem 20.42), D,(ln*)=D A x'
23.2 Show that dx = In 1*1 + C for x * 0
Case 1. x> 0. Then D,(ln|*| + C) = D,(ln*) = Case 2. x<0. Then D,(ln \x\ + C) =
D,[ln (-*)] = • D,(-x) = - -(-!) =
In Problems 23.3-23.9, find the derivative of the given function.
23.3 In (4* - 1).
By the chain rule, D,[ln(4*-l)] = •D,(4*-l) =
23.4 (In x)3.
By the chain rule, DA.[(ln x)3] = 3(ln x)- • Dx(\n x) = 3(ln x)2 •
23.5 VhT7.
By the chain rule, D^VHTx) = D,[(ln x)"2] = ^(Inx)'1'2 • Df(\nx)= ^Inx)'1'2 • - =
23.6 In (In*).
By the chain rule, DJln (In x)] = •D,(lnx) =
23.7 x2 In x.
By the product rule, Dx(x2 In x) = x2 • Dx(\n x) + In x • Dv(x2) = x2 • + In x • (2.v) = x + 2x In .v = .v(l +
2 In*).
23.8 In
By the chain rule,
In
23.9 ln|5*-2|.
By the chain rule, and Problem 23.2, D,(ln \5x - 2|) = •D,(5*-2) =
In Problems 23.10-23.19, find the indicated antiderivative.
23.10 dx.
dx = dx=$ \n\x\ + C.
185
, 186 CHAPTER 23
23.11 dx.
Let u = 7x-2, du=l dx. Then dx = du= % ln| M l + C = \\n\lx-2\ + C.
23.12 dx.
Let u = x4 — 1, du=4x*dx. Then dx = du = | In \u\ + C = | In \x4 - 1| + C.
23.13 J cot x dx.
Let u = sin x, du = cos x dx. Then J cot x dx = dx = dw = ln|w| + C = ln|sinA-| + C
23.14
Let w = In x, du = dx. Then dx = rf« = ln|w| + C = ln(|lnjc|)+C. [Compare Prob-
lem 23.6.]
23.15
Let w = 1 — sin 2x, du = —2 cos 2x dx Then
dx = du = -{ \n\u\ + C= -| I n | l - s i n 2 x | + C
23.16
rfx = 3 J A-2 rfx + 2 dx - 3 J x^ 3 dx = jc3 + 2 In |*| + l^r" 2 + C.
23.17
Let u - tan x, du = sec2 x dx. Then
dx = du = In |M| + C = In |tan x\ + C
23.18 dx.
Let u = \-Vx, du = -(l/2Vx)dx. Then
dx = -2 du = -2 In |«| + C = -2 In |1 - Vx\ + C
23.19 dx.
dx = (In*) dx={(lnx)2 + C.
23.20 Show that dx = \n\g(x)\ + C.
Let M = g(x), du = g'(x) dx. Then
dx = du = ln\u\ + C = \n\g(x)\ + C
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