Logarithmic differentiation
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Logarithmic differentiation

[From: ] [author: ] [Date: 12-10-13] [Hit: ]
com/29lh6p2.jpg-Let y be a function of x,ln y = ln √[ (x + 1)^9 / (2x + 1)^5 ] ... Take the natural log of both sides.......
I need help solving these two problems. Please!
http://i48.tinypic.com/29lh6p2.jpg

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Let y be a function of x, such that:

y = √[ (x + 1)^9 / (2x + 1)^5 ]

ln y = ln √[ (x + 1)^9 / (2x + 1)^5 ] ... Take the natural log of both sides.

ln y = ln [ (x + 1)^9 / (2x + 1)^5 ]^(1/2) ... Rewrite as a power.

ln y = (1/2) ln [ (x + 1)^9 / (2x + 1)^5 ] ... Properties of logarithms.

ln y = (1/2) ln (x + 1)^9 - ln (2x + 1)^5 ... Properties of logarithms.

ln y = (1/2)9 ln (x + 1) - 5 ln (2x + 1) ... Properties of logarithms.

Now (implicitly) differentiate both sides with respect to x:

D[ ln y ]/dx = D[ (1/2)9 ln (x + 1) - 5 ln (2x + 1) ]/dx

(1/y) y' = (9/2)[1/(x + 1)] - 5/(2x + 1)

y' = y(9/2)[1/(x + 1)] - 5/(2x + 1) ... Isolate y'.

y' = √[ (x + 1)^9 / (2x + 1)^5 ] [ 9/2(x + 1) - 5/(2x + 1) ] ... Substitute.

dy/dx = √[ (x + 1)^9 / (2x + 1)^5 ] [ 9/2(x + 1) - 5/(2x + 1) ]

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Let f(x) = x^(1/3)

Let L(x) be the linearization of f(x) at x = -8

Note that linearization simply means linear approximation. Finding the linear approximation means finding the equation of the tangent line to the function, which has a formula (based on the point-slope formula for a line).

L(x) - f(a) = f '(a) (x - a)

a = -8

f(a) = (-8)^(1/3) = -2

f '(a) = (1/3)x^(-2/3) | x = -8

f '(a) = (1/3)[ 1/(-2)^2 ]

f '(a) = (1/3)(1/4)

f '(a) = 12

Substitute:

L(x) - (-2) = (12) (x - (-8))

L(x) + 2 = 12(x + 8)

L(x) = 12(x + 8) - 2

L(x) = 12x + 96 - 2

L(x) = 12x + 94

~~~

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what is the definition of "lineaization"?
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keywords: differentiation,Logarithmic,Logarithmic differentiation
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