I need help solving these two problems. Please!
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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) ]
~~~
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
~~~
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) ]
~~~
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"?