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Math Help - Help needed finding the second derivative.

  1. #1
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    Help needed finding the second derivative.

    Find the second derivative
     <br />
f(x) = ((x-2)^3)/(x^2)<br />

    Attempt:

    f'(x)/g(x) = (f'*g-g'*f)/(g^2)
     <br />
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)<br />

    f(x)/g(x) = (f'*g-g'*f)/(g^2)
    f(x)*d(x) = f'd + d'f

    f''(x) = ((6(x-2)*x^2+2x*(3*(x-2)^2) -(2*((x-2)^3)+2x*(3(x-2)^2))*(x^4)-(4x^3)*((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^8)
    *Sorry the math thing can't fit the size of this

    Simplifying doesnt give me the right answer. The correct answer is 24(x-2)/(x^4).
    How do i do this?
    Last edited by Fallen186; January 7th 2009 at 06:02 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Fallen186 View Post
    Find the second derivative
     <br />
f(x) = ((x-2)^3)/(x^2)<br />

    Attempt:

    f'(x)/g(x) = (f'*g-g'*f)/(g^2)
     <br />
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)<br />

    I do it again

    f(x)/g(x) = (f'*g-g'*f)/(g^2)
    f(x)*d(x) = f'd + d'f

    f''(x) = ((6(x-2)*x^2+2x*(3*(x-2)^2) -(2*((x-2)^3)+2x*(3(x-2)^2))*(x^4)-(4x^3)*((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^8)
    *Sorry the math thing can't fit the size of this

    Simplifying doesnt give me the right answer. The correct answer is 24(x-2)/(x^4).
    How do i do this?
    This looksl like a prime candidate for logarthimic differenatiation.

    Since f(x)=\frac{(x-2)^3}{x^2}\implies \ln f(x)=\ln\left(\frac{(x-2)^3}{x^2}\right)=3\ln(x-2)-2\ln(x)

    Differentiating both sides gives

    \frac{f'(x)}{f(x)}=\frac{3}{x-2}-\frac{2}{x}\implies f'(x)=f(x)\left\{\frac{3}{x-2}-\frac{2}{x}\right\}

    But remember that f(x)=\frac{(x-2)^3}{x^2}

    So f'(x)=\frac{(x-2)^3}{x^2}\left\{\frac{3}{x-2}-\frac{2}{x}\right\}
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Fallen186 View Post
    Find the second derivative
     <br />
f(x) = ((x-2)^3)/(x^2)<br />

    Attempt:

    f'(x)/g(x) = (f'*g-g'*f)/(g^2)
     <br />
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)<br />
    That is correct, but you should simplify it before differentiating it again.

    f'(x) = \frac{3(x-2)^2x^2-2x(x-2)^3}{x^4} = \ldots = \frac{(x-2)^2(x+4)}{x^3}

    Now differentiate, either by using the quotient rule again, or (preferably) by logarithmic differentiation as in Mathstud28's suggestion.
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