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Math Help - TEST derivative using y= lnx

  1. #1
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    TEST derivative using y= lnx

    Hi please help deriving this question:see the image for the equation. deriv.bmp
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    Re: TEST derivative using y= lnx

    i do know that one of the rules involving natural logarithms is: y=lnx which derived is y'= 1/x . But I'mstill not sure. This is my teacher's solution:y'= -5-3x^2 / (-5-5x-x^3)
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    Re: TEST derivative using y= lnx

    y =ln(-5-5x-x^3)

    derivative of ln(x) is y' = 1/(x)*(x)'

    y'= (ln(-5-5x-x^3))'*(-5-5x-x^3)' so, you have 1/(-5-5x-x^3)*(5x-3x^2) =(5x-3x^2)/(-5-5x-x^3)

    dokrbb
    Last edited by dokrbb; April 17th 2013 at 12:03 PM. Reason: added the rule
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    Re: TEST derivative using y= lnx

    If y is the log of a function of x

    y=ln(g(x))

    Substitute for the function in the log and use the chain rule.
    Let u= g(x), then
    y=ln(u)

    \frac{du}{dx}=g'(x)

    \frac{dy}{du}=\frac{1}{u}=\frac{1}{g(x)}

    \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=\frac{1}{g(x)} g'(x)

    So for your problem f'(x)=\frac{-3x^2-5}{-5-5x-x^3}
    Last edited by Shakarri; April 17th 2013 at 01:33 PM.
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    Re: TEST derivative using y= lnx

    thank you
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