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Math Help - Distributing the log function

  1. #1
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    Distributing the log function

    Is distributing the natural log function legal?

    EX: (d/dx) ln(x^2((cosx)^2)e^-x^2)

    = (d/dx) ln(x^2) (d/dx) ln(cos(x)^2) (d/dx) -x^2
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  2. #2
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    Not like that. You can use the log addition rule: \ln(x^2 \cos^2(x) e^{-x^2}) = \ln(x^2) + \ln(\cos^2(x)) + \ln(e^{-x^2})

    All those can be simplified further using other log rules and can be differentiated separately but don't forget the chain rule when differentiating!"
    Last edited by e^(i*pi); April 8th 2011 at 08:21 AM. Reason: fixing a bracket
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  3. #3
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    Ah forgot about the log rules. Could someone check my work on this question.

    d/dx ln(x^2((cos(x)^2)e^-x^2)

    =(2x/x^2)+(1/((cos(x))^2))(-2cos(x)sin(x))+ e^-x^2

    =((2x)-(2cos(x)sin(x))-2x(e^-x^2))/(x^2)((cos(x))^2)

    Can't see a way I could simplify more, or even if the answer is right.
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  4. #4
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    Quote Originally Posted by ppark View Post
    Ah forgot about the log rules. Could someone check my work on this question.

    d/dx ln(x^2((cos(x)^2)e^-x^2)

    =(2x/x^2)+(1/((cos(x))^2))(-2cos(x)sin(x))+ e^-x^2
    ln(e^{-x^2})= -x^2. Its derivative is -2x.

    =((2x)-(2cos(x)sin(x))-2x(e^-x^2))/(x^2)((cos(x))^2)
    I am not sure how you have combined these. \frac{a}{b}+ \frac{c}{d} is NOT equal to \frac{a+ c}{bd}.

    Can't see a way I could simplify more, or even if the answer is right.
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