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Math Help - Derivatives of log function!

  1. #1
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    Derivatives of log function!

    The problem is f(x)=(ln x^2)^2
    I looked at the answer key and the answer is 4(ln x)^2/x
    I do not get how the numerator can be squared.
    I only got 4(ln x).
    and how the denominator becomes x, instead of 2x?

    Thanks lots!
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  2. #2
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    f(x)=(ln x^2)^2 = (2lnx)^2 = 4 (lnx)^2
    f'(x) = 4*2(lnx)*(1/x) = 4ln(x^2)/x

    OR

    Take it as a compound function, like:
    g(u(x))=u(x)^2 where u(x)=ln x^2.

    In that case:
    f'(x) = 2u*u'(x) = 2u*(1/x^2)*(x^2)'=(2u/x^2)*2x
    =4u/x
    =4(lnx^2)/x

    Either your answer key is wrong OR You are typing it wrongly here. OR I may be wrong

    -O
    4(ln x)^2/x
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  3. #3
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    Quote Originally Posted by oswaldo View Post
    f(x)=(ln x^2)^2 = (2lnx)^2 = 4 (lnx)^2
    [snip]
    \ln x^2 \neq 2 \ln x. \ln x^2 = 2 \ln |x|.
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  4. #4
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    Thank you for the correction. The result should remain the same though.

    One more detail. Answer key: 4(ln x)^2/x is definitely wrong! Cause it is not defined for negative numbers. However f(x) is defined everywhere except zero, and differentiable everywhere except zero.

    -O
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