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Derivative of Ln

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data

    f(x) = e^(x^3)

    2. Relevant equations

    3. The attempt at a solution

    lnf(x) = ln(e) + 3ln(x)
    1 / f(x) f'(x) = 1 + 3x^2 / x^3
    f'(x) = ( e^(x^3) * 3X^2 ) / x^3

    However, my book says the answer is 3x^2*e^(x^3).

    Any help would be greatly appreciated! I've been stuck on this question for ages!
  2. jcsd
  3. Dec 9, 2007 #2


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    Science Advisor

    You don't need to take the logarithm, just differentiate it using the chain rule. Let y=x^3, then [tex]\frac{df}{dx}=\frac{d}{dy}e^y\cdot\frac{dy}{dx}[/tex]
  4. Dec 9, 2007 #3


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    You are going about the whole problem the long way. Just use the chain rule. But if you do want to go the long way, ln(e^(x^3))=x^3*ln(e)=x^3. So ln(f(x))=x^3.
  5. Dec 9, 2007 #4
    Thanks a lot!!
  6. Dec 9, 2007 #5
    I need help with another question.

  7. Dec 10, 2007 #6
    I assume you mean:


    Use the same method, first the chain rule for the square root, giving you:

    [tex]\frac{dy}{dx}=\frac{1}{2}\left(x^x\right)^{-\frac{1}{2}}\cdot \frac{d}{dx}\left(x^x\right)[/tex]

    After this consider the general formula:

    [tex]\frac{d}{dx}\left[f(x)\right]^{g(x)}=\left[f(x)\right]^{g(x)}\cdot ln[f(x)]\cdot \frac{d}{dx}[g(x)]+g(x) \cdot \left[f(x)\right]^{g(x)-1}\cdot \frac{d}{dx}[f(x)][/tex]

    In this case f(x)=x and g(x)=x

    Do not try to remember the formula, instead try to remember how to derive it. You can apply it then to any case you need. It is proven using the chain rule as well, after taking the logarithm.
  8. Dec 10, 2007 #7


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    To differentate [itex]u= x^x[/itex], you might want to use "logarithmic differentiation":
    ln(y)= x ln(x).
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