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Thread: Natural log derivatives? How?

  1. #1
    Newbie
    Joined
    Jan 2009
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    Question Natural log derivatives? How?

    This is the equation I have:

    Let $\displaystyle f(x) = -4 \ln(5 x) $
    Find $\displaystyle f'(x) = $

    My Work Sheet says the answer is:
    $\displaystyle
    -4/x
    $

    My question is, shouldn't it be somthing like:
    $\displaystyle
    -20/5x
    $
    ...when you derive $\displaystyle ln(5 x)$ shouldn't it become $\displaystyle 1/5x$ and from there don't you have to use the chain rule on $\displaystyle 5x$ and get 5?

    It seems to me you would only get $\displaystyle -4/x $ if $\displaystyle f(x) = -4 \ln(x)$

    What is it I don't understand?! Thank You
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  2. #2
    Junior Member
    Joined
    Sep 2008
    Posts
    28
    the thing with natural log derivatives is that (d/dx) ln(x) is actually (1/x)*x' (so 1 divided by the inside, then times the derivative of the inside)

    -4ln(5x) becomes -4*(1/5x)*5 = -4/x
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  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
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    Wow -20/5x = -4/x...

    Im dumb... haha I guess that's what doing hours upon hours of math does... just scrambles the brain.
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