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Math Help - Find the derivative

  1. #1
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    Find the derivative

    f(x)=ln(ln(1-x))

    Find f '(x)
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  2. #2
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    Hi

    The derivative of \ln(u(x)) is \frac{u'(x)}{u(x)}

    Here u(x) = \ln(1-x)

    Can you get it from here ?
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  3. #3
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    well the thing is i already have to correct answer. its -1/(1-x)ln(1-x). but depending on the methods i have been using, i get both that and -1/(1-x)^2ln(1-x)
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  4. #4
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    i get the -1/(x-1)^2ln(x-1) when i use the chain method
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  5. #5
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    The derivative of \ln(u(x)) is \frac{u'(x)}{u(x)}

    Here u(x) = \ln(1-x) = \ln(v(x)) where v(x) = 1-x

    u'(x) = \frac{v'(x)}{v(x)} = \frac{-1}{1-x}

    Therefore f'(x) = \frac{\frac{-1}{1-x}}{\ln(1-x)} = -\frac{1}{(1-x)\ln(1-x)}
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  6. #6
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    Quote Originally Posted by emory View Post
    i get the -1/(x-1)^2ln(x-1) when i use the chain method
    Could you explain how you get this result ?
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  7. #7
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    take the derivative of the outside, which is inside over derivative of inside. (-1/(1-x))/ln(1-x). Then take the derivative of the inside which is -1/(x-1) then derivative of that which is -1
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  8. #8
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    Quote Originally Posted by emory View Post
    take the derivative of the outside, which is inside over derivative of inside. (-1/(1-x))/ln(1-x).
    This first part is actually 1/ln(1-x).

    (fog)' = f'og . g'

    f(x) = ln(x) and g(x) = ln(1-x)
    f'(x) = 1/x and g'(x) = -1/(1-x)

    f'og(x) = f'(g(x)) = 1/g(x) = 1/ln(1-x)

    (fog)'(x) = f'og(x) . g'(x) = 1/ln(1-x) . (-1)/(1-x) = -1/[(1-x)ln(1-x)]
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  9. #9
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    oh ok gotcha thanks a bunch
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