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Using implicit differentiation to differentiate log_a (x)

  1. Jun 22, 2006 #1
    Could someone please make sure I'm doing this right.
    I want to find the derivative of the logarithm to the base a of x, using implicit differentiation.

    Let [tex]y = \log_{a} x[/tex]

    [tex]a^y = x[/tex]

    [tex]\frac{d}{dx} (a^y) = 1[/tex] (implicit differentiation)

    [tex]\frac{d}{dx} (e^{\ln a})^y = 1[/tex]

    [tex]\frac{d}{dx} (e^{(\ln a)y}) = 1[/tex]

    [tex]e^{(\ln a)y} \frac{d}{dx} ((\ln a)(y)) = 1[/tex]

    [tex](e^{\ln a})^y \frac{d}{dy} ((\ln a)(y)) \frac{dy}{dx} = 1[/tex] (did I use the chain rule correctly here?)

    [tex](a^y)(\ln a) \frac{dy}{dx} = 1[/tex]

    [tex]x \ln a \frac{dy}{dx} = 1[/tex]

    [tex]\frac{dy}{dx} = \frac{1}{x \ln a}[/tex]
     
    Last edited: Jun 22, 2006
  2. jcsd
  3. Jun 22, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yep, very nicely done!
     
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