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Math Help - proof of derivative of a^x

  1. #1
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    proof of derivative of a^x

    how do you prove that the derivative of a^x = (a^x)(ln a)?

    i tried using logarithmic differentiation:
    y = a^x
    ln y = ln (a^x)
    ln y = x (ln a)
    d/dx (ln y) = d/dx [x (ln a)]
    (1/y) (dy/dx) = (1)(ln a) + (x)(1/a)
    dy/dx = y [(ln a) + (x/a)]
    so if i substitute
    dy/dx = a^x [(ln a) + (x/a)]

    did i do something wrong? how come it's not coming out correctly? i used logarithmic differentiation to prove that the derivative of e^x = e^x and it worked. how come it doesn't work for proving this?
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  2. #2
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    Quote Originally Posted by oblixps View Post
    y = a^x
    ln y = ln (a^x)
    ln y = x (ln a)
    d/dx (ln y) = d/dx [x (ln a)]
    (1/y) (dy/dx) = (1)(ln a) + (x)(1/a)
    dy/dx = y [(ln a) + (x/a)]
    so if i substitute
    dy/dx = a^x [(ln a) + (x/a)]
    ?
    You're differentiating with respect to x. So, \ln a is just simply a constant which you can pull out, i.e.: \frac{d}{dx} \left(x \ln a\right) = \left(\ln a\right) \left[ \frac{d}{dx} \big( x \big)\right]
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  3. #3
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    oh wow i can't believe i didn't see that...
    thanks!!
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  4. #4
    MHF Contributor matheagle's Avatar
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    i tried using logarithmic differentiation:
    y = a^x
    ln y = ln (a^x)
    ln y = x (ln a)
    d/dx (ln y) = d/dx [x (ln a)]
    (1/y) (dy/dx) = (1)(ln a) + (x)(1/a)
    dy/dx = y [(ln a) + (x/a)]
    so if i substitute
    dy/dx = a^x [(ln a) + (x/a)]

    did i do something wrong? how come it's not coming out correctly?
    YES you did.
    Your technique is correct.
    BUT a is a constant, so there's no product rule here
    {d\over dx}(x\ln a)=\ln a
    Now continue...
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