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Math Help - Differentiation

  1. #1
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    Differentiation

    Given that y=ln[\frac{1}{2}(1+e^{-x})], show that \frac{dy}{dx}=\frac{1}{2}e^{-y}-1
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Differentiation

    You can do it again this way:
    y=\ln\left(\frac{1+e^{-x}}{2}\right)
    \frac{dy}{dx}=\frac{d\ln\left(\frac{1}{2}\right)}{  dx} +\frac{d\ln\left(1+e^{-x}\right)}{dx}=\frac{-e^{-x}}{1+e^{-x}}

    If we enter the given information into the equation we get:
    \frac{-e^{-x}}{e^{-x}+1} =\frac{1}{2}e^{-\ln\left(\frac{1+e^{-x}}{2}}\right)}-1

    Because e^{\ln(x)}=x we get:
    \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{2}\left(\frac{2}{e^{-x}+1}\right)-1
    \Leftrightarrow \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{e^{-x}+1}-\frac{e^{-x}+1}{e^{-x}+1}
     \Leftrightarrow \frac{e^{-x}}{e^{-x}+1}=\frac{-e^{-x}}{e^{-x}+1}
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  3. #3
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    Re: Differentiation

    Quote Originally Posted by Siron View Post
    You can do it again this way:
    y=\ln\left(\frac{1+e^{-x}}{2}\right)
    \frac{dy}{dx}=\frac{d\ln\left(\frac{1}{2}\right)}{  dx} +\frac{d\ln\left(1+e^{-x}\right)}{dx}=\frac{-e^{-x}}{1+e^{-x}}

    If we enter the given information into the equation we get:
    \frac{-e^{-x}}{e^{-x}+1} =\frac{1}{2}e^{-\ln\left(\frac{1+e^{-x}}{2}}\right)}-1

    Because e^{\ln(x)}=x we get:
    \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{2}\left(\frac{2}{e^{-x}+1}\right)-1
    \Leftrightarrow \frac{-e^{-x}}{e^{-x}+1}=\frac{1}{e^{-x}+1}-\frac{e^{-x}+1}{e^{-x}+1}
     \Leftrightarrow \frac{e^{-x}}{e^{-x}+1}=\frac{-e^{-x}}{e^{-x}+1}
    Sorry but arent we supposed to prove from left to right?
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