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Calculus I

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Find the derivative of
y=ln((1+e^x)/(1-e^x))

I know that the derivative of y=ln(1+e^x) would be (e^x)/(1+e^x), but I'm not sure what to do with the 1-e^x on the bottom here. Can someone help please? Thanks!

  • Calculus I - ,

    In google type:
    wolfram alpha

    When you see lis of results click on:

    Wolfram Alpha:Computational Knowledge Engine

    When page be open in rectangle type:

    derivartive ln((1+e^x)/(1-e^x))

    and click option =

    After few secons you will see result.

    Then clic option Show steps

  • Calculus I - ,

    first of all recall that
    ln (A/B) = ln A - ln B
    so
    y = ln((1+e^x)/(1-e^x) )
    = ln (1+e^x) - ln (1-e^x)
    dy/dx = e^x/(1+e^x) - (-e^x)/(1 - e^x)

    simplify as needed

  • Calculus I - ,

    Okay, thanks guys!

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