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Math Help - derivative of Ln and exponential function

  1. #1
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    derivative of Ln and exponential function

    y=Ln(e^x + e^-x) , so far I got y=1/e^x-1/e^-x . I Dont think its correct though. please help
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    MHF Contributor Amer's Avatar
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    Quote Originally Posted by bgonzal8 View Post
    y=Ln(e^x + e^-x) , so far I got y=1/e^x-1/e^-x . I Dont think its correct though. please help
    in general y=ln(f(x)) \Rightarrow y'=\frac{f'(x)}{f(x)} so


    y=\ln (e^x + e^{-x}) \Rightarrow = \frac{e^{x} - e^{-x}}{e^x + e^{-x}}
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  3. #3
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    Quote Originally Posted by Amer View Post
    in general y=ln(f(x)) \Rightarrow y'=\frac{f'(x)}{f(x)} so


    y=\ln (e^x + e^{-x}) \Rightarrow = \frac{e^{x} - e^{-x}}{e^x + e^{-x}}

    Thank you
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  4. #4
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    Quote Originally Posted by Amer View Post
    in general y=ln(f(x)) \Rightarrow y'=\frac{f'(x)}{f(x)} so


    y=\ln (e^x + e^{-x}) \Rightarrow = \frac{e^{x} - e^{-x}}{e^x + e^{-x}}
    = tanh(x)
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