derivative of Ln and exponential function

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Oct 3rd 2009, 12:29 PM
bgonzal8

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
Oct 3rd 2009, 12:33 PM
Amer

Quote:

Originally Posted by bgonzal8

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

$y=\ln (e^x + e^{-x}) \Rightarrow = \frac{e^{x} - e^{-x}}{e^x + e^{-x}}$
Oct 3rd 2009, 03:44 PM
bgonzal8

Quote:

Originally Posted by Amer

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
Oct 3rd 2009, 04:50 PM
HallsofIvy

Quote:

Originally Posted by Amer

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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