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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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

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in general $\displaystyle y=ln(f(x)) \Rightarrow y'=\frac{f'(x)}{f(x)}$ so


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

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Originally Posted by Amer

in general $\displaystyle y=ln(f(x)) \Rightarrow y'=\frac{f'(x)}{f(x)}$ so


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

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Thank you

Quote:

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Originally Posted by Amer

in general $\displaystyle y=ln(f(x)) \Rightarrow y'=\frac{f'(x)}{f(x)}$ so


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

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= tanh(x)