# Math Help - derivative of Ln and exponential function

1. ## 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

2. 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

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

3. 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

4. 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)