# derivative of Ln and exponential function

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

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

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

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

Thank you
• Oct 3rd 2009, 04:50 PM
HallsofIvy
Quote:

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

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

= tanh(x)