derivative of Ln and exponential function

Printable View

October 3rd 2009, 01:29 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
October 3rd 2009, 01:33 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

$y=\ln (e^x + e^{-x}) \Rightarrow = \frac{e^{x} - e^{-x}}{e^x + e^{-x}}$
October 3rd 2009, 04:44 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
October 3rd 2009, 05:50 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)

All times are GMT -8. The time now is 03:57 AM.

Copyright © 2005-2015 Math Help Forum. All rights reserved.

Copyright © 2005-2014 Math Help Forum. All rights reserved.