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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- October 3rd 2009, 12:29 PMbgonzal8derivative 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, 12:33 PMAmer
- October 3rd 2009, 03:44 PMbgonzal8
- October 3rd 2009, 04:50 PMHallsofIvy