y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

So I extract the 1/2 so it becomes

y = 1/2 ln (e^x + e^-x)

I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)

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- December 5th 2009, 06:09 PMArchduke01chain rule issue with derivative involving ln and e
y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

So I extract the 1/2 so it becomes

y = 1/2 ln (e^x + e^-x)

I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x) - December 5th 2009, 06:14 PMJhevon
- December 5th 2009, 06:16 PMVonNemo19
- December 5th 2009, 06:19 PMJhevon
- December 5th 2009, 06:22 PMArchduke01
- December 5th 2009, 06:24 PMVonNemo19
- December 5th 2009, 06:29 PMArchduke01

how did ln 1/2 become - ln 2?

- December 5th 2009, 06:30 PMVonNemo19
- December 5th 2009, 06:35 PMArchduke01
- December 5th 2009, 06:41 PMJhevon
- December 5th 2009, 06:43 PMArchduke01
- December 5th 2009, 06:44 PMVonNemo19