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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- Dec 5th 2009, 05:09 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) - Dec 5th 2009, 05:14 PMJhevon
- Dec 5th 2009, 05:16 PMVonNemo19
- Dec 5th 2009, 05:19 PMJhevon
- Dec 5th 2009, 05:22 PMArchduke01
- Dec 5th 2009, 05:24 PMVonNemo19
- Dec 5th 2009, 05:29 PMArchduke01
$\displaystyle = \ln (e^x + e^{-x}) - \ln 2

$

how did ln 1/2 become - ln 2?

- Dec 5th 2009, 05:30 PMVonNemo19
- Dec 5th 2009, 05:35 PMArchduke01
- Dec 5th 2009, 05:41 PMJhevon
- Dec 5th 2009, 05:43 PMArchduke01
- Dec 5th 2009, 05:44 PMVonNemo19
$\displaystyle \ln2$ is a constant. Constants dissapear when differentiating.

Regarding the chain rule...

You know that $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ right?

Well, like I said before (if you forget about that nasty sign mistake that I made), let $\displaystyle u=e^x+e^{-x}$, then by the chain rule we have

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}{\frac{du}{dx}}=\frac{1 }{u}\frac{du}{dx}=\frac{1}{e^x+e^{-x}}\cdot(e^x-e^{-x})$.

See what I'm sayin' brominator?