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)
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)
$\displaystyle \ln2$ 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?