Answer
$\dfrac{-4 }{(e^x-e^{-x})^2}$
Work Step by Step
We have: $g(x)=\dfrac{e^x+e^{-x}}{e^x-e^{-x}}$
We differentiate both sides with respect to $x$.
$g^{\prime}(x)=\dfrac{d}{dx} [\dfrac{e^x+e^{-x}}{e^x-e^{-x}}]$
Use rules: $\displaystyle \frac{d}{dx}[a^{u}]=a^u \ln (a) \frac{du}{dx}$ and $\displaystyle \frac{d}{dx}[x^{n}]=nx^{n-1}$
Now, $g^{\prime}(x)=\dfrac{(e^x-e^{-x}) (e^x-e^{-x}) -(e^x+e^{-x}) (e^x+e^{-x}) }{(e^x-e^{-x})^2}$
Simplify to obtain:
$g^{\prime}(x)=\dfrac{-4 }{(e^x-e^{-x})^2}$
