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