Answer
$$0$$
Work Step by Step
We have: $g(x)=e^{-x+3} e^{2x-1} e^{-x+11}=e^{13}$
We differentiate both sides with respect to $x$.
$g^{\prime}(x)=\dfrac{d}{dx} [e^{13}]$
Use rule: $\displaystyle \frac{d}{dx}[k]=0$; where $k$ is a constant number.
Now, $g^{\prime}(x)=0$
