Answer
$s^{\prime}(x)=\displaystyle \frac{e^{4x-1}(4x^{3}-4-3x^{2})}{(x^{3}-1)^{2}}$
Work Step by Step
$s(x) $ is a quotient$ \displaystyle \frac{u(x)}{v(x)}$ where
$ u(x)=e^{4x-1},\qquad$
$u^{\prime}(x)= \displaystyle \frac{d}{dx}[e^{w}]=e^{u}\cdot\frac{dw}{dx}$,
with $w=4x-1\displaystyle \quad\frac{dw}{dx}=4$
$u^{\prime}(x)=\displaystyle \frac{d}{dx}[e^{4x-1}]=e^{4x-1}\cdot(4)=4e^{4x-1}$
$v(x)=x^{3}-1$
$v^{\prime}(x)=3x^{2}$
So
$s^{\prime}(x)=\displaystyle \frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{[v(x)]^{2}}$
$=\displaystyle \frac{4e^{4x-1}(x^{3}-1)-e^{4x-1}(3x^{2})}{[x^{3}-1]^{2}}$
$s^{\prime}(x)=\displaystyle \frac{e^{4x-1}(4x^{3}-4-3x^{2})}{(x^{3}-1)^{2}}$
