Answer
$h^{\prime}(x)=-[(2x-1)(x-1)]^{-4/3}\cdot(4x-3)$
Work Step by Step
h(x) is a composite function multiplied with a constant, $3$.
$u(v(x))=[(2x-1)(x-1)]^{-1/3}$
The last calculation in this composition would be a power to ($-1/3$), so
$u(x)=x^{-1/3},\displaystyle \qquad \frac{du}{dx}=-\frac{1}{3}x^{-4/3}$
$v(x)=(2x-1)(x-1)$ , which is a product,
... $(fg)^{\prime}=f^{\prime}g+fg^{\prime}$
$\displaystyle \frac{dv}{dx}=(2)(x-1)+(2x-1)(1)$
$=2x-2+2x-1$
$=4x-3$
Now,
$h(x)=3u(v(x)), $
$h^{\prime}(x)=3\displaystyle \frac{du}{dv}\frac{dv}{dx}=$
$=3(-\displaystyle \frac{1}{3}v^{-4/3})\cdot(4x-3)$
$h^{\prime}(x)=-[(2x-1)(x-1)]^{-4/3}\cdot(4x-3)$
