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