Answer
$h^{\prime}(x)=6.2(3.1x-2)+6.2(3.1x-2)^{-3}$
Work Step by Step
Let $h_{1}(x)=(3.1x-2)^{2},\quad h_{2}(x)=(3.1x-2)^{-2}.$
Then $h(x)=h_{1}(x)+h_{2}(x)$
and $h^{\prime}(x)=h_{1}^{\prime}(x)-h_{2}^{\prime}(x).$
Apply the Generalized Power Rule on $h_{1}$ and $h_{2}$
$\displaystyle \frac{d}{dx}[u^{n}]=nu^{n-1}\frac{du}{dx}$
$u(x)=3.1x-2, \displaystyle \quad \frac{du}{dx}=3.1.$
$h_{1}^{\prime}(x)= \displaystyle \frac{d}{dx}[u^{2}]$
$=2u^{2-1}\cdot(3.1)=6.2(3.1x-2)$
$h_{2}^{\prime}(x)= \displaystyle \frac{d}{dx}[u^{-2}]$
$=-2u^{-2-1}\cdot(3.1)=-6.2(3.1x-2)^{-3}$
Now, $h^{\prime}(x)=h_{1}^{\prime}(x)-h_{2}^{\prime}(x)$,
$h^{\prime}(x)=6.2(3.1x-2)-[-6.2(3.1x-2)^{-3}]$
$h^{\prime}(x)=6.2(3.1x-2)+6.2(3.1x-2)^{-3}$
