Answer
$(0.7x^{-0.3}-4)\cdot(x^{-1}+x^{-2})-(x^{0.7}-4x-5)\cdot(x^{-2}+2x^{-3})$
Work Step by Step
$f(x)=x^{0.7}-4x-5, \ \ \quad g(x)=x^{-1}+x^{-2},\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=0.7x^{-0.3}-4$
$g^{\prime}(x)=-x^{-2}-2x^{-3}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=(0.7x^{-0.3}-4)\cdot(x^{-1}+x^{-2})+(x^{0.7}-4x-5)\cdot(-x^{-2}-2x^{-3})$
$=(0.7x^{-0.3}-4)\cdot(x^{-1}+x^{-2})-(x^{0.7}-4x-5)\cdot(x^{-2}+2x^{-3})$
... we do not need to expand the answer...