Answer
$(x^{-0.5}+4)\cdot(x-x^{-1})+(2x^{0.5}+4x-5)\cdot(1+x^{-2})$
Work Step by Step
$f(x)=$2$x^{0.5}+4x-5, \ \ \quad g(x)=x-x^{-1},\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=2(0.5x^{-0.5})+4=x^{-0.5}+4$
$g^{\prime}(x)=1-(-x^{-2})=1+x^{-2}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=(x^{-0.5}+4)\cdot(x-x^{-1})+(2x^{0.5}+4x-5)\cdot(1+x^{-2})$
... we do not need to expand the answer...