Answer
(a)$f(x)=g(x).h(x)=(-3x^2+4x-5).(0.5x^{-2}-2x^{0.5})$
(b)$f^{'}(x)=(-3x^2+4x-5)[-x^{-3}-x^{-0.5}]+(0.5x^{-2}-2x^{0.5})[(-6x+4 ] $
Work Step by Step
$g(x)=-3x^2+4x-5;h(x)=0.5x^{-2}-2x^{0.5}$
(a)
Let $f(x) =g(x)h(x)$
$f(x)=g(x).h(x)=(-3x^2+4x-5).(0.5x^{-2}-2x^{0.5})$
(b)
Taking derivative with respect to $x$ of $f(x)$
$f^{'}(x)= \frac{d[ (-3x^2+4x-5).(0.5x^{-2}-2x^{0.5})]}{dx}$
Using the product rule of differentiation
$f^{'}(x)=(-3x^2+4x-5)\frac{d(0.5x^{-2}-2x^{0.5}) }{dx}+(0.5x^{-2}-2x^{0.5})\frac{d(-3x^2+4x-5)}{dx}$
$f^{'}(x)=(-3x^2+4x-5)[ \frac{d(0.5x^{-2})}{dx}-\frac{d(2x^{0.5})}{dx}]+(0.5x^{-2}-2x^{0.5})[\frac{d(-3x^2)}{dx}+\frac{d(4x)}{dx}-\frac{d(5)}{dx} ] $
$f^{'}(x)=(-3x^2+4x-5)[0.5 \frac{d(x^{-2})}{dx}-2\frac{d(x^{0.5})}{dx}]+(0.5x^{-2}-2x^{0.5})[(-3)\frac{d(x^2)}{dx}+4\frac{d(x)}{dx}-0 ] $
$f^{'}(x)=(-3x^2+4x-5)[0.5 (-2x^{-3})-2(0.5)x^{0.5-1}]+(0.5x^{-2}-2x^{0.5})[(-3)(2x)+4 ] $
$f^{'}(x)=(-3x^2+4x-5)[-x^{-3}-x^{-0.5}]+(0.5x^{-2}-2x^{0.5})[(-6x+4 ] $
