Answer
\[(a)\: y'=2xf(x)+x^2f'(x)\]
\[(b)\: y'=\frac{x^2f'(x)-2xf(x)}{x^4}\]
\[(c) \: y'=\frac{2xf(x)-x^2f'(x)}{[f(x)]^2}\]
\[(d)\: y'=\frac{2x^2f'(x)+xf(x)-1}{2x^{\frac{3}{2}}}\]
Work Step by Step
$(a)\;y=x^2f(x)$
Differentiate with respect to $x$ using product rule
$y'=(x^2)'f(x)+x^2f'(x)$
$\Rightarrow y'=2xf(x)+x^2f'(x)$
$(b)\;y=\frac{f(x)}{x^2}$
Differentiate with respect to $x$ using quotient rule
$y'=\frac{f'(x)(x^2)-f(x)(x^2)'}{(x^2)^2}$
$\Rightarrow y'=\frac{x^2f'(x)-2xf(x)}{x^4}$
$(c)\;y=\frac{x^2}{f(x)}$
Differentiate with respect to $x$ using quotient rule
$y'=\frac{(x^2)'f(x)-x^2f'(x)}{[f(x)]^2}$
$\Rightarrow y'=\frac{2xf(x)-x^2f'(x)}{[f(x)]^2}$
(d) $\;y=\frac{1+xf(x)}{\sqrt x}$
$y=\frac{1}{\sqrt x}+\frac{x}{\sqrt x}f(x)$
$y=x^{\frac{-1}{2}}+\sqrt{x}f(x)$
Differentiating with respect to $x$ using product rule
$y'=\frac{-1}{2}x^{\frac{-3}{2}}+\left[(\sqrt x)'f(x)+\sqrt x f'(x)\right]$
$y'=\frac{-1}{2x^{\frac{3}{2}}}+\frac{f(x)}{2\sqrt x}+\sqrt x f'(x)$
$\Rightarrow y'=\frac{2x^2f'(x)+xf(x)-1}{2x^{\frac{3}{2}}}$
