Answer
\[(a) y'=g(x)+xg'(x)\]
\[(b) y'=\frac{g(x)-xg'(x)}{[g(x)]^2}\]
\[(c) y'=\frac{xg'(x)-g(x)}{x^2}\]
Work Step by Step
$(a)\; y=xg(x)$
Differentiating with respect to $x$ by product rule
$y'=(x)'g(x)+xg'(x)$
$y'=g(x)+xg'(x)$
Hence $\;y'=g(x)+xg'(x)$.
$(b) \; y=\frac{x}{g(x)}$
Differentiating with respect to $x$ by quotient rule
$y'=\frac{(x)'g(x)-xg'(x)}{[g(x)]^2}$
$y'=\frac{g(x)-xg'(x)}{[g(x)]^2}$
Hence $\;y'=\frac{g(x)-xg'(x)}{[g(x)]^2}$.
$(c) \; y=\frac{g(x)}{x}$
Differentiating with respect to $x$ by quotient rule
$y'=\frac{xg'(x)-g(x)(x)'}{x^2}$
$y'=\frac{xg'(x)-g(x)}{x^2}$
Hence $\;y'=\frac{xg'(x)-g(x)}{x^2}$.