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
Since $g(x)$ is a differentiable function, $g'(x)$ exists.
(a) $y=xg(x)$
Apply the Product Rule, we have $$y'=x'g(x)+xg'(x)$$ $$y'=g(x)+xg'(x)$$
(b) $y=\frac{x}{g(x)}$
Apply the Quotient Rule, we have $$y'=\frac{x'g(x)-xg'(x)}{[g(x)]^2}$$ $$y'=\frac{g(x)-xg'(x)}{[g(x)]^2}$$
(c) $y=\frac{g(x)}{x}$
Apply the Quotient Rule, we have $$y'=\frac{g'(x)x-g(x)x'}{x^2}$$ $$y'=\frac{xg'(x)-g(x)}{x^2}$$
