## Calculus: Early Transcendentals 8th Edition

(a) $$y'=g(x)+xg'(x)$$ (b)$$y'=\frac{g(x)-xg'(x)}{[g(x)]^2}$$ (c) $$y'=\frac{xg'(x)-g(x)}{x^2}$$
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 $$y'=\frac{g'(x)x-g(x)x'}{x^2}$$ $$y'=\frac{xg'(x)-g(x)}{x^2}$$