Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 69

If $g$ is a differentiable function, find an express for the derivative of each of the following functions.

$\begin{equation} \begin{aligned} \text{a.) } y &= xg(x) && \text{b.) } y = \frac{x}{g(x)}\\ \text{c.) } y &= \frac{g(x)}{x} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{a.) } y &= xg(x)\\ \\ y'&= x g'(x) + g(x) \frac{d}{dx} (x)\\ \\ y'&= x g'(x) + g(x) (1)\\ \\ y'&= x g'(x) + g(x) \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{b.) } y &= \frac{x}{g(x)}\\ \\ y'&= \frac{g(x) \frac{d}{dx}(x) - g'(x)}{[g(x)]^2}\\ \\ y'&= \frac{g(x)(1) - x g'(x)}{[g(x)]^2}\\ \\ y'&= \frac{g(x) - xg'(x)}{[g(x)]^2} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{c.) } y &= \frac{g(x)}{x}\\ \\ y'&= \frac{xg'(x) - g(x) \frac{d}{dx}(x)}{x^2}\\ \\ y'&= \frac{xg'(x) - g(x) (1) }{x^2}\\ \\ y'&= \frac{xg'(x) - g(x)}{x^2} \end{aligned} \end{equation}$