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}
$