Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 70

If $f$ is a differentiable function, find an expression for the derivative of the following functions:

$
\begin{equation}
\begin{aligned}
\text{a. ) } y &= x^2 f(x) &&& \text{b. ) } y &= \frac{f(x)}{x^2}\\
\\
\text{c. ) } y &= \frac{x^2}{f(x)} &&& \text{d. ) } y &= \frac{1+xf(x)}{\sqrt{x}}
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{a.) } y &= x^2 f(x)\\
\\
y'&= x^2 \frac{d}{dx}[f(x)] + f(x) \frac{d}{dx}(x^2) && \text{Using Product Rule}\\
\\
y'&= x^2 f'(x) + f(x)(2x) && \text{Simplify}\\
\\
\end{aligned}
\end{equation}\\
\quad \boxed{y'=x^2f'(x)+2xf(x)}
$



$
\begin{equation}
\begin{aligned}
\text{b.) } y &= \frac{f(x)}{x^2}\\
\\
y' &= \frac{x^2 \frac{d}{dx} [f(x)] - \left[ f(x) \frac{d}{dx} (x^2) \right]}{(x^2)^2}
&& \text{Using Quotient Rule}\\
\\
y' &= \frac{x^2 f'(x) - f(x)(2x)}{x^4}
&& \text{Factor } x \text{in the equation}\\
\\
y' &= \frac{\cancel{x}[xf'(x)-2f(x)]}{\cancel{(x)}(x^3)}
&& \text{Simplify the equation}\\
\\
\end{aligned}
\end{equation}\\
\quad \boxed{\displaystyle y' = \frac{xf'(x)-2f(x)}{x^3}}
$



$
\begin{equation}
\begin{aligned}
\text{c.) } y &= \frac{x^2}{f(x)}\\
\\
y'&= \frac{f(x) \frac{d}{dx}(x^2)- \left[ x^2 \frac{d}{dx} f(x)\right]}{[f(x)]^2}
&& \text{Using Quotient Rule}\\
\\
y'&= \frac{f(x)(2x)-x^2f'(x)}{[f(x)]^2}
&& \text{Simplify the equation}\\
\\

\end{aligned}
\end{equation}\\
\quad \boxed{\displaystyle y'= \frac{2xf(x) - x^2 f'(x)}{[f(x)]^2}}
$



$
\begin{equation}
\begin{aligned}
\text{d.) } y &= \frac{1+xf(x)}{\sqrt{x}}\\
\\
y'&= \frac{(x)^{\frac{1}{2}}\frac{d}{dx}[1+xf(x)]-\left[ (1+xf(x)) \frac{d}{dx}(x^{\frac{1}{2}})\right]}{(\sqrt{x})^2}
&& \text{Using Quotient Rule}\\
\\
y'&= \frac{(\sqrt{x})[0+xf'(x)+f(x)(1)] - [ 1+xf(x)] \left(\frac{1}{2\sqrt{x}}\right)}{x}
&& \text{Simplify the equation}\\
\\
y'&= \frac{2x^2f'(x)+2xf(x)-1+xf(x)}{2x\sqrt{x}}
&& \text{Combine like terms}\\
\\
\end{aligned}
\end{equation}\\
\quad \boxed{\displaystyle y'= \frac{2x^2f'(x)+3xf(x)-1}{2x\sqrt{x}}}
$