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:

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



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



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



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



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