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%  Differentiation Rules: Products and Quotients                   %
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\problemgrouptitle{Differentiation Rules: Products and Quotients}

\begin{problemblock}[2]{Differentiate the following functions involving products and quotients.  (Any value that is not the function variable should be considered a constant.)}

\item $f(x)=\left(x^4-3x^2+2\right)\left(x^{\frac{1}{3}}-x\right)$
\answer{$f'(x)=\left(4x^3-6x\right)\left(x^\frac{1}{3}-x\right)+\left(x^4-3x^2+2\right)\left(\displaystyle\frac{1}{3}x^{-\frac{2}{3}}-1\right)$}

\item $h(x)=\left(x^3+\pi x+2\right)\left(2+\displaystyle\frac{1}{x^3}\right)$	\answer{$f'(x)=\left(3x^2+\pi\right)\left(2+x^{-3}\right)+\left(x^3+\pi x+2\right)\left(-3x^{-4}\right)$}

\item $y=\left(x^2-1\right)\left(x^3+2\right)\left(2x^2+\sqrt{x}\right)$
\answer{$\displaystyle\frac{dy}{dx}=\left(2x\right)\left(x^3+2\right)\left(2x^2+\sqrt{x}\right)+\left(x^2-1\right)\left(3x^2\right)\left(2x^2+\sqrt{x}\right)+\left(x^2-1\right)\left(x^3+2\right)\left(4x+\frac{1}{2}x^{-\frac{1}{2}}\right)$}

\item $\displaystyle f(x)=\frac{x-4}{x-6}$
\answer{$\displaystyle \frac{df}{dx}=-\frac{2}{(x-6)^2}$}

\item $f(\theta)=\displaystyle\frac{\theta^2+3\theta-4}{\theta^2-7}$
\answer{$f'(\theta)=\displaystyle\frac{(2\theta+3)\left(\theta^2-7\right)-\left(\theta^2+3\theta-4\right)(2\theta)}{\left(\theta^2-7\right)^2}=-\frac{3\theta^2+6\theta+21}{(\theta^2-7)^2}$}

\item $g(x)=\displaystyle\frac{1+x^2}{\sqrt{x}}$ ; Also find $g'(4)$ .
\answer{$g'(x)=\displaystyle-\frac{1}{2}x^{-\frac{3}{2}}+\frac{3}{2}x^{\frac{1}{2}}$ ; $g'(4)=\frac{47}{16}$}

\item $\displaystyle f(v)=\frac{(2v+3)(v+4/v)}{v^2+v}$
\answer{$f'(v)=\frac{[(2)(v+4/v)+(2v+3)(1-4/v^2)](v^2+v)-(2v+3)(v+4/v)(2v+1)}{\left(v^2+v\right)^2}$}

\item $\displaystyle h(x)=cx^2+\left(3\sqrt{x}+2\right)\left(2x^2+x\right)$
\answer{$\displaystyle h'(x)=2cx+\frac{3}{2\sqrt{x}}\left(2x^2+x\right)+\left(3\sqrt{x}+2\right)\left(4x+1\right)$}

\end{problemblock}
