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%  Differentiation Rules: Chain Rule                               %
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\problemgrouptitle{Differentiation Rules: Chain Rule}

\begin{problemblock}[2]{Differentiate the following functions requiring use of the Chain Rule. (Any value that is not the function variable should be considered a constant.)}

\item $f(x)=\left(x^8-3x^4+2\right)^{12}$
\answer{$f'(x)=12\left(x^8-3x^4+2\right)^{11}\left(8x^7-12x^3\right)$}

\item $g(x)=\displaystyle\sqrt{3x^2+2}$ ; Also find $g'(2)$ .
\answer{$g'(x)=\displaystyle\frac{1}{2}\left(3x^2+2\right)^{-\frac{1}{2}}(6x)=\frac{3x}{\sqrt{3x^2+2}}$ ; $g'(2)=\frac{6}{\sqrt{14}}$}

\item $f(\theta)=\sin(\theta^2)$
\answer{$f'(\theta)=2\theta\cos(\theta^2)$}

\item $h(\theta)=\cot^2\theta$
\answer{$h'(\theta)=-2\cot\theta\csc^2\theta$}

\item $\displaystyle f(x)=\sec\left[\left(x^3+3\right)\left(\sqrt{x}+x\right)\right]$
\answer{$\displaystyle f'(x)=\sec\left[\left(x^3+3\right)\left(\sqrt{x}+x\right)\right]\tan\left[\left(x^3+3\right)\left(\sqrt{x}+x\right)\right]\left[\left(3x^2\right)\left(\sqrt{x}+x\right)+\left(x^3+3\right)\left(\frac{1}{2\sqrt{x}}+1\right)\right]$}

\item $y=4\cos\sqrt[3]{x}$
\answer{$\displaystyle y'=\left(-4\sin\sqrt[3]{x}\right)\cdot\left(\frac{1}{3}x^{-\frac{2}{3}}\right)=-\frac{4\sin\sqrt[3]{x}}{3\sqrt[3]{x^2}}$}

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


\item $y=\left(\csc x+2\right)^5+x^2+x$
\answer{$\displaystyle \frac{dy}{dx}=-5\left(\csc x+2\right)^4\csc x\cot x+2x+1$}
\item $y=\pi\tan\theta+\tan(\pi\theta)$
\answer{$\displaystyle y'=\pi\sec^2\!\theta+\pi\sec^2(\pi\theta)$}

\item $g(x)=\left(\displaystyle\sqrt{x+\sqrt{x}}\right)\left(x^4-1\right)^7$
\answer{$\displaystyle g'(x)=
\frac{1}{2}\left(x+\sqrt{x}\right)^{-\frac{1}{2}}\left(1+\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x^4-1\right)^7+\left(\sqrt{x+\sqrt{x}}\right)(7)\left(x^4-1\right)^6\left(4x^3\right)\newline 
=\frac{\left(2\sqrt{x}+1\right)\left(x^4-1\right)^7}{4\sqrt{x}\sqrt{x+\sqrt{x}}}+28x^3\left(x^4-1\right)^6\sqrt{x+\sqrt{x}}$}

\item $f(x)=\left(\displaystyle\frac{x-3}{x+1}\right)^3$
\answer{$\displaystyle f'(x)=3\left(\frac{x-3}{x+1}\right)^2\cdot\frac{4}{(x+1)^2}=12\frac{(x-3)^2}{(x+1)^4}$}

\item $A(t)=\cos(\omega t+\phi)$ ; Also find $\displaystyle\left.\frac{dA}{dt}\right|_{t=0}$~.
\answer{$\displaystyle\frac{dA}{dt}=-\omega\sin(\omega t+\phi)$ ; $\displaystyle\left.\frac{dA}{dt}\right|_{t=0}=-\omega\sin(\phi)$}

\item $f(x)=\sin\left[\cos\left(x^2+x\right)\right]$
\answer{$\displaystyle f'(x)=\left(\cos\left[\cos(x^2+x)\right]\right)\cdot\left(-\sin\left(x^2+x\right)\right)\cdot\left(2x+1\right)$}

\end{problemblock}

\item For the curve $y=5x+3\sin(2x)-2\cos(3x)$ and point $P(0,-2)$\,,\\[.3em]
  \begin{inparaenum}\item Confirm the point $P$ lies on the curve. \item Find the equation of the tangent line at $P$\,.\end{inparaenum}
\answer{\begin{enumerate}
\item $-2=5(0)+3(0)-2(1)$
\item $y=11x-2$
\end{enumerate}
}

\item Find the value(s) of $\theta$ for which the curve $f(\theta)=\cos^2\theta -\sin\theta $ has a horizontal tangent line.
\answer{$\displaystyle \left\{\frac{7\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\bigcup\left\{\frac{11\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\bigcup\left\{\frac{\pi}{2}+n\pi\ |\ n\ \mbox{an integer}\right\}$}

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