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\markright{Chapter 3: Answers 4}
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\Bf{Chapter 3: Answers 4 \hfill Jack K. Cohen \hfill Colorado School of Mines}

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\item  Use \Mma\ and if you still don't see the light, ask your friends or the instructor for help.

\item  Use the fact that  $D \sin x = \cos x$ to find the derivatives of the following functions:
	\begin{enumerate}
	\item $D \sin (x^3) = \cos ( \cdots ) \cdot D( \cdots ) = \cos (x^3) \cdot Dx^3 = 3 x^2 \cos x^3$.
	\item $D (\sin x)^3 = 3 ( \cdots )^2 \cdot D( \cdots ) = 3 (\sin x)^2 \cdot D \sin x = 3 \sin^2 x \cos x$.
	\item $D\sin( \sin x) = \cos ( \cdots ) \cdot D( \cdots ) = \cos ( \sin x) \cdot D \sin x = \cos (\sin x) \cos x$.
	\item $D\sin( \sin ( \sin x)) = \cos ( \cdots ) \cdot D( \cdots ) = \cos (\sin (\sin x)) \cdot D \sin (\sin x) = \cos (\sin (\sin x)) \cos ( \cdots ) \cdot D( \cdots ) = \cos (\sin (\sin x)) \cos (\sin x) \cdot D \sin x = \cos (\sin (\sin x)) \cos (\sin x) \cos x$.  (Whew!)
	\end{enumerate}
\item The area of a circle in terms of the radius is $A = \pi r^2$.
	\begin{enumerate}
	\item $\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} = 2 \pi r \frac{dr}{dt}$.
	\item (3.3.49)  $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt} = 2 \pi  \cdot 10 \cdot 2 = 40 \pi $ cm$^2$/sec.
	\item From part (a), $\frac{dr}{dt} = \frac{1}{2 \pi r} \frac{dA}{dt}$.  Use $r = \sqrt{A/ \pi}$ and simplify to get  $\frac{dr}{dt} = \frac{1}{2 \sqrt{\pi A}} \frac{dA}{dt}$.
	\item (3.3.50) $- \sqrt{3}/ 15$ cm/sec.
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