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\title{MATH 2640\\Homework \#0}
\author{James A. Swenson}
\date{January 28, 2014}
\maketitle

\begin{prob}[\S3.5 \#3]
Write the composite function $y=(1-x^2)^{10}$ in the form $f(g(x))$.  [Identify the inner function $u=g(x)$ and the outer function $y=f(u)$.]
Then find the derivative $dy/dx$.
\end{prob}

\begin{soln}
If $u=1-x^2$, then $y=u^{10}$.  In other words, $y=f(g(x))$ when $g(x)=1-x^2$ and $f(u)=u^{10}$.  Therefore, by the Chain Rule,
\[ \frac{dy}{dx} = f'(g(x))g'(x) = 10(g(x))^9(-2x) = \boxed{-20x(1-x^2)^9}. \]
\end{soln}

\begin{prob}[\S7.8 \#168]
Evaluate $\displaystyle\lim_{x\to 0}\frac{\ln(1-3x)}{x}$.% Try taking out the \displaystyle -- what changes?
\end{prob}

\begin{soln}
\begin{eqnarray*}
    \lim_{x\to 0}\frac{\ln(1-3x)}{x}    & \stackrel{\text{L'H\^op}\atop{(0/0)}}{=}  & \lim_{x\to 0}\left(\frac{ \frac{1}{1-3x} (-3)}{1}\right)\\
                                        & =                                         & \frac{1}{1-0}(-3)\\
                                        & =                                         & \boxed{-3}.\vphantom{\int}
                                                                                    % An invisible integral sign improves the vertical spacing!
\end{eqnarray*}
\end{soln}

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