%% Math 131, Fall, 2003
\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,epic,eepic,pstcol}
\usepackage{math131}

\newcommand{\ddx}[1]{\frac{d}{d#1}}

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\begin{document}

\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Quiz \#7 Solutions}
Professor Hwang \\
\end{center}

\be
\item Find $\ds\frac{d}{dx} e^x\sin x$

  \soln By the product rule,
  \[
  \ddx{x}e^x\sin x=e^x\ddx{x}\sin x + \sin x \ddx{x}e^x
  = (\cos x+\sin x)e^x
  \]

\sep

\item Find $\ds\frac{d}{dt}(t^3-3t)^{24}$

  \soln The chain rule, with outer function $v=u^{24}$ and $u=t^3-3t$,
  gives 
  \[
  \ddx{t}(t^3-3t)^{24} = 24(t^3-3t)^{23}(3t^2-3).
  \]
  This answer can be written $72(t^3-3t)^{23}(t^2-1)$, or even
  factored further, but there's no real need to.


\sep

\item Let $x$~be a function of~$t$, and suppose
  $\ds\frac{dx}{dt}=\sqrt{x^2+1}$. Compute $\ds\frac{d^2x}{dt^2}$~as a
  function of~$x$.
  
  \soln Write $\ds\frac{dx}{dt}=\sqrt{x^2+1}=(x^2+1)^{1/2}$,
  then use the chain rule to differentiate:
  \begin{align*}
  \frac{d^2x}{dt^2} 
    &= \ddx{t}\Bigl(\frac{dx}{dt}\bigr) \\
    &= \ddx{t}(x^2+1)^{1/2} \\
    &= \frac{1}{2}(x^2+1)^{-1/2}(2x)\frac{dx}{dt} \\
    &= \frac{1}{2}(x^2+1)^{-1/2}(2x)(x^2+1)^{1/2} \\
    &= x.
  \end{align*}

\ee
\eject

\end{document}
