%% Math 131, Fall, 2003
\documentclass[11pt]{article}
\usepackage{latexsym,amsmath}

\usepackage{math131}

\pagestyle{empty}
\newcommand{\ddx}[1]{\ds\frac{d}{d#1}}

\begin{document}

\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
Solutions to Worksheet for Thursday, October~23
\end{center}

Compute the indicated derivatives:

\be

\item $\ddx{x}xe^{-x}$,\qquad $\ddx{x} x^2e^{-x}$,\qquad $\ddx{x} x^ne^{-x}$

\soln It is enough to find the general formula:
\[
\frac{d}{dx}(x^ne^{-x}) = (nx^{n-1})e^{-x} - x^n e^{-x}
= x^{n-1}(-x+n)e^{-x}.
\]

\item $\ddx{u}\frac{e^u}{u^2+5}$,\qquad $\ddx{t}\frac{e^u}{u^2+5}$
  (assume~$u$ is a function of~$t$)
  
  \soln The quotient and chain rules give:
\[
\frac{d}{dt}\frac{e^u}{u^2+5}
= \frac{(u^2+5)e^u-e^u(2u)}{(u^2+5)^2}\frac{du}{dt}
= e^u\frac{u^2-2u+5}{(u^2+5)^2}\frac{du}{dt}
\]

\item $\ddx{w}\frac{\cos w}{w^2+5}$,\qquad $\ddx{v}\frac{v^2+5}{\sin v}$

\soln The quotient rule gives
\begin{align*}
\frac{d}{dw}\frac{\cos w}{w^2+5}
&= \frac{(w^2+5)(-\sin w)-(\cos w)(2w)}{(w^2+5)^2}
= -\frac{(w^2+5)\sin w + 2w\cos w}{(w^2+5)^2} \\
& \\
\frac{d}{dv}\frac{v^2+5}{\sin v}
&= \frac{(\sin v)(2v) - (v^2+5)(\cos v)}{\sin^2 v}
\end{align*}

\item $\ddx{x}\ln(x^2)$,\qquad $\ddx{x}\ln x^5$,\qquad $\ddx{x}(x^n)$

\soln Again, it is enough to find the general formula. By the chain
rule, 
\[
\frac{d}{dx}\ln(x^n) = \frac{nx^{n-1}}{x^n} = \frac{n}{x}
\]
This is ``not surprising'' because $\ln(x^n)=n\ln x$.

\item $\ddx{s}\ln(s^2+1)$,\qquad $\ddx{x}\ln(\cos x)$

\soln Using the chain rule, we find that
\[
\frac{d}{ds}\ln(s^2+1)] = \frac{2s}{s^2+1},\qquad
\frac{d}{dx}\ln(\cos x) = -\frac{\sin x}{\cos x} = -\tan x.
\]

\ee
\eject

\end{document}
