\def\whichexam{Exam I}
\def\examdate{February 3, 2004}

\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\bigskip}

\let\ds=\displaystyle
\font\sm=cmr7
\magnification1200
\centerline{\whichexam}
\centerline{\examdate}
\def\Int{\displaystyle\int}
\def\Lim{\displaystyle\lim}
\def\frac#1#2{{{#1}\over{#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\bigskip
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\mbox#1{\hbox{\rm#1}}

%:12
\newp 12.
Use logarithmic differentiation to find $f^\prime(x)$ if
$$f(x) = \frac{x\ \sin^2(x + 7)}{(2x^3 + 2)^8}\ .$$

$$\ln \bigl( f(x)\bigr) = 
\ln(x) + 2 \ln\bigl( \sin(x+7)\bigr) - 8\ln(2x^3+2)
$$
so
$$\frac{ f^\prime(x)}{f(x)} = 
\frac{1}{x} + \frac{2\cos(x+7)}{\sin(x+7)} - \frac {8(6x^2)}{2x^3+2}
$$
and
$$f^\prime(x) = \biggl(\frac{x\ \sin^2(x + 7)}{(2x^3 + 2)^8}\biggr)
\biggl(\frac{1}{x} + \frac{2\cos(x+7)}{\sin(x+7)} - \frac {48x^2}{2x^3+2}\biggr)
$$
\endp

%:13
\newp 13.  
Evaluate the following limit.
$$ \lim_{x \rightarrow \infty} \frac{(\ln x)^2}{x}\ .$$

Use l'Hospital's Rule twice:
$$\matrix{%
\displaystyle\lim_{x \rightarrow \infty} \frac{(\ln x)^2}{x}
  & = & \displaystyle\lim_{x \rightarrow \infty} \frac{2 (\ln x) (1/x)}{1} 
& \mbox{[l'Hospital for ``$\infty/\infty$'']} \cr\noalign{\vskip10pt}
  & = &\displaystyle \lim_{x \rightarrow \infty} \frac{2 \ln x}{x} \cr\noalign{\vskip10pt}
  & = & \displaystyle\lim_{x \rightarrow \infty} \frac{2 (1/x)}{1} 
& \mbox{[l'Hospital for ``$\infty/\infty$'']} \cr\noalign{\vskip10pt}
  & = & \displaystyle\lim_{x \rightarrow \infty} \frac{2}{x} \cr\noalign{\vskip10pt}
  & = & 0\  .\cr
}$$
\endp
%: 14
\newp 14.
To evaluate 
$\ds\int x \ln x\ dx$ apply integration by parts.
There are several ways to proceed.
{\baselineskip=18pt
\itemitem{Choice 1:} $dv = dx$.  Then $u=x \ln x$; $v = x$ and 
$du = \bigl(\ln x +x \frac{1}{x}\bigr) dx =  \bigl(\ln x + 1\bigr) dx$, so
$\ds\int x \ln x\ dx = x^2 \ln x - \int x\bigl(\ln x + 1\bigr) dx = 
x^2 \ln x - \int x \ln x dx - \int x\ dx = x^2 \ln x - \int x \ln x dx -\frac 12 x^2$.
Now solve for the integral:
$\ds2\int x \ln x\ dx =  x^2 \ln x  -\frac 12 x^2 + C$, so
$$\ds \int x \ln x\ dx = \frac 12 x^2 \ln x  -\frac 14 x^2 + C$$
}

{\baselineskip=18pt
\itemitem{Choice 2:} $dv = x\ dx$.  
Then $u= \ln x$; $v = \frac12 x^2$ and 
$du =  \frac{1}{x}\ dx$, so
$\ds\int x \ln x\ dx =  \frac12 x^2 \ln x - \int \frac12 x^2\frac 1x\ dx =
 \frac12 x^2 \ln x - \frac12 \int x\ dx =  \frac12 x^2 \ln x -\frac14 x^2 + C$.
}

\bigskip
{\baselineskip=18pt
\itemitem{Choice 3:}
$dv = \ln x\ dx$. Then $u = x$; $v = x \ln x - x$ and $du = dx$, so
$\ds\int x \ln x\ dx = x^2 \ln x - x^2 - \int\bigl( x \ln x - x\bigr)dx = 
 x^2 \ln x - x^2 - \int  x \ln x\ dx - \int x\ dx =  x^2 \ln x - x^2 + \int  x \ln x\ dx + \frac 12 x^2  =
 x^2 \ln x - \frac 12 x^2 + \int  x \ln x\ dx$.
 Solve for the integral as in Choice 1.

}

\endp

\end

