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\begin{document}
\begin{center}

\textbf{College of the Holy Cross} \\
\textbf{MATH 135, section 1}\\
\textbf{Exam 3 Solutions -- Friday, November 22}
\end{center}

\vskip 10pt
\noindent
I.  For each of the following functions find the derivative and simplify. 
\begin{enumerate}
\item[A.] (5) $\displaystyle f(x) = e^{2x} \cos(4x)$
\item[] {\it Solution:}  By the product and chain rules, 
$$f'(x) = -4 e^{2x} \sin(4x) + 2e^{2x}\cos(4x) = 2e^{2x}(\cos(4x) - 2\sin(4x)).$$
\item[B.] (5) $\displaystyle g(x)= \frac{x^2 + 2}{\ln(x)}$ 
\item[] {\it Solution:} By the quotient rule,
$$g'(x) = \frac{2x \ln(x) - (x^2 + 2)\frac{1}{x}}{(\ln(x))^2} = \frac{2x^2\ln(x) - x^2 - 2}{x(\ln(x))^2}.$$
\item[C.] (5) $\displaystyle h(x)=\tan^{-1}(2x + 1)$
\item[] {\it Solution:}  By the derivative rule for inverse tangent and the chain rule,
$$h'(x) = \frac{2}{1 + (2x + 1)^2}.$$
\item[D.] (10) $\displaystyle k(x) = x^{\sin(x)}$
\item[] {\it Solution:}  Since $x$ appears in the base and the exponent, we must use logarithmic differentiation.
Writing $y = x^{\sin(x)}$, $\ln(y) = \sin(x) \ln(x)$.  So then differentiating implicitly,
$$\frac{y'}{y} = \cos(x) \ln(x) + \frac{\sin(x)}{x}$$
and 
$$y' = x^{\sin(x)}\left(\cos(x) \ln(x) + \frac{\sin(x)}{x}\right).$$
\item[E.] (10) Find $y'$ by implicit differentiation if $\displaystyle x^3y^2 + 2y = 3x$.  \\ 
\item[] {\it Solution:}
$$3x^2y^2 + x^3\cdot 2yy' + 2y' = 3,$$
so 
$$y' = \frac{3 - 3x^2y^2}{2x^3y + 2}.$$
\end{enumerate}
\vskip 10pt

\noindent
II. (20)  A rocket is launched vertically and is tracked by a ground
station $4$ miles from the launch pad.  What is the vertical speed 
of the rocket when its height above the ground is $3$ miles and 
its distance to the ground station is increasing at $3300$ miles per
hour?   
\vskip 10pt
\noindent
{\it Solution:}  Let $y$ be the height of the rocket above the launch pad as a function of $t$
and $z$ be the distance to the ground station as a function of $t$.  At each time the 
position of the rocket, the launch pad, and the ground station form a right triangle, so by the 
Pythagorean theorem we have
$$16 + y^2 = z^2.$$
Differentiating with respect to $t$:
$$2 y \frac{dy}{dt} = 2z \frac{dz}{dt}.$$
The vertical speed of the rocket that we want is $\frac{dy}{dt}$.  We 
know that at the time $t$ when $y = 3$, $z = 5$ 
and $\frac{dz}{dt} = 3300$.  Therefore 
$$\frac{dy}{dt} = \frac{5\cdot 3300}{3} = 5500 \rm{m/hr}.$$
\noindent
\noindent
III.  All parts of this question refer to the function $\displaystyle f(x) = -2x^3 + 24x + 11$.
\begin{enumerate}
\item[A.] (10) Find the critical numbers of $f(x)$.
\item[] {\it Solution:}  We have $f'(x) = -6x^2 + 24 = 0$ when $x = \pm 2$.  $f'(x)$ exists for
all real $x$, so the only critical numbers are $x = \pm 2$.
\item[B.] (10) What does the Second Derivative Test tell you about the behavior of $f$ at each of
these critical numbers?
\item[] {\it Solution:}  $f''(x) = -12x$. We see $f''(2) = -24 < 0$, so by the Second Derivative Test,
$f$ has a local maximum at $x = 2$.
Similarly, $f''(-2) = 24 > 0$.  Therefore $f$ has a local minimum at $x = -2$.  
\end{enumerate}
\vskip 10pt
\noindent
IV.  All parts of this question refer to the function $\displaystyle f(x) = \frac{x}{(2x + 1)^2}$,
for which the first two derivatives are, in simplified form:
\begin{eqnarray*}
f'(x) &=& \frac{1 - 2x}{(2x + 1)^3}\\
f''(x) &=& \frac{8x - 8}{(2x + 1)^4}
\end{eqnarray*} 
\bigskip
\begin{enumerate}
\item[A.] (5) What is the domain of $f(x)$?
\item[] {\it Solution:}  Domain of $f$ is all $x \ne -1/2$, or $(-\infty,-1/2) \cup (-1/2,\infty)$.  
\item[B.] (10) Find all critical numbers and determine where $f(x)$ is
increasing and decreasing.
\item[] {\it Solution:}  From the formula for $f'(x)$, we see $f'(x) = 0$ when $x = 1/2$, and $f'(x)$
exists for all $x \ne -1/2,1/2$.  Technical point:  $x = -1/2$ is \emph{not} a critical number for $f$,
since critical numbers must be in the domain of the function.  So the only critical number is 
$x = 1/2$.  Looking at the sign of $f'(x)$, we see $f'(x) < 0$ for $x$ in $(-\infty,-1/2)$, $f'(x) > 0$
for $x$ in $(-1/2,1/2)$, and $f'(x) < 0$ for $x$ in $(1/2,\infty)$.  Therefore $f$ is decreasing 
on $(-\infty,-1/2)$ and $(1/2,\infty)$ and increasing on $(-1/2,1/2)$.   

\item[C.] (5) What is the absolute minimum value of $f(x)$ on the interval $[0,3]$?
\item[] {\it Solution:} The critical number $1/2$ is in this interval. We have $f(1/2) = \frac{1}{8}$, $f(0) = 0$
and $f(3) = \frac{3}{49}$.  The absolute minimum value is $f(0) = 0$.  (Note that by the First Derivative Test,
$f$ has a local maximum at $x = 1/2$, so the minimum must be at one of the endpoints.)
\vskip 10pt
\item[D.] (5) Determine the concavity of $f(x)$, and find all inflection points.
\item[] {\it Solution:}  $f''(x) > 0$ on $(1,\infty)$ and $f''(x) < 0$ on $(-\infty, -1/2)$ and $(-1/2,1)$.  
The concavity changes at $x = 1$, so that is the only inflection point.  



\end{enumerate}





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