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\centerline{MATH 135 -- Calculus 1}
\centerline{``Derivative Practice''}
\centerline{October 23, 2019}
\vskip 10pt
\noindent
{\it Background}
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We have now seen the sum and product rules for derivatives. The goals for today
are:
\begin{enumerate}
\item[(1)] To introduce another rule called the \emph{quotient rule}.
\item[(2)] To practice using these rules, and
\item[(3)] To think about some of the information about a function that we can get 
from the derivative.
\end{enumerate}
The quotient rule for derivatives says:  If $f,g$ are differentiable and
$g(x) \ne 0$ then $f/g$ is differentiable at $x$ and 
$$\frac{d}{dx}\left( \frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$$
(``bottom times derivative of top minus top times derivative of bottom, all over
bottom squared'').  For instance
$$\frac{d}{dx} \left(\frac{x^2 + 3}{x^4 + x}\right) = \frac{(x^4 + x)(2x) - (x^2 + 3)(4x^3)}{(x^4+x)^2} = 
\frac{-2x^5-4x^3+2x^2}{(x^4+x)^2}$$
(This could be simplified even farther by factoring $x^2$ out of the numerator and
denominator.   But notice that $x = 0$ is not in the domain of the
original function, so it's not in the domain of the derivative either.  Even though
you could cancel $x^2$ between the top and bottom in the derivative, our function is
not differentiable at $x = 0$.)
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\noindent
{\it Questions}
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\begin{enumerate}
\item[(1)]  Differentiate each of these with respect to the indicated variable.  Note:
you will want to think first about which rule(s) you need to apply, and then apply them.
{\it Don't worry too much about simplifying your answers -- any correct form is OK for this.}
\begin{enumerate}
\item[(a)]  $\displaystyle{f(x) = \frac{x^2 + e^x}{\sqrt{x}}}$
\item[] {\it Solution:} Write $\sqrt{x} = x^{1/2}$.  By the quotient rule:
\begin{eqnarray*}
f'(x) &=& \frac{x^{1/2}(2x + e^x) - (x^2 + e^x)\frac{1}{2}x^{-1/2}}{(x^{1/2})^2}\\
&=& \frac{3x^2 + 2x e^x - e^x}{2x^{3/2}}
\end{eqnarray*} 
(Note that this function could also be rewritten as $f(x) = x^{3/2} + x^{-1/2}e^x$ and
differentiated using the power, product, and exponential rules.  That's somewhat easier!)
\item[(b)]  $\displaystyle{g(t) = e^t\left(1 + \frac{t^2}{1 + t^2}\right)}$
\item[] {\it Solution:}  Use the product and quotient rules:
$$g'(t) = e^t\left(\frac{(1+t^2)(2t) - t^2(2t)}{(1 + t^2)^2}\right) + e^t \left(1 + \frac{t^2}{1 + t^2}
\right).$$
\item[(c)]  $\displaystyle{h(z) = \frac{3}{z^{2/3}} - z(e^z + 4z)}$
\item[] {it Solution:}  It's easier to write the first part of this as a power instead of using the
quotient rule:
$$h(z) = 3z^{-2/3} - z(e^z + 4z)$$
so 
$$h'(z) = -2 z^{-5/3} - z(e^z + 4) - (e^z + 4z) = -2z^{-5/3} - ze^z - e^z - 8z.$$
\end{enumerate}
\item[(2)]  Section 3.4 in our book builds on the way we motivated the study of derivatives
by considering instantaneous velocities and slopes of tangent lines.  If $f$ is any function 
and $f'(a)$ exists, then we can think of $f'(a)$ as an \emph{(instantaneous) rate of change} of 
$f$ with respect to the variable in $f$, at $a$.  The \emph{units} of an instantaneous rate
of change are always (units of $f$-values)/(units of the input variable in $f$).  For instance,
if we had a function $P(R)$ giving the electrical power (in units of watts) delivered to a device
by a battery, as a function of the resistance of the device (in units of ohms), then the units
of $P'(R)$ would be watts/ohm.  So suppose we have a battery delivering power to a device with 
$$P(R) = \frac{2.25 R}{(R + .5)^2} = \frac{2.25 R}{R^2 + R + .25}$$
where $R \ge 0$.   
\begin{enumerate}
\item[(a)]  What is the instantaneous rate of change of the power with respect to resistance
when $R = 3$ ohms?  (Give your answer with correct units.)
\item[] {\it Solution:}  First we compute by the quotient rule:
$$P'(R) = \frac{(R^2 + R + .25)(2.25) - 2.25R(2R + 1)}{(R^2 + R + .25)^2}
= \frac{-2.25R^2 + .5625}{(R^2 + R + .25)^2}.$$
We want $P'(3) = \frac{-2.25\cdot 3^2 + .5625}{(3^2 + 3 + .25)^2} \doteq -.13$ watts/ohm.
(This indicates that the power is decreasing as the resistance increases near $R = 3$.)
\item[(b)]  What is the power delivered to a device with $R = 5$ ohms?  What is the instantaneous rate of change of the power with respect to resistance
when $R = 5$ ohms?  Give each answer with the correct units.)
\item[] {\it Solution:}  The power is $P(5) \doteq .37$ watts.  The rate of change 
of power with respect to resistance is $P'(5) \doteq -.06$ watts/ohm.
\item[(c)]  Is the instantaneous rate of change of the power with respect to resistance ever 
equal to zero?  What does that mean?  Generate a sketch of the graph of $P(R)$ ($R$ on the 
horizontal axis, $P$ on the vertical axis) and show any points where $P'(R) = 0$.  
\item[] {\it Solution:}  Yes, the rate of change of power with respect to resistance
is zero when 
$$0 = P'(R) \Rightarrow -2.25R^2 - .5626 = 0 \Rightarrow R = \pm .5$$
(I'm using the simplified form of $P'(R)$ from part (a) to do this easily.)
Since we only want $R > 0$, the positive root $R = .5$ is the only one.
The graph of $P(R)$ for positive $R$ indicates what is happening:
\begin{figure}
\begin{center}
\includegraphics[width=.4\linewidth]{Power.png}
\caption{The graph $P = \frac{2.25 R}{(R + .5)^2}$ for $0 \le R \le 10$} 
\end{center}
\end{figure}
\end{enumerate} 
\end{enumerate}
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\noindent
{\it Assignment}
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One writeup of solutions to these problems from each group, due at the end of class.
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