\magnification1200\footline{}
\input epsf.def
\gdef\localbox#1{\epsfscale=800\vtop to1.5in{\null\vskip
-1in\epsfbox{#1.eps}\vss}}

\def\frac#1#2{{{#1}\over{#2}}}
\def\equiz{\par\noindent\vrule width\hsize height 0pt depth .1 pt
\par}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\dquiz{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}

\long\def\stp\centerline#1 \centerline#2 \vskip10pt \noindent{%
\par\bigskip\noindent{#1} \qquad{ #2}\par\medskip\noindent}

\input epsf.def
\gdef\localbox#1{\epsfscale=1600\vtop to1.5in{\epsfbox{#1.eps}\vss}}


%%%%%%%%%%%%%%%
%################
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\newcount\cc\cc=1
\def\np{%
\par\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}



\stp
\centerline{\bf Math.~126 Quiz \#1}
\centerline{January  21, 2003}
\vskip10pt
\noindent
\itemitem{1.}
Using your technique from Chapter 4, show that the
function $$f(x) = x^3 - 3x^2 - 9x+1$$ is decreasing on the interval
$[\ -1, 3\ ]$.
\itemitem{2.}
Let $g(x)$ be the inverse function for $f(x)$, $x$ in $[\ -1, 3\ ]$.
The equation $f(x)=1$ has solutions
$x=0$ and $x=\frac{3 \pm 3\sqrt{5\ }}{2}$.
Use this information to find $g^\prime(1)$.
\itemitem{3.} Write an equation for the tangent line to the graph of
$y=g(x)$ at the point $x=1$.
\dquiz
\item{1.} Compute $f^\prime(x)=3x^2-6x-9$.
$f^\prime(x)=0$ if and only if $3(x^2-2x-3)=3(x-3)(x+1)=0$
if and only if $x=3$ or $x=-1$. The function $f^\prime$ 
is continuous everywhere, so $f^\prime<0$ on $(-1,3)$.
Hence $f$ is decreasing on $[-1,3]$.
\item{2.}
Since $f(x)=1$ has solutions, $x=0$ and $x=\frac{3 \pm 3\sqrt{5\ }}{2}$, 
$g(1)$ is either $0$ or one of $\frac{3 \pm 3\sqrt{5\ }}{2}$.
Since $0$ is in the interval $[-1,3]$, $g(1)=0$.
({\bf Note:} You weren't asked but the theory also guarantees 
that neither $\frac{3 \pm 3\sqrt{5\ }}{2}$ is in the interval
$[-1,3]$.)

\itemitem{}The basic formula says 
$$g^\prime(1)=\frac{1}{f^\prime(0)}$$
and $f^\prime(0) = 3\cdot 0^2 - 6\cdot 0 - 9 = -9$, so
$g^\prime(1) = \frac{1}{-9}= -\frac{1}{9}$.
\item{3.}
The tangent line to the graph of $g(x)$ and $x=1$ has slope $-\frac19$
and goes through the point $\bigl(1,g(1)\bigr)=(1,0)$, so
an equation for it is $y-0 = -\frac{1}{9} (x-1)$.
\equiz

\stp
\centerline{\bf Math.~126 Quiz \#2}
\centerline{January  28, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
\itemitem{1.} Solve the equation $8^{x^2}=9$ for $x$.
An answer such as $\ln(6)-\sqrt{\ln 8}$ or whatever is fine.
\itemitem{2.}Find the derivative with respect to $x$ of
$(2x)^{3x}\ .$
\dquiz
\item{1.} Take $\ln$ of both sides:
$\ln\bigl( 8^{x^2}\bigr) = \ln(9)$.
Simplify the left hand side:
$x^2\cdot \ln(8)=\ln(9)$, or
$x^2=\frac{\ln(9)}{\ln(8)}$
or
$x=\pm\sqrt{\frac{\ln(9)}{\ln(8)}\ }$.
\item{2.}
Easiest is to let $y=(2x)^{3x}$ and use logarithmic differentiation:
$\ln y = \ln\bigl((2x)^{3x}\bigr)=(3x) \ln(2x)$.
Hence
$\frac{y^\prime}{y}=(3) \ln(2x) + (3x) \bigl(\frac{2}{2x}\bigr)=
3\ln(2x) + 3$.
Hence
$y^\prime = y\bigl(3\ln(2x) + 3\bigr)=
(2x)^{3x} \bigl(3\ln(2x) + 3\bigr)$.

Another way:
\item{}$(2x)^{3x} = e^{^{3x \ln(2x)}}$, so
$\frac{d\ (2x)^{3x}}{dx} = e^{^{3x \ln(2x)}} \ \frac{d 3x \ln(2x)}{dx}$, 
or
$\frac{d\ (2x)^{3x}}{dx} = e^{^{3x \ln(2x)}} \bigl(
(3)\ln(2x) + (3x) \frac{2}{2x}\bigr) =
(2x)^{3x}\bigl( 3\ln(2x) + 3\bigr)$.
\item{}
\equiz

\stp
\centerline{\bf Math.~126 Quiz \#3}
\centerline{February  11, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
A charged rod of length $L$ beginning at the origin and
lying along the $x${--}axis produces an electric field at a
point $(a,b)$ in the plane given by the integral
$$E(a,b)=
\int_{-a}^{L-a}\ 
\frac{\lambda\> b}{4\pi \varepsilon_{_0} \bigl(x^2 + b^2 \bigr)^{3/2}}
\ dx$$
where $\lambda$ is the charge density per unit length on the rod and
$\varepsilon_{_0}$ is the free space permittivity.
Assume that the charge density, $\lambda$, and the free space permittivity, 
$\varepsilon_{_0}$, are constant.

Evaluate the integral to determine an expression for the electric field
in terms of $a$, $b$, $L$, $\lambda$ and $\varepsilon_{_0}$.
\dquiz
We make the substitution $x=b \tan \theta$, 
$-\frac \pi2 < \theta < \frac\pi2$, so that
$x^2 + b^2= b^2\bigl(\tan^2\theta +1 \bigr) = b^2 \sec^2\theta$.
Then $\frac{d x}{d \theta} = b \sec^2 \theta$ so
$dx = b \sec^2\theta d\theta$.

We want to change the limits of integration in this definite integral
so, since $\theta = \arctan \bigl(\frac{x}{b}\bigr)$, let
$\theta_1 = \arctan\bigl(\frac{-a}{b} \bigr)$ and
$\theta_2 = \arctan\bigl(\frac{L-a}{b} \bigr)$. 
If you are writing this out by hand instead of cut-and-paste, let
$c=\frac{\lambda}{4\pi\varepsilon_0}$.
Then $E(a,b)= \displaystyle\int_{\theta_1}^{\theta_2} 
\frac{c b}{ \bigl(b^2\sec^2\theta\bigr)^{3/2}} b \sec^2\theta\> d\theta$, 
or
$E(a,b)= \displaystyle\int_{\theta_1}^{\theta_2} 
\frac{c b^2}{\vert b\vert^3} \frac{\sec^2\theta}{\sec^3\theta}\ d\theta$.
Since $b$ may be positive or negative, we must use $\vert b \vert$ 
where we did, but since $\theta = \arctan \bigl(\frac{x}{b}\bigr)$,
it follows that $\theta$ is between $-\frac\pi2$ and $+\frac\pi2$
and hence $\sec \theta > 0$, so we do not need the absolute value
of $\sec\theta$.

Hence $E(a,b)= \displaystyle
\frac{c}{\vert b \vert}\int_{\theta_1}^{\theta_2} 
\frac{1}{\sec\theta}\> d\theta =
\displaystyle
\frac{c}{\vert b \vert}\int_{\theta_1}^{\theta_2} 
\cos\theta\> d\theta = 
\displaystyle
\frac{c}{\vert b \vert} \sin\theta\bigg\vert_{\theta_1}^{\theta_2}
= \frac{c}{\vert b \vert}\bigl( \sin\theta_2 - \sin\theta_1\bigr)$.

Analyzing the triangles with $b>0$ shows that
$\sin\theta_1 = \displaystyle \frac{-a}{\sqrt{ a^2 + b^2\ }}$ and
$\sin\theta_2 = \displaystyle \frac{L-a}{\sqrt{ (L-a)^2 + b^2\ }}$.
Hence

$$E(a,b)=
\frac{\lambda}{4\pi\varepsilon_0\> b }
\Biggl( \frac{L-a}{\sqrt{ (L-a)^2 + b^2\ }} +
\frac{a}{\sqrt{ a^2 + b^2\ }}\Biggr)\ .$$

\localbox{triangles}

Since clearly $E(a,-b) = -E(a,b)$, the formula holds for all $b\neq 0$.
For $b=0$, clearly $E(a,0) = 0$.

\noindent{\bf Remark:} If $0\leq a \leq L$ and $b=0$ 
then the formula is probably physically meaningless 
since we are actually on the rod.
However, if $a<0$ or $a>L$ and $b=0$, then $E(a,0)=0$ 
probably is physically meaningful.
You can use l'Hopital's Rule to check that for $a<0$ or $a>L$,
$\displaystyle\lim_{b\to 0} E(a,b)=0$.
\equiz

\stp
\centerline{\bf Math.~126 Quiz \#5}
\centerline{February  25, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent

\hbox{\hskip 1.8in%
\raise .75 true in\vtop{\hsize=1.5in \parindent=0pt Find the centroid of the shaded region.}
}
\par
\def\localbox#1{\epsfscale=1000\vtop to1.5in{\epsfbox{#1.eps}\vss}}

\null\hskip -20pt\localbox{region}
\dquiz
The mass, or equivalently the area, is given by
$$A = \int_a^b (\sin x - \cos x) dx\ .$$
where $a$ is the $x${--}coordinate of the left{--}hand intersection 
point and $b$ is the $x${--}coordinate of the 
right{--}handed intersection point.

The point $a$ satisfies $\sin(a)=\cos(a)$ or $\tan (a)=1$ or
$a = \frac{\pi}{4}$.
The point $b$ satisfies the same equation and so $b=a+\pi$ or
$b = \frac{5\pi}{4}$.
You could also eye{--}ball the points from the graph.

Hence $A= -\cos(x) - \sin(x) \Big
\vert_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = 
-\bigl(-\frac{\sqrt 2}{2}\bigr) - \bigl(-\frac{\sqrt 2}{2}\bigr) -
\Bigl( -\frac{\sqrt2}{2} - \frac{\sqrt2}{2}\Bigr)=
4\frac{\sqrt2}{2}=2\sqrt 2$.

The moment about the $x${--}axis, $M_x$ is given by
$$M_x=\int_a^b \frac{\sin x + \cos x}{2}(\sin x - \cos x) dx$$
where $a$ and $b$ are as above.
Now $M_x=\frac{1}{2}\displaystyle\int_a^b \sin^2(x) - \cos^2(x)\ dx
= \frac{1}{2}\displaystyle\int_a^b \frac{-\cos(2x)}{2} dx=
\frac{-1}{8}\sin(2x)\Big\vert_a^b=
\frac{-1}{8}\sin\Bigl(\frac{5\pi}{2}\Bigr) - 
\frac{-1}{8}\sin\Bigl(\frac{\pi}{2}\Bigr)=
\frac{-1}{8}(1) - \frac{-1}{8}(1)
=0$.

The moment about the $y${--}axis, $M_y$ is given by
$$M_y=\int_a^b x(\sin x - \cos x) dx$$
where $a$ and $b$ are as above.
Integrate by parts: $dv = (\sin x - \cos x) dx$, $u=x$, so
$v=-\cos x - \sin x$ and $du =dx$ so
$\displaystyle
M_y=-(\cos x + \sin x)x\Big\vert_a^b + \int_a^b \sin x + \cos x\ dx=
-(\cos x + \sin x)x\Big\vert_a^b - \cos x + \sin x \Big\vert_a^b=
-\cos x(x+1) +\sin x(1-x)\Big\vert_a^b=
-2x\cos x\Big\vert_a^b$ since $\sin(x)=\cos(x)$ for $x=a$ and for $x=b$.

Hence $M_y=\displaystyle 
-2\Bigl(\frac{5\pi}{4}\Bigr)\Bigl(-\frac{\sqrt2}{2}\Bigr) +
2\Bigl(\frac{\pi}{4}\Bigr)\Bigl(\frac{\sqrt2}{2}\Bigr)=
\frac{6\pi\sqrt 2}{4}=\frac{3\pi\sqrt 2}{2}$.

Therefore, the center of mass is at
$\bigl(\bar x, \bar y\bigr)=\displaystyle\Bigl(\frac{M_y}{A},
\frac{M_x}{A}\Bigr)=\Bigl(
\frac{\frac{3\pi\sqrt 2}{2}}{2\sqrt2}
, 0 \Bigr)=\Bigl(\frac{3\pi}{4},0\Bigr)$. 

Looking at the graph, this is not unreasonable.
\equiz

\stp
\centerline{\bf Math.~126 Quiz \#6}
\centerline{March  4, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Solve the initial value problem
$$x\frac{d y}{d x} =
x^2{\tan x} +y$$
$$y\bigl(\pi/4\bigr)=1$$
\dquiz
The equation is linear but not in standard form.
The standard form is
$$\frac{d y}{d x} + \frac{-1}{x} y = x \tan x\ .$$
Hence $P(x)=\frac{_1}{x}$ and $\int P dx = -\ln \vert x\vert + C$.
Hence we may use $I = e^{-\ln\vert x\vert}= \frac{1}{\vert x \vert}$
as an integrating factor, and even use $I=\frac{1}{x}$.

Check: $\frac{d \frac{1}{x}{y}}{dx} = \frac{1}{x} y^\prime
 -\frac{1}{x^2} y= \frac {1}{x}\Bigl( y^\prime - \frac{1}{x} y\Bigr)$,
so $I = \frac{1}{x}$ is an integrating factor.

Hence $y = \frac{1}{I}\int I \cdot Q dx$, where $Q(x)=x \tan x$, and
$I \cdot Q = \tan x$.
Therefore 
$\int I\cdot Q dx = \int \tan x dx = \ln \vert \sec(x)\vert +C$
and $\frac{1}{x} y =  \ln \vert \sec(x)\vert +C$, or
$y = x (\ln\vert\sec x\vert + C)$.

To solve the initial value problem, note $y(\pi/4)=1$ and
$y(\pi/4)=\frac{\pi}{4}(\ln\vert\sec(\pi/4)\vert + C)$, so
$1=\frac{\pi}{4}(\ln\sqrt2 + C)$, or
$C= \frac{4}{\pi} - \ln\sqrt 2$.
Hence
$$y =  x \Bigl(\ln\vert\sec x\vert + \frac{4}{\pi} - \ln\sqrt 2\ \Bigr)
\ .$$
\equiz

\stp
\centerline{\bf Math.~126 Quiz \#7}
\centerline{March  26, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Find the area of the region inside the polar curve
$r = 3\cos\theta$ and outside the polar curve
$r = 2 - \cos\theta$.
\vskip.25in
\localbox{shadedpolarregion}
\dquiz
We have shaded the required region.
We start by locating the initial and terminal angles.
We {\sl hope\/} that they satisfy the equation
$3\cos \theta = 2 - \cos \theta$.
This equation is equivalent to $4 \cos\theta = 2$ or
$\cos \theta = \frac 12$ and hence $\theta = \pm \frac{\pi}{3}$.
We are looking for two angles and we have found two angles.

It now follows that the area is given by
$$\eqalign{
A =& \frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}
\Bigl( \bigl(3 \cos \theta\bigr)^2 - 
\bigl(2 - \cos \theta \bigr)^2\Bigr)\ d\theta\cr
=& \frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}
\Bigl( 9 \cos^2 \theta - 
\bigl(4 - 4\cos \theta +\cos^2 \theta\bigr)\Bigr)\ d\theta\cr
=& \frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}
\Bigl( 8 \cos^2 \theta +
4\cos \theta - 4\Bigr)\ d\theta\ .\cr
}$$

\item{}
$\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1\ d\theta=
\frac{2\pi}{3}$\ ;
\item{}
$\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 
\cos\theta \ d\theta= \sin \theta 
\Big\vert_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=
\frac{2\sqrt{3}}{2}= \sqrt3$\ ;
\item{}$\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 
\cos^2\theta \ d\theta= 
\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 
\frac{1+\cos(2 \theta)}{2}\ d\theta=
\frac{\theta}{2}+\frac{\sin(2 \theta)}{4}
\Big\vert_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=
2\frac{\pi}{6} + 2\frac{\frac{\sqrt{3}}{2}}{4}=
\frac{\pi}{3}+ \frac{\sqrt{3}}{4}$\ .

Hence 
$$A= 4 \Bigl(\frac{\pi}{3}+ \frac{\sqrt{3}}{4}\Bigr) +
2 \Bigl(\sqrt{3}\Bigr) -
2\Bigl(\frac{2\pi}{3}\Bigr) = 3\sqrt 3\ .$$

\equiz
\stp
\centerline{\bf Math.~126 Quiz \#8}
\centerline{April  2, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Consider the ellipse in the $xy$-plane which has
foci at $(3,2)$ and $(3,8)$ with eccentricity $\frac12$.
\medskip
\item{1.}What are the coordinates of the center?
\item{2.}What are the coordinates of the vertices?
\item{3.}What is the equation of the ellipse?
\item{4.}What are the equations of the directrixes?
\dquiz
The center is at $\Bigl(\frac{3+3}{2}, \frac{2+8}{2}\Bigr)=
(3,5)$.
The major axis is parallel to the $y$ axis.

The distance from the center to a focus is $c=8-5=3$ or
$c=5-2=3$.

Since $\displaystyle e=\frac{c}{a}$, 
$\displaystyle \frac{1}{2}=\frac{3}{a}$, so $a=6$:
the vertices are at $(3, 5 \pm 6)$ or at $(3,11)$ and $(3,-1)$.

Since $b^2 = a^2 - c^2$ for an ellipse, $b^2 = 36 - 9 =27$.
An equation for our ellipse is
$$\frac{(x-3)^2}{27} + \frac{(y-5)^2}{36} = 1\ .$$

The distance from the vertex $(3,-1)$ to the focus $(3,2)$
is $3$.
If $k$ is the distance from the vertex $(3,-1)$ to the
associated directrix, then 
$\displaystyle  e = \frac{3}{k}$, so $k=6$ and so the directixes
are the lines $y = -1 - 6 = -7$ and
$y = 11 + 6 =17$.
\equiz


\stp
\centerline{\bf Math.~126 Quiz \#9}
\centerline{April  8, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\noindent
Which of the following series converge and which diverge?
Why?
\bigskip
\item{A.}$\displaystyle\isum{1}\frac1{n\>!}$:
\medskip
\item{}Recall $n\>!=n(n-1)\cdots2\cdot1$
\item{B.}$\displaystyle\isum{2} \frac{n}{1+\sqrt{n}}$:
\item{C.}$\displaystyle\isum{2} \Bigl(\frac{2^n}{3^n} + 
\frac{1}{n^2}\Bigr)$:

\item{D.}$\displaystyle\isum{2} \Bigl(\frac{\pi^n}{3^n} + 
\frac{1}{n^2}\Bigr)$:
\dquiz
\item{A.}
By the Ratio test
$\displaystyle\lim_{n\to\infty}\Bigg\vert
\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}\Bigg\vert=
\lim_{n\to\infty}\frac{1}{n+1}=0$ and $0<1$ so series converges.

\item{B.}$\displaystyle\lim_{n\to\infty} \frac{n}{1+\sqrt{n}}=
\lim_{n\to\infty}\frac{\sqrt{n}}{\frac{1}{\sqrt{n}}+1}=\infty$.
Since $\lim_{n\to\infty}a_n \neq 0$, the series diverges.
An equally good calculation is
$\displaystyle\lim_{n\to\infty} \frac{n}{1+\sqrt{n}}=
\lim_{n\to\infty}\frac{n}{\sqrt{n}}=\lim_{n\to\infty}\sqrt{n} = \infty$.

\item{C.}$\displaystyle\isum{2} \frac{2^n}{3^n}$ is 
a geometric series with $r=\frac{2}{3} < 1$ and so converges:
$\displaystyle\isum{2} \frac{1}{n^2}$ is a $p${--}series with $p=2>1$
and so it converges.
Therefore the sum of the two convergent series converges.

\item{D.}This time 
$\displaystyle\isum{2} \frac{\pi^n}{3^n}$ is a geometric series
with $r=\frac{\pi}{2}>1$ and so it diverges.
This suggests using the Comparison Test:
$\frac{\pi^n}{2^n} < \frac{\pi^n}{3^n} + \frac{1}{n^2}$ and
$\displaystyle\isum{2} \frac{\pi^n}{3^n}$ diverges: therfore
$\displaystyle\isum{2} \Bigl(\frac{\pi^n}{3^n} + 
\frac{1}{n^2}\Bigr)$ diverges.
\equiz

\stp
\centerline{\bf Math.~126 Quiz \#10}
\centerline{April  15, 2003}
\vskip10pt
\noindent

Consider the function $f(x)$ defined by the power series
$$f(x) = \sum_{n=1}^{\infty}
\frac{x^n}{n^{\frac{3}{2}}}
$$
\item{1.}
Find the interval of convergence of the power series for $f$.
\item{2.}
Write down the power series for $f^\prime(x)$.
\item{3.}
Find the interval of convergence of the power series for $f^\prime$.
\dquiz
\item{1.}
Via the Ratio Test:
$\displaystyle\Bigg\vert \frac{\frac{x^{n+1}}{(n+1)^{3/2}}}
{\frac{x^{n}}{n^{3/2}}}\Bigg\vert = 
\bigg\vert\frac{x^{n+1} n^{3/2}}{x^n (n+1)^{3/2}}\bigg\vert=
\vert x\vert \frac{n^{3/2}}{(n+1)^{3/2}}$.

In the limit,
$\displaystyle\lim_{n\to\infty} \vert x\vert \frac{n^{3/2}}{(n+1)^{3/2}}=
\vert x \vert$, since we can compute the limit of an algebraic
quotient by examining the highest power of $n$ in the numerator and
the highest power of $n$ in the denominator.
It follows that the radius of convergence of this power series
is $1$.
When $x=\pm1$,
$\displaystyle \sum_{n=1}^{\infty}
\bigg\vert\frac{(\pm1)^n}{n^{\frac{3}{2}}}\bigg\vert$
is a $p${--}series with $p=\frac{3}{2}>1$ and hence convergent.
Thus the original two series,
$\displaystyle \sum_{n=1}^{\infty}
\frac{(\pm1)^n}{n^{\frac{3}{2}}}$, are absolutely convergent,
hence convergent.
Therefore the domain of $f$ is the interval $[-1,1]$ (and the
convergence is absolute in the entire interval).
\item{2.}
By our theorem,
$f^\prime(x)=\displaystyle \sum_{n=1}^{\infty}
\frac{n x^{n-1}}{n^{\frac{3}{2}}}=
\sum_{n=1}^{\infty}
\frac{x^{n-1}}{n^{\frac{1}{2}}}
$ or $\displaystyle\sum_{n=0}^{\infty}
\frac{x^{n}}{(n+1)^{\frac{1}{2}}}$.

\item{3.}
The radius of convergence for $f^\prime$ is the same as for $f$,
and hence it is $1$.
This time the two series we need to examine are
$\displaystyle\sum_{n=0}^{\infty}
\frac{(\pm1)^{n}}{(n+1)^{\frac{1}{2}}}$.
When $x=+1$, we have a $p${--}series with $p = \frac{1}{2}<1$ so
the series diverges.
When $x=-1$ the series is alternating (obvious), the terms go to
$0$ (also obvious) and the terms are decreasing (
$g(x) = x^{-1/2}$, so $g^\prime(x)= -\frac{1}{2} x^{-3/2}$ and
when $x>0$, $x^{-3/2}>0$ so $g^\prime(x)<0$ and the terms
are decreasing).
Hence this series converges conditionally and the domain of
$f^\prime(x)$ is $[-1,1)$.

\equiz
\end