\documentclass[11pt]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{amssymb}
\usepackage[nomarginpar]{geometry}
\usepackage{color}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{fancyhdr}
\usepackage{multicol}
\usepackage{hyperref}
\usepackage{boxedminipage}
\usepackage{setspace}

\setcounter{MaxMatrixCols}{10}
%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=5.00.0.2570}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{Created=Wednesday, July 12, 2006 00:27:03}
%TCIDATA{LastRevised=Saturday, February 04, 2017 03:44:15}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="Title" CONTENT="Derivatives of Trig Functions">}
%TCIDATA{<META NAME="DocumentShell" CONTENT="Scientific Notebook\Booklet #1 - with Instructions">}
%TCIDATA{CSTFile=40 LaTeX article.cst}
%TCIDATA{PageSetup=72,72,72,72,1}
%TCIDATA{ComputeGeneralSettings=0,15,15,0,0,0,0}
%TCIDATA{ComputePlot2DSettings=0,Line,Solid,Thin,Dot,[flat::RGB:0000000000],Normal,0}
%TCIDATA{Counters=arabic,1}
%TCIDATA{<META NAME="PrintViewPercent" CONTENT="100">}
%TCIDATA{ComputeDefs=
%$f\left( x\right) =\sqrt{x^{2}-1}$
%}

%TCIDATA{AllPages=
%H=36
%F=36,\PARA{038<p type="texpara" tag="Body Text" >\hfill \hfill }
%}


\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\geometry{left=0.5in,right=0.4in,top=0.6in,bottom=0.4in}
\pagestyle{fancy}
\lhead{\color{blue} \large Lecture Notes}
\chead{\color{black} \Large  Differentiating Trigonometric Functions}
\rhead{ page   \ \thepage}
\cfoot{}
\lfoot{\small   \copyright $\;$  Hidegkuti,  Powell,  2012}
\rfoot{\small Last revised: January 19, 2017}
\textwidth 7.5in 
\textheight 9.6in 
\setlength{\headheight}{25pt}
\setlength{\parindent}{0in}
\setlength{\parskip}{0.09in}
\setlength{\columnsep}{0.3in}

\begin{document}


\bigskip \bigskip

%TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}%
%BeginExpansion
\begin{multicols}{2}%
%EndExpansion

\begin{enumerate}
\item $\dfrac{d}{dx}\left( \sin x\right) =\cos x\medskip \smallskip $%
\bigskip $\medskip $

\item $\dfrac{\smallskip d}{dx}\left( \cos x\right) =-\sin x\medskip
\bigskip \smallskip $

\item $\dfrac{d}{dx}\left( \tan x\right) =\sec ^{2}x=\tan ^{2}x+1\medskip
\bigskip \smallskip \medskip $

\item $\dfrac{d}{dx}\left( \cot x\right) =-\csc ^{2}x=-\cot ^{2}x-1\bigskip
\medskip \smallskip $

\item $\dfrac{d}{dx}\left( \sec x\right) =\dfrac{\sin x}{\cos ^{2}x}=\sec
x\tan x\medskip \smallskip $\bigskip

\item $\dfrac{d}{dx}\left( \csc x\right) =-\dfrac{\cos x}{\sin ^{2}x}=-\csc
x\cot x\bigskip \medskip \smallskip $

\item $\dfrac{d}{dx}\left( \sin ^{-1}x\right) =\dfrac{1}{\sqrt{1-x^{2}}}%
\medskip $\bigskip

\item $\dfrac{d}{dx}\left( \cos ^{-1}x\right) =-\dfrac{1}{\sqrt{1-x^{2}}}%
\bigskip \medskip $

\item $\dfrac{d}{dx}\left( \tan ^{-1}x\right) =\dfrac{1}{x^{2}+1}\medskip $%
\bigskip

\item $\dfrac{d}{dx}\left( \cot ^{-1}x\right) =-\dfrac{1}{x^{2}+1}\bigskip
\medskip $

\item $\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{\left\vert
x\right\vert \sqrt{x^{2}-1}}\bigskip \medskip $

\item $\dfrac{d}{dx}\left( \csc ^{-1}x\right) =-\dfrac{1}{\left\vert
x\right\vert \sqrt{x^{2}-1}}\medskip $ \ \ 
\end{enumerate}

%TCIMACRO{\TeXButton{m col end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion
\pagebreak

\begin{center}
{\LARGE Proofs}\vspace{0.16in}
\end{center}

\fbox{Theorems 1 and 2: \ $\dfrac{d}{dx}\left( \sin x\right) =\cos x$ \ and $%
\dfrac{d}{dx}\left( \cos x\right) =-\sin x$}\vspace{0.1in}

Claim 1.) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$

Proof: \ This theorem and the next one are necessary for differentiating $%
\sin x$ and $\cos x$. \ Recall a theorem: \ Let $\ r$ be the radius of a
circle. \ If $\alpha $ is measured in radians, then the area of a sector
with a central angle of $\alpha $ is $A_{\text{sector}}=\dfrac{\alpha r^{2}}{%
2}$. \ (Notation: \ $\overline{AB}$ will denote the length of line segment $%
AB$.)

Let $x$ be a very small positive angle, measured in radians, drawn into a
unit circle as shown on the picture below. \ Let $B$ be the point where the
unit circle intersects the ray determined by $x$. \ We then draw a tangent
line to the circle at point $B$. \ Let $A$ be the point where the tangent
line intersects the $x-$axis. \ We also draw a vertical line through $B.$ \
Let $D$ be the point where this vertical line intersects the $x-$axis. \
Finally, let us denote by $E$ the point with coordinates $\left( 0,1\right) $%
.\FRAME{dtbpF}{2.776in}{2.2399in}{0pt}{}{}{sinx_xbetter.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 2.776in;height 2.2399in;depth
0pt;original-width 2.7337in;original-height 2.2001in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'sinx_xbetter.bmp';file-properties
"XNPEU";}}

The proof will be based on the following fact: because they include each
other, the following three areas can be easily compared: 
\begin{equation*}
\text{Area of triangle }CDB\text{ }\leq \text{ Area of sector }CEB\text{ }%
\leq \text{ Area of triangle }ABC
\end{equation*}%
Area of triangle $CDB$: \ the horizontal side, $\overline{CD}=\cos x$ and
the vertical side, $\overline{DB}=\sin x$. \ Since this is a right triangle,
the area is: $A_{CDB}=\dfrac{1}{2}\sin x\cos x$.

Area of sector $CEB$: $A_{\text{sector}}=\dfrac{1^{2}x}{2}=\dfrac{x}{2}$.

Area of triangle $ABC$: there is a right angle at point $B$ because the
tangent line drawn to a circle is perpendicular to the radius drawn to the
point of tangency. \ So the area is $A_{ABC}=\dfrac{1}{2}\overline{AB}\cdot 
\overline{BC}$. \ Clearly $\overline{BC}=1$. \ To compute $\overline{AB}$,
in triangle $ABC$, \ $\tan x=\dfrac{\overline{AB}}{1}$ and so $\overline{AB}%
=\tan x$.

Area of triangle $ABC$: \ $\dfrac{1}{2}\left( 1\right) \left( \tan x\right) =%
\dfrac{\tan x}{2}$ or $\dfrac{\sin x}{2\cos x}$. \ So now 
\begin{equation*}
\text{Area of triangle }CDB\text{ }\leq \text{ Area of sector }CEB\text{ }%
\leq \text{ Area of triangle }ABC
\end{equation*}%
translates to 
\begin{equation*}
\dfrac{1}{2}\sin x\cos x\leq \dfrac{x}{2}\leq \dfrac{\sin x}{2\cos x}
\end{equation*}

Let us divide all three sides by $\dfrac{\sin x}{2}$. \ Because $x$ is small
and positive, $\dfrac{\sin x}{2}$ is positive and so we do not need to
reverse the inequality signs.%
\begin{equation*}
\cos x\leq \dfrac{x}{\sin x}\leq \dfrac{1}{\cos x}
\end{equation*}%
Suppose now that $x$ approaches zero. \ Then both $\cos x$ and $\dfrac{1}{%
\cos x}$ approach $1$. \ By the sandwich principle, $\dfrac{x}{\sin x}$, the
quantity locked in between those two must also approach $1.$%
\begin{equation*}
\begin{array}{ccccc}
\cos x & \leq & \dfrac{x}{\sin x} & \leq & \dfrac{1}{\cos x} \\ 
\downarrow &  &  &  & \downarrow \\ 
1 &  &  &  & 1%
\end{array}%
\end{equation*}%
If $\dfrac{x}{\sin x}$ approaches $1,$ so is its reciprocal, $\dfrac{\sin x}{%
x}$.

So far, we have proven the statement for positive values of $x$, that is, $%
\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin x}{x}=1$. \ A similar argument
works for negative values of $x$. \vspace{0.1in}

Claim 2.) \ $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x}=0\medskip $

Proof: 
\begin{eqnarray*}
\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x} &=&\lim\limits_{x\rightarrow
0}\dfrac{\cos x-1}{x}\cdot 1=\lim\limits_{x\rightarrow 0}\left( \dfrac{\cos
x-1}{x}\cdot \dfrac{\cos x+1}{\cos x+1}\right) =\lim\limits_{x\rightarrow 0}%
\dfrac{\cos ^{2}x-1}{x\left( \cos x+1\right) }=\lim\limits_{x\rightarrow 0}%
\dfrac{-\left( 1-\cos ^{2}x\right) }{x\left( \cos x+1\right) } \\
&=&\lim\limits_{x\rightarrow 0}\dfrac{-\sin ^{2}x}{x\left( \cos x+1\right) }%
=\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \dfrac{-\sin x}{\cos x+1}%
=\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot
\lim\limits_{x\rightarrow 0}\dfrac{-\sin x}{\cos x+1}=1\cdot 0=0
\end{eqnarray*}%
We are now ready to prove that \ $\dfrac{d}{dx}\left( \sin x\right) =\cos x$
\ and \ $\dfrac{d}{dx}\left( \cos x\right) =-\sin x$

Proof:%
\begin{eqnarray*}
\dfrac{d}{dx}\left( \sin x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{\sin
\left( x+h\right) -\sin x}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sin x\cos
h+\cos x\sin h-\sin x}{h} \\
&=&\lim\limits_{h\rightarrow 0}\left( \dfrac{\sin x\cos h-\sin x}{h}+\dfrac{%
\cos x\sin h}{h}\right) \medskip \medskip =\lim\limits_{h\rightarrow 0}%
\dfrac{\sin x\left( \cos h-1\right) }{h}+\lim\limits_{h\rightarrow 0}\cos x%
\dfrac{\sin h}{h} \\
&=&\sin x\lim\limits_{h\rightarrow 0}\dfrac{\cos h-1}{h}+\cos
x\lim\limits_{h\rightarrow 0}\dfrac{\sin h}{h}=\sin x\cdot 0+\cos x\cdot
1=\cos x
\end{eqnarray*}%
\begin{eqnarray*}
\dfrac{d}{dx}\left( \cos x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{\cos
\left( x+h\right) -\cos x}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\cos x\cos
h-\sin x\sin h-\cos x}{h} \\
&=&\lim\limits_{h\rightarrow 0}\left( \dfrac{\cos x\cos h-\cos x}{h}-\dfrac{%
\sin x\sin h}{h}\right) =\lim\limits_{h\rightarrow 0}\dfrac{\cos x\left(
\cos h-1\right) }{h}-\lim\limits_{h\rightarrow 0}\dfrac{\sin x\sin h}{h} \\
&=&\cos x\lim\limits_{h\rightarrow 0}\dfrac{\cos h-1}{h}-\sin
x\lim\limits_{h\rightarrow 0}\dfrac{\sin h}{h}=\cos x\cdot 0-\sin x\cdot
1=-\sin x
\end{eqnarray*}%
\bigskip \pagebreak

\fbox{Theorem 3 and 4: \ $\dfrac{d}{dx}\left( \tan x\right) =\sec ^{2}x=\tan
^{2}x+1$ and $\dfrac{d}{dx}\left( \cot x\right) =-\csc ^{2}x=-\cot ^{2}x-1$.}%
\bigskip

Proof: \ We write $\tan x=\dfrac{\sin x}{\cos x}$ and apply the quotient
rule.%
\begin{equation*}
\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x}\right) =\dfrac{\left( \dfrac{d}{dx%
}\sin x\right) \cos x-\left( \dfrac{d}{dx}\cos x\right) \sin x}{\cos ^{2}x}=%
\dfrac{\cos x\cos x-\left( -\sin x\right) \sin x}{\cos ^{2}x}=\dfrac{\cos
^{2}x+\sin ^{2}x}{\cos ^{2}x}=\dfrac{1}{\cos ^{2}x}=\sec ^{2}x
\end{equation*}%
We will now prove $\dfrac{1}{\cos ^{2}x}=\tan ^{2}x+1,$ which is a very
important connection. \ Looking at the previous computation, 
\begin{equation*}
\dfrac{1}{\cos ^{2}x}=\dfrac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=\dfrac{\cos
^{2}x}{\cos ^{2}x}+\dfrac{\sin ^{2}x}{\cos ^{2}x}=1+\tan ^{2}x
\end{equation*}%
The proof for$\dfrac{d}{dx}\left( \cot x\right) =-\cot ^{2}x-1=-\csc ^{2}x$
is very similar. \ We apply the quotient rule.%
\begin{eqnarray*}
\dfrac{d}{dx}\left( \cot x\right) &=&\dfrac{d}{dx}\left( \dfrac{\cos x}{\sin
x}\right) =\dfrac{\left( \dfrac{d}{dx}\cos x\right) \sin x-\cos x\left( 
\dfrac{d}{dx}\sin x\right) }{\sin ^{2}x}=\dfrac{-\sin x\sin x-\cos x\left(
\cos x\right) }{\sin ^{2}x}=\dfrac{-\sin ^{2}x-\cos ^{2}x}{\sin ^{2}x} \\
&=&-\dfrac{1}{\sin ^{2}x}=-\csc ^{2}x
\end{eqnarray*}%
Also, 
\begin{equation*}
\dfrac{-\sin ^{2}x-\cos ^{2}x}{\sin ^{2}x}=\dfrac{-\sin ^{2}x}{\sin ^{2}x}-%
\dfrac{\cos ^{2}x}{\sin ^{2}x}=-1-\cot ^{2}x
\end{equation*}%
\bigskip

\fbox{Theorems \ 5 and 6: $\dfrac{d}{dx}\left( \sec x\right) =\sec x\tan x$
\ and \ $\dfrac{d}{dx}\left( \csc x\right) =-\csc x\cot x$}\bigskip

Proof: \ We write $\sec x=\dfrac{1}{\cos x}=\left( \cos x\right) ^{-1}$ and
apply the chain rule.%
\begin{equation*}
\dfrac{d}{dx}\left( \sec x\right) =\dfrac{d}{dx}\left( \left( \cos x\right)
^{-1}\right) =-1\left( \cos x\right) ^{-2}\left( \dfrac{d}{dx}\left( \cos
x\right) \right) =\dfrac{-1}{\cos ^{2}x}\left( -\sin x\right) =\dfrac{\sin x%
}{\cos ^{2}x}=\dfrac{1}{\cos x}\cdot \dfrac{\sin x}{\cos x}=\sec x\tan x
\end{equation*}

The proof for\ $\dfrac{d}{dx}\csc x$ is virtually identical: \ we apply the
chain rule.%
\begin{equation*}
\dfrac{d}{dx}\left( \csc x\right) =\dfrac{d}{dx}\left( \left( \sin x\right)
^{-1}\right) =-1\left( \sin x\right) ^{-2}\left( \dfrac{d}{dx}\sin x\right) =%
\dfrac{-1}{\sin ^{2}x}\cos x=-\dfrac{\cos x}{\sin ^{2}x}=-\dfrac{\cos x}{%
\sin x}\dfrac{1}{\sin x}=-\cot x\csc x
\end{equation*}%
\bigskip

Note: why do we prefer the form $\sec x\tan x$ over the form $\dfrac{\sin x}{%
\cos ^{2}x}$? \ One of the reasons is the adventage we'll see in
differentiating the inverse functions $\sec ^{-1}x$ and $\csc ^{-1}x$%
.\pagebreak

\fbox{Theorems 7 and 8: \ \ $\dfrac{d}{dx}\left( \sin ^{-1}x\right) =\dfrac{1%
}{\sqrt{1-x^{2}}}$ and $\dfrac{d}{dx}\left( \cos ^{-1}x\right) =-\dfrac{1}{%
\sqrt{1-x^{2}}}$}$\medskip $

Proof: \ Recall that when we compose a function $f$ with its inverse $%
f^{-1,}\,$the result is always the same function, (also called the identity
function, $id\left( x\right) =x$)%
\begin{equation*}
f\left( f^{-1}\left( x\right) \right) =x
\end{equation*}%
We will state this fact for $f\left( x\right) =\sin x$ and differentiate
both sides of the equation. \ For the left-hand side, we use the chain rule.%
\begin{eqnarray*}
\sin \left( \sin ^{-1}x\right) &=&x \\
\cos \left( \sin ^{-1}x\right) \cdot \dfrac{d}{dx}\sin ^{-1}x &=&1\text{ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ divide by }\cos \left( \sin ^{-1}x\right) \\
\dfrac{d}{dx}\sin ^{-1}x &=&\dfrac{1}{\cos \left( \sin ^{-1}x\right) }
\end{eqnarray*}%
We now need to simplify $\cos \left( \sin ^{-1}x\right) $. \ \ We will
present two methods to simplify this expression.

Method 1. \ We first introduce a new variable, $\beta $. \ Let $\beta =\sin
^{-1}x$. \ This means that $-\dfrac{\pi }{2}<\beta <\dfrac{\pi }{2}$, and $%
\sin \beta =x$. \ We need to simplify $\cos \underset{\beta }{\underbrace{%
\left( \sin ^{-1}x\right) }}=\cos \beta $. \ Since $\sin ^{2}\beta +\cos
^{2}\beta =1,$%
\begin{equation*}
\cos \beta =\pm \sqrt{1-\sin ^{2}\beta }=\pm \sqrt{1-x^{2}}
\end{equation*}%
Since $-\dfrac{\pi }{2}<\beta <\dfrac{\pi }{2}$, $\cos \beta $ is positive
and so $\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}$. \ 

Method 2. \ We first introduce a new variable, $\beta $. \ Let $\beta =\sin
^{-1}x$. \ This means that $-\dfrac{\pi }{2}<\beta <\dfrac{\pi }{2}$, and $%
\sin \beta =x$. \ Now the goal is to simplify $\cos \underset{\beta }{%
\underbrace{\left( \sin ^{-1}x\right) }}=\cos \beta $. \ 

%TCIMACRO{%
%\TeXButton{2col begin}{\begin{multicols}{2}
%}}%
%BeginExpansion
\begin{multicols}{2}
%
%EndExpansion
%TCIMACRO{%
%\TeXButton{box start - 3.75in}{\begin{boxedminipage}{3.75in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}
%}}%
%BeginExpansion
\begin{boxedminipage}{3.75in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}
%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

We will use a right triangle to find the expression - up to its sign. $\ $We
first draw a right triangle in which $\sin \beta =x=\dfrac{x}{1}$.\FRAME{%
dtbpF}{1.9182in}{0.9132in}{0pt}{}{}{referencetriangle1a.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.9182in;height 0.9132in;depth
0pt;original-width 2.4405in;original-height 1.1467in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle1a.bmp';file-properties "XNPEU";}}

Next we use the Pythagoream Theorem to find the \newline
missing side to be $\sqrt{1-x^{2}}$.\vspace{0.16in}%
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}} }%
%BeginExpansion
\end{boxedminipage}
%EndExpansion
\ 

%TCIMACRO{%
%\TeXButton{box start - 3.75in}{\begin{boxedminipage}{3.75in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}}}%
%BeginExpansion
\begin{boxedminipage}{3.75in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

The adventage of this method is that now we can read \textit{any}
trigonometric function value of $\beta =\sin ^{-1}x$ using this right
triangle.\FRAME{dtbpF}{1.6907in}{0.972in}{0pt}{}{}{referencetriangle1b.bmp}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.6907in;height 0.972in;depth
0pt;original-width 2.0133in;original-height 1.1467in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle1b.bmp';file-properties "XNPEU";}}From the triangle, 
\begin{equation*}
\cos \beta =\cos \left( \sin ^{-1}x\right) =\dfrac{\sqrt{1-x^{2}}}{1}=\sqrt{%
1-x^{2}}
\end{equation*}
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}%
%BeginExpansion
\end{boxedminipage}%
%EndExpansion
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

The answer at this point is really $\pm \sqrt{1-x^{2}}$ as the triangle gave
us the answer only up to a sign. \ For the sign, we need to argue using the
location of $\beta $ on the unit circle. \ Since $\beta =\sin ^{-1}x$, $-%
\dfrac{\pi }{2}\leq \beta \leq \dfrac{\pi }{2}$. \ Thus $\beta $ is in the
first or in the fourth quadrant. \ In both cases, cosine is positive, thus $%
\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}$.

Thus $\dfrac{d}{dx}\left( \sin ^{-1}x\right) =\dfrac{1}{\cos \left( \sin
^{-1}x\right) }=\dfrac{1}{\sqrt{1-x^{2}}}$.\pagebreak

The proof for \ $\dfrac{d}{dx}\left( \cos ^{-1}x\right) =-\dfrac{1}{\sqrt{%
1-x^{2}}}$ is virtually identical.

Proof: \ Recall that when we compose a function $f$ with its inverse $%
f^{-1,}\,$the result is always the same function.%
\begin{equation*}
f\left( f^{-1}\left( x\right) \right) =x
\end{equation*}%
We will state this fact for $f\left( x\right) =\cos x$ and differentiate
both sides of the equation. \ For the left-hand side, we use the chain rule.%
\begin{eqnarray*}
\cos \left( \cos ^{-1}x\right) &=&x \\
-\sin \left( \cos ^{-1}x\right) \cdot \dfrac{d}{dx}\left( \cos ^{-1}x\right)
&=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }\sin \left( \cos
^{-1}x\right) \\
\dfrac{d}{dx}\left( \cos ^{-1}x\right) &=&-\dfrac{1}{\sin \left( \cos
^{-1}x\right) }
\end{eqnarray*}%
We now need to simplify the expression $\sin \left( \cos ^{-1}x\right) $. \
\ We will present two methods for this.

Method 1. \ Let $\alpha =\cos ^{-1}x$. \ Then $x=\cos \alpha $ and $\alpha $
is between $0$ and $\pi $. \ 
\begin{equation*}
\sin \left( \underset{\alpha }{\underbrace{\cos ^{-1}x}}\right) =\sin \alpha
=\pm \sqrt{1-\cos ^{2}\alpha }=\pm \sqrt{1-x^{2}}
\end{equation*}%
Since $\alpha $ is between $0$ and $\pi $, $\sin \alpha $ is positive and so 
$\sin \alpha =\sqrt{1-x^{2}}$.

Method 2. $\ $Let $\alpha =\cos ^{-1}x$. \ Then $x=\cos \alpha $ and $\alpha 
$ is between $0$ and $\pi $. \ 

%TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}%
%BeginExpansion
\begin{multicols}{2}%
%EndExpansion
%TCIMACRO{%
%\TeXButton{box start - 3.75in}{\begin{boxedminipage}{3.75in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}}}%
%BeginExpansion
\begin{boxedminipage}{3.75in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

We first draw a triangle in which $\cos \alpha =x=\dfrac{x}{1}$. \ Please
note that every time we approach such a trigonometric question using a right
triangle, our answer would be accurate up to sign - for the sign we would
have to argue separately.\FRAME{dtbpF}{1.772in}{0.7455in}{0pt}{}{}{%
referencetriangle2a.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
1.772in;height 0.7455in;depth 0pt;original-width 2.4604in;original-height
1.0196in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'referencetriangle2a.bmp';file-properties "XNPEU";}}

We find the missing side via the Pythagoream Theorem: $\sqrt{1-x^{2}}$. \ 
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}%
%BeginExpansion
\end{boxedminipage}%
%EndExpansion

%TCIMACRO{%
%\TeXButton{box start - 3.75in}{\begin{boxedminipage}{3.75in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}}}%
%BeginExpansion
\begin{boxedminipage}{3.75in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

Now we can read \textit{any} trigonometric function value using this
triangle.\FRAME{dtbpF}{2.2018in}{0.9132in}{0pt}{}{}{referencetriangle2b.bmp}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 2.2018in;height 0.9132in;depth
0pt;original-width 2.4933in;original-height 1.0196in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle2b.bmp';file-properties "XNPEU";}}

Now we read sine from the triangle: 
\begin{equation*}
\sin \alpha =\sin \left( \cos ^{-1}x\right) =\dfrac{\sqrt{1-x^{2}}}{1}=\sqrt{%
1-x^{2}}
\end{equation*}%
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}%
%BeginExpansion
\end{boxedminipage}%
%EndExpansion
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

The answer at this point is really $\pm \sqrt{1-x^{2}}$ as the triangle gave
us the answer only up to a sign.\vspace{0.02in} \ For the sign, we need to
argue using the location of $\alpha $ on the unit circle. \ Since $\alpha
=\cos ^{-1}x$, $0\leq \alpha \leq \pi $. \ Thus $\alpha $ is in the\vspace{%
0.02in} first or in the second quadrant. \ In both cases, sine is positive,
thus $\sin \left( \cos ^{-1}x\right) =\sqrt{1-x^{2}}$.

Consequently, $\dfrac{d}{dx}\left( \cos ^{-1}x\right) =-\dfrac{1}{\sin
\left( \cos ^{-1}x\right) }=-\dfrac{1}{\sqrt{1-x^{2}}}$

%TCIMACRO{%
%\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}}}%
%BeginExpansion
\begin{boxedminipage}{7.5in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

\FRAME{itbpF}{0.7922in}{0.6287in}{0.1807in}{}{}{tree.jpg}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"USEDEF";valid_file "F";width 0.7922in;height 0.6287in;depth
0.1807in;original-width 4.4685in;original-height 3.531in;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'../../../../../../aaaaa_book143/tree.jpg';file-properties "XNPEU";}} 
{\Large Enrichment} \ 

If $\dfrac{d}{dx}\left( \sin ^{-1}x\right) $ and $\dfrac{d}{dx}\left( \cos
^{-1}x\right) $ are opposites, then what can be said about the function $%
f\left( x\right) =\sin ^{-1}x+\cos ^{-1}x$?%
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}%
%BeginExpansion
\end{boxedminipage}%
%EndExpansion
\pagebreak 

\fbox{Theorems 9 and 10: $\ \dfrac{d}{dx}\left( \tan ^{-1}x\right) =\dfrac{1%
}{x^{2}+1}$ and $\dfrac{d}{dx}\left( \cot ^{-1}x\right) =-\dfrac{1}{x^{2}+1}$%
}\vspace{0.16in}

Proof: \ Recall that $\dfrac{d}{dx}\left( \tan x\right) =\sec ^{2}x=\tan
^{2}x+1$. \ Also recall that when we compose a function $f$ with its inverse 
$f^{-1,}\,$the result is always the same function.%
\begin{equation*}
f\left( f^{-1}\left( x\right) \right) =x
\end{equation*}%
We will state this fact for $f\left( x\right) =\tan x$ and differentiate
both sides of the equation. \ For the left-hand side, we use the chain rule.%
\begin{eqnarray*}
\tan \left( \tan ^{-1}x\right) &=&x \\
\sec ^{2}\left( \tan ^{-1}x\right) \cdot \dfrac{d}{dx}\left( \tan
^{-1}x\right) &=&1 \\
\left( \tan ^{2}\left( \tan ^{-1}x\right) +1\right) \cdot \dfrac{d}{dx}%
\left( \tan ^{-1}x\right) &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ }\tan \left( \tan ^{-1}x\right) =x \\
\left( x^{2}+1\right) \cdot \dfrac{d}{dx}\left( \tan ^{-1}x\right) &=&1\text{
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }x^{2}+1 \\
\dfrac{d}{dx}\left( \tan ^{-1}x\right) &=&\dfrac{1}{x^{2}+1}
\end{eqnarray*}

The proof for $\dfrac{d}{dx}\left( \cot ^{-1}x\right) =-\dfrac{1}{x^{2}+1}$
is virtually identical. \ We compose the function $\cot x$ with its inverse $%
\cot ^{-1}x\,$ and differentiate. \ Recall that $\dfrac{d}{dx}\cot x=-\cot
^{2}x-1$%
\begin{eqnarray*}
\cot \left( \cot ^{-1}x\right) &=&x \\
\left( -\cot ^{2}\left( \cot ^{-1}x\right) -1\right) \cdot \dfrac{d}{dx}%
\left( \cot ^{-1}x\right) &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }%
\cot \left( \cot ^{-1}x\right) =x \\
\left( -x^{2}-1\right) \cdot \dfrac{d}{dx}\left( \cot ^{-1}x\right) &=&1%
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }-x^{2}-1 \\
\dfrac{d}{dx}\left( \cot ^{-1}x\right) &=&\dfrac{1}{-x^{2}-1}=-\dfrac{1}{%
x^{2}+1}
\end{eqnarray*}%
$\bigskip \bigskip $

\fbox{Theorem 11 and 12: $\ \dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1%
}{\left\vert x\right\vert \sqrt{x^{2}-1}}$ \ and \ $\dfrac{d}{dx}\left( \csc
^{-1}x\right) =-\dfrac{1}{\left\vert x\right\vert \sqrt{x^{2}-1}}$}

Proof: \ We compose the function $\sec x$ with its inverse $\sec ^{-1}x$ and
differentiate. \ Recall that \newline
$\dfrac{d}{dx}\left( \sec x\right) =\sec x\tan x$.%
\begin{eqnarray*}
\sec \left( \sec ^{-1}x\right) &=&x \\
\sec \left( \sec ^{-1}x\right) \tan \left( \sec ^{-1}x\right) \cdot \dfrac{d%
}{dx}\left( \sec ^{-1}x\right) &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ }\sec \left( \sec ^{-1}x\right) =x \\
x\tan \left( \sec ^{-1}x\right) \cdot \dfrac{d}{dx}\left( \sec ^{-1}x\right)
&=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }x\tan \left(
\sec ^{-1}x\right) \\
\dfrac{d}{dx}\left( \sec ^{-1}x\right) &=&\dfrac{1}{x\tan \left( \sec
^{-1}x\right) }
\end{eqnarray*}%
We now just need to simplify the expression $\tan \left( \sec ^{-1}x\right) $%
.

Method 1. \ To simplify $\tan \left( \sec ^{-1}x\right) $, we introduce a
new variable $\alpha $. \ Let $\alpha =\sec ^{-1}x$. \ Then we have $\tan 
\underset{\alpha }{\left( \underbrace{\sec ^{-1}x}\right) }=\tan \alpha $
where $\sec \alpha =x$ and $\alpha $ is between $0$ and $\pi $. \ Recall
that $\sec ^{2}\alpha =\tan ^{2}\alpha +1$. \ If we don't have this formula
memorized, we can easily derive it from the Pythagorean identity.%
\begin{eqnarray*}
\sin ^{2}\alpha +\cos ^{2}\alpha &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide
by }\cos ^{2}x \\
\dfrac{\sin ^{2}\alpha }{\cos ^{2}\alpha }+\dfrac{\cos ^{2}\alpha }{\cos
^{2}\alpha } &=&\dfrac{1}{\cos ^{2}\alpha } \\
\tan ^{2}\alpha +1 &=&\sec ^{2}\alpha \\
\tan \alpha &=&\pm \sqrt{\sec ^{2}\alpha -1}
\end{eqnarray*}%
Method 2. \ \ To simplify $\tan \left( \sec ^{-1}x\right) $, we introduce a
new variable $\alpha $. \ Let $\alpha =\sec ^{-1}x$. \ Then $\sec \alpha =x$
and $\alpha $ is between $0$ and $\pi .$ \ Then we need to compute $\tan
\alpha $. \ 

%TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}%
%BeginExpansion
\begin{multicols}{2}%
%EndExpansion
%TCIMACRO{%
%\TeXButton{box start - 3.75in}{\begin{boxedminipage}{3.75in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}}}%
%BeginExpansion
\begin{boxedminipage}{3.75in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

We draw a right triangle in which $\sec \alpha =x=\dfrac{x}{1}$.\FRAME{dtbpF%
}{1.932in}{0.8518in}{0pt}{}{}{referencetriangle3a.bmp}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"USEDEF";valid_file "F";width 1.932in;height 0.8518in;depth
0pt;original-width 2.4405in;original-height 1.0603in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle3a.bmp';file-properties "XNPEU";}}

We find the missing side via the Pythagoream Theorem: $\sqrt{x^{2}-1}$. \ 
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}%
%BeginExpansion
\end{boxedminipage}%
%EndExpansion

%TCIMACRO{%
%\TeXButton{box start - 3.75in}{\begin{boxedminipage}{3.75in}
%\setlength{\fboxrule}{15pt}
%\setlength{\fboxsep}{15pt}}}%
%BeginExpansion
\begin{boxedminipage}{3.75in}
\setlength{\fboxrule}{15pt}
\setlength{\fboxsep}{15pt}%
%EndExpansion
%TCIMACRO{\TeXButton{white}{\color{white}}}%
%BeginExpansion
\color{white}%
%EndExpansion
.%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

Now we can read \textit{any} trigonometric function value using this
triangle.\FRAME{dtbpF}{1.9597in}{0.864in}{0pt}{}{}{referencetriangle3b.bmp}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.9597in;height 0.864in;depth
0pt;original-width 2.4405in;original-height 1.0603in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle3b.bmp';file-properties "XNPEU";}}

Now we read from the triangle: 
\begin{equation*}
\tan \alpha =\tan \left( \sec ^{-1}x\right) =\dfrac{\sqrt{x^{2}-1}}{1}=\sqrt{%
x^{2}-1}
\end{equation*}%
%TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}%
%BeginExpansion
\end{boxedminipage}%
%EndExpansion
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

The answer at this point is really $\pm \sqrt{x^{2}-1}$ as the triangle gave
us the answer only up to a sign.\vspace{0.02in} \ Thus the derivative is 
\begin{equation*}
\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{x\left( \pm \sqrt{x^{2}-1}%
\right) }=\pm \dfrac{1}{x\sqrt{x^{2}-1}}
\end{equation*}%
We now need to figure out the sign of the derivative. \ From the graph of $%
\sec ^{-1}x$ we can see that it is strictly increasing on both intervals
making up its domain, thus the derivative is always positive. \ If $x$ is
positive, then $\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{x\sqrt{%
x^{2}-1}}$ and if $x$ is negative, then $\dfrac{d}{dx}\left( \sec
^{-1}x\right) =-\dfrac{1}{x\sqrt{x^{2}-1}}$. \ \ This can be expressed in a
shorter form as 
\begin{equation*}
\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{\left\vert x\right\vert 
\sqrt{x^{2}-1}}
\end{equation*}%
The proof for $\dfrac{d}{dx}\left( \csc ^{-1}x\right) =-\dfrac{1}{\left\vert
x\right\vert \sqrt{x^{2}-1}}$ is virtually identical. \ \ As before, we
compose the function $\csc x$ with its inverse and differentiate. \ Recall
that $\dfrac{d}{dx}\left( \csc x\right) =-\csc x\cot x$%
\begin{eqnarray*}
\csc \left( \csc ^{-1}x\right) &=&x \\
-\csc \left( \csc ^{-1}x\right) \cot \left( \csc ^{-1}x\right) \cdot \dfrac{d%
}{dx}\left( \csc ^{-1}x\right) &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ }\csc \left( \csc ^{-1}x\right) =x \\
-x\cot \left( \csc ^{-1}x\right) \cdot \dfrac{d}{dx}\left( \csc
^{-1}x\right) &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }%
-x\cot \left( \csc ^{-1}x\right) \\
\dfrac{d}{dx}\left( \csc ^{-1}x\right) &=&-\dfrac{1}{x\cot \left( \csc
^{-1}x\right) }
\end{eqnarray*}

We need to simplify $\cot \left( \csc ^{-1}x\right) $. \ Let $\alpha =\csc
^{-1}x$. \ \ $\cot \underset{\alpha }{\left( \underbrace{\csc ^{-1}x}\right) 
}=\cot \alpha $ where $\csc \alpha =x$ and $\alpha $ is between $-\dfrac{\pi 
}{2}$ and $\dfrac{\pi }{2}$.

Method 1. \ We start with the Pythagorean identity and divide both sides by $%
\sin ^{2}\alpha $.

\begin{eqnarray*}
\sin ^{2}\alpha +\cos ^{2}\alpha &=&1\text{ \ \ \ \ \ \ \ \ divide by }\sin
^{2}x \\
\dfrac{\sin ^{2}\alpha }{\sin ^{2}\alpha }+\dfrac{\cos ^{2}\alpha }{\sin
^{2}\alpha } &=&\dfrac{1}{\sin ^{2}\alpha }\text{ } \\
1+\cot ^{2}\alpha &=&\csc ^{2}\alpha \\
\cot ^{2}\alpha &=&\csc ^{2}\alpha -1 \\
\cot \alpha &=&\pm \sqrt{\csc ^{2}\alpha -1}
\end{eqnarray*}

Method 2. \ We draw a right triangle in which $\csc \alpha =x=\dfrac{x}{1}$.
\ \FRAME{dtbpF}{1.7513in}{0.7593in}{0pt}{}{}{referencetriangle4a.bmp}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.7513in;height 0.7593in;depth
0pt;original-width 2.0799in;original-height 0.8864in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle4a.bmp';file-properties "XNPEU";}}We find the missing side
using the Pythagorean Theorem and read the desired trigonometric function
value.\FRAME{dtbpF}{1.8619in}{1.1026in}{0pt}{}{}{referencetriangle4b.bmp}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.8619in;height 1.1026in;depth
0pt;original-width 2.0799in;original-height 1.2194in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'referencetriangle4b.bmp';file-properties "XNPEU";}}

Thus the derivative is 
\begin{equation*}
\dfrac{d}{dx}\left( \csc ^{-1}x\right) =-\dfrac{1}{x\left( \pm \sqrt{x^{2}-1}%
\right) }=\pm \dfrac{1}{x\sqrt{x^{2}-1}}
\end{equation*}%
From the graph of $\csc ^{-1}x$ we can see that it is strictly decreasing on
both intervals of its domain, thus the derivative is always negative. \ If $x
$ is positive, then $\dfrac{d}{dx}\left( \csc ^{-1}x\right) =-\dfrac{1}{x%
\sqrt{x^{2}-1}}$ and if $x$ is negative, then $\dfrac{d}{dx}\left( \csc
^{-1}x\right) =\dfrac{1}{x\sqrt{x^{2}-1}}$. \ \ This can be expressed in a
shorter form as 
\begin{equation*}
\dfrac{d}{dx}\left( \csc ^{-1}x\right) =-\dfrac{1}{\left\vert x\right\vert 
\sqrt{x^{2}-1}}
\end{equation*}

$\vspace{0.3in}$

\href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html%
}{For more documents like this, visit our page at\
http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail
questions or comments to mhidegkuti@ccc.edu.}

\end{document}