\magnification=1100
\overfullrule0pt

\input amssym.def
\input prepictex
\input pictex
\input postpictex


% ********************* Definitions ************************************

%\def\widetilde{\mathaccent"0365 }

\def\CC{{\Bbb C}}
\def\FF{{\Bbb F}}
\def\HH{{\Bbb H}}
\def\NN{{\Bbb N}}
\def\OO{{\Bbb O}}
\def\QQ{{\Bbb Q}}
\def\RR{{\Bbb R}}
\def\ZZ{{\Bbb Z}}

\def\cA{{\cal A}}

\def\cB{{\cal B}}
\def\cR{{\cal R}}
\def\cZ{{\cal Z}}
\def\cC{{\cal C}}
\def\cR{{\cal R}}

\def\fb{\frak{b}}
\def\fg{\frak{g}}
\def\fh{\frak{h}}
\def\fn{\frak{n}}

\def\Card{\hbox{Card}}
\def\End{\hbox{End}}
\def\Hom{\hbox{Hom}}
\def\Ind{\hbox{Ind}}
\def\Id{\hbox{Id}}
\def\codim{\hbox{codim}}

\def\diag{\hbox{diag}}
\def\id{\hbox{id}}
\def\im{\hbox{im}}
\def\tr{\hbox{tr}}
\def\Tr{\hbox{Tr}}

\def\sech{\hbox{sech}}
\def\csch{\hbox{csch}}
\def\coth{\hbox{coth}}

\def\fbox{}


% ********************* FONTS ************************************

\font\smallcaps=cmcsc10
\font\titlefont=cmr10 scaled \magstep1
\font\titlefontbold=cmbx10 scaled \magstep1
\font\titlesubfont=cmr10 scaled \magstep1
\font\sectionfont=cmbx10
\font\tinyrm=cmr10 at 8pt

% ******************** SECTION HEADERS ***************************

\newcount\sectno
\newcount\subsectno
\newcount\resultno

\def\section #1. #2\par{
\sectno=#1
\resultno=0
\bigskip\noindent{\sectionfont #1.  #2}~\medbreak}

\def\subsection #1\par{\bigskip\noindent{\it  #1} \medbreak}

%******************* MATHEMATICAL LABELS **************************

\def\prop{ \global\advance\resultno by 1
\medskip\noindent{\bf Proposition \the\sectno.\the\resultno. }\sl}
\def\lemma{ \global\advance\resultno by 1
\medskip\noindent{\bf Lemma \the\sectno.\the\resultno. }
\sl}
\def\fact{ \global\advance\resultno by 1
\medskip\noindent{\bf Fact \the\sectno.\the\resultno. }
\sl}

\def\remark{ \global\advance\resultno by 1
\medskip\noindent{\bf Remark \the\sectno.\the\resultno. }}
\def\example{ \global\advance\resultno by 1
\medskip\noindent{\bf Example \the\sectno.\the\resultno. }\sl}
\def\cor{ \global\advance\resultno by 1
\medskip\noindent{\bf Corollary \the\sectno.\the\resultno. }\sl}
\def\thm{ \global\advance\resultno by 1
\medskip\noindent{\bf Theorem \the\sectno.\the\resultno. }\sl}
\def\defn{ \global\advance\resultno by 1
\medskip\noindent{\it Definition \the\sectno.\the\resultno. }\slrm}
\def\endthm{\rm\medskip}
\def\thmend{\rm\medskip}
\def\endlemma{\rm\medskip}
\def\endfact{\rm\medskip}
\def\endexample{\rm\medskip}
\def\endprop{\rm\medskip}
\def\endcor{\rm\medskip}
\def\pf{\rm\smallskip\noindent{\it Proof. }}
\def\endpf{\qed\hfil\medskip}
\def\pfend{\qed\hfil\medskip}
\def\note{\smallbreak\noindent{Note:}}
\def\enddefn{\rm\medskip}

%Homemade Struts:
\newbox\strutAbox
\setbox\strutAbox=\hbox{\vrule height 12pt depth6pt width0pt}
\def\strutA{\relax\copy\strutAbox}
\newbox\strutBbox
\setbox\strutBbox=\hbox{\vrule height 10pt depth5pt width0pt}
\def\strutB{\relax\copy\strutBbox}
\newbox\strutDbox
\setbox\strutDbox=\hbox{\vrule height 11pt depth5pt width0pt}
\def\strutD{\relax\copy\strutDbox}
%high strut:
\newbox\strutHbox
\setbox\strutHbox=\hbox{\vrule height 11pt depth1pt width0pt}
\def\strutH{\relax\copy\strutHbox}
%low strut:
\newbox\strutLbox
\setbox\strutLbox=\hbox{\vrule height 1pt depth5pt width0pt}
\def\strutL{\relax\copy\strutLbox}



% hack to ignore lots of typed stuff....
\def\ignore#1{\relax}

% ******************  QED SIGNS  *********************************

\def\qed{\hbox{\hskip 1pt\vrule width4pt height 6pt depth1.5pt \hskip 1pt}}

\def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
\hbox{\vrule width.#2pt height#1pt \kern#1pt
\vrule width.2pt}
\hrule height.2pt}}}}
\def\square{\mathchoice\sqr54\sqr54\sqr{3.5}3\sqr{2.5}3}
\def\whiteslug{\bf $ \square $ \rm}  % open square


%*************** EQUATIONS WITH NUMBERS **************

\def\formula{\global\advance\resultno by 1
\eqno{(\the\sectno.\the\resultno)}}
\def\formulano{\global\advance\resultno by 1 (\the\sectno.\the\resultno)}
\def\tableno{\global\advance\resultno by 1
\the\sectno.\the\resultno. }
\def\lformula{\global\advance\resultno by 1
\leqno(\the\sectno.\the\resultno)}

%************Commutative diagrams**********************

\def\mapright#1{\smash{\mathop
        {\longrightarrow}\limits^{#1}}}

\def\mapleftright#1{\smash{\mathop
        {\longleftrightarrow}\limits^{#1}}}


\def\mapsrightto#1{\smash{\mathop
        {\longmapsto}\limits^{#1}}}

\def\mapleft#1{\smash{
   \mathop{\longleftarrow}\limits^{#1}}}

\def\mapdown#1{\Big\downarrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}

\def\lmapdown#1{{\hbox{$\scriptstyle#1$}}
\llap {$\vcenter{\hbox{\Big\downarrow}}$} }

\def\mapup#1{\Big\uparrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\mapne#1{\Big\nearrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\mapse#1{
%{$\vcenter{
\hbox{$\scriptstyle#1$}
%$}
\rlap{ $\vcenter{\hbox{$\searrow$}}$ }  }
\def\mapnw#1{\Big\nwarrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\mapsw#1{
%\Big
\swarrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}


%********** DATING ******************************************
\def\monthname {\ifcase\month\or January\or February\or March\or April\or
May\or June\or
July\or August\or September\or October\or November\or December\fi}

\newcount\mins  \newcount\hours  \hours=\time \mins=\time
\def\now{\divide\hours by60 \multiply\hours by60 \advance\mins by-\hours
     \divide\hours by60         % NOTE: \divide only gives integer answers.
     \ifnum\hours>12 \advance\hours by-12
       \number\hours:\ifnum\mins<10 0\fi\number\mins\ P.M.\else
       \number\hours:\ifnum\mins<10 0\fi\number\mins\ A.M.\fi}
\def\today {\monthname\ \number\day, \number\year}


%**************** PAGE HEADERS *************************

\nopagenumbers
\def\runningtitle{\smallcaps inverse functions}
\headline={\ifnum\pageno>1\eoheadline\else\firstheadline\fi}
\def\names{\smallcaps a.\ ram}
%\def\firstheadline{\noindent Preliminary Draft \hfill  \today}
\def\firstheadline{}
\def\eoheadline{\ifodd\pageno\oddheadline\else\evenheadline\fi}
\def\oddheadline{\tenrm\hfil\runningtitle\hfil\folio}
\def\evenheadline{\tenrm\folio\hfil{\names}\hfil}


%**************** TITLE *************************
\vphantom{$ $}  %My kludge to get the first page to move down a bit
\vskip.75truein
\centerline{\titlefont Inverse functions and their derivatives}
\bigskip
\centerline{\rm Arun Ram}
%${}^\ast$ 
\centerline{Department of Mathematics}
\centerline{University of Wisconsin, Madison}
\centerline{Madison, WI 53706 USA}
\centerline{{\tt ram@math.wisc.edu}}
\medskip
\centerline{Version: \today}

%\footnote{}{\tinyrm 
%${}^\ast$ 
%Research partially supported by the National Security Agency
%and by EPSRC Grant GR K99015 at the Newton Institute for
%Mathematical Sciences.}
%\footnote{}{\tinyrm
%\noindent {Keywords:} exponential functions, trigonometric functions}

\bigskip

%**************** ABSTRACT *************************
%\noindent{\bf Abstract.}



\bigskip\bigskip\noindent
$\sqrt{x}$ is the function that undoes $x^2$.  This means that
$$\sqrt{x^2}=x\qquad\hbox{and}\qquad (\sqrt{x})^2 = x.$$
$\ln x$ is the function that undoes $e^x$.  This means that
$$\ln(e^x)=x\qquad\hbox{and}\qquad e^{\ln x} = x.$$
$\sin^{-1} x$ is the function that undoes $\sin x$.  This means that
$$\sin^{-1}(\sin x)=x\qquad\hbox{and}\qquad \sin(\sin^{-1} x) = x.$$
$\cos^{-1} x$ is the function that undoes $\cos x$.  This means that
$$\cos^{-1}(\cos x)=x\qquad\hbox{and}\qquad \cos(\cos^{-1} x) = x.$$
$\tan^{-1} x$ is the function that undoes $\tan x$.  This means that
$$\tan^{-1}(\tan x)=x\qquad\hbox{and}\qquad \tan(\tan^{-1} x) = x.$$
$\cot^{-1} x$ is the function that undoes $\cot x$.  This means that
$$\cot^{-1}(\cot x)=x\qquad\hbox{and}\qquad \cot(\cot^{-1} x) = x.$$
$\sec^{-1} x$ is the function that undoes $\sec x$.  This means that
$$\sec^{-1}(\sec x)=x\qquad\hbox{and}\qquad \sec(\sec^{-1} x) = x.$$
$\csc^{-1} x$ is the function that undoes $\csc x$.  This means that
$$\csc^{-1}(\csc x)=x\qquad\hbox{and}\qquad \csc(\csc^{-1} x) = x.$$
$\log_a x$ is the function that undoes $a^x$.  This means that
$$\log_a(a^{\sqrt{7}\pi i \sin 32})=
\sqrt{7}\pi i \sin 32\qquad\hbox{and}\qquad 
a^{\log_a(\sqrt{7}\pi i \sin 32)} = \sqrt{7}\pi i \sin 32.$$

\smallskip\noindent
{\bf WARNING:} $\sin^{-1} x$ is VERY DIFFERENT from $(\sin x)^{-1}$.
For example,
$$\sin^{-1} 0 = \sin^{-1}(\sin 0) = 0,
\qquad\hbox{BUT}\qquad
(\sin 0)^{-1} = {1\over \sin 0} = {1\over 0} = \hbox{UNDEFINED}.
$$

\bigskip\noindent
{\bf Example:}  Explain why $\ln 1 = 0$.
$$\ln 1 = \ln(e^0)=0.$$

\bigskip\noindent
{\bf Example:}  Explain why $\ln(ab) = \ln a+ \ln b$.
$$\ln(ab) = \ln(e^{\ln a}\cdot e^{\ln b}) 
=\ln(e^{\ln a+\ln b}) = \ln a+\ln b.$$

\bigskip\noindent
{\bf Example:}  Explain why 
$\displaystyle{\ln\left({1\over a}\right) = -\ln a }$.
$$\ln\left({1\over a}\right) 
= \ln\left({1\over e^{\ln a}}\right) 
= \ln\left(e^{-\ln a}\right) 
= -\ln a.$$

\bigskip\noindent
{\bf Example:}  Explain why $\displaystyle{\ln\big(a^b\big) = b\ln a }$.
$$\ln(a^b) 
= \ln\left(\big(e^{\ln a}\big)^b\right) 
= \ln\left(e^{b\ln a}\right) 
= b\ln a.$$

Thus
$$\matrix{
e^0=1 &\quad &\hbox{turns into} &\quad &\ln 1=0,  \hfill\cr
\cr
e^xe^y=e^{x+y} &&\hbox{turns into} &&\ln(ab)=\ln a+\ln b,  \hfill\cr
\cr
\displaystyle{e^{-x}={1\over e^x}} 
&&\hbox{turns into} &&\displaystyle{\ln\left({1\over a}\right)=-\ln a},
\quad\hbox{and} \hfill \cr
\cr
\left(e^x\right)^y=e^{yx} &&\hbox{turns into} &&\ln(a^b)=b\ln a.  \hfill\cr
}$$

\bigskip\noindent
{\bf Example:}  Explain why $\displaystyle{ {d\ln x\over dx} = {1\over x} }$.
\bigskip
Since\quad $e^{\ln x} = x$,\qquad
$\displaystyle{ {d e^{\ln x}\over dx} = {dx\over dx} }$.
\medskip
So \quad $\displaystyle{
e^{\ln x}\, {d\ln x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{ x\, {d\ln x\over dx} = 1 }$.
\qquad\quad
So\quad $\displaystyle{ {d\ln x\over dx} = {1\over x} }$.



\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {d\sin^{-1} x\over dx} }$.
\bigskip\noindent
Since\quad $\sin(\sin^{-1} x) = x$,\qquad
$\displaystyle{ {d \sin(\sin^{-1} x)\over dx} = {dx\over dx} }$.
\medskip\noindent
So \quad $\displaystyle{
\cos(\sin^{-1} x)\, {d\sin^{-1} x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{{d\sin^{-1} x\over dx} = 
{1\over \cos(\sin^{-1} x)} }$.
\smallskip\noindent
So we would like to ``simplify'' $\cos(\sin^{-1} x)$.
\smallskip\noindent
Since $1-\cos^2(\sin^{-1} x) = \sin^2(\sin^{-1} x)$, 
\qquad 
$1-\big(\cos (\sin^{-1} x)\big)^2 = \big(\sin(\sin^{-1} x)\big)^2$. 
\smallskip\noindent
So \quad
$1-\big(\cos (\sin^{-1} x)\big)^2 = x^2$. 
\qquad\quad
So \quad
$1-x^2 =  \big(\cos (\sin^{-1} x)\big)^2$. 
\smallskip\noindent
So \quad
$\cos (\sin^{-1} x) = \sqrt{1-x^2}$. 
\qquad\quad
So \quad
$\displaystyle{{d\sin^{-1} x\over dx} = 
{1\over \cos(\sin^{-1} x)}
= {1\over \sqrt{1-x^2} } }$. 

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {d\cos^{-1} x\over dx} }$.
\bigskip\noindent
Since\quad $\cos(\cos^{-1} x) = x$,\qquad
$\displaystyle{ {d \cos(\cos^{-1} x)\over dx} = {dx\over dx} }$.
\medskip\noindent
So \quad $\displaystyle{
-\sin(\cos^{-1} x)\, {d\cos^{-1} x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{{d\cos^{-1} x\over dx} = 
{-1\over \sin(\cos^{-1} x)} }$.
\smallskip\noindent
So we would like to ``simplify'' $\sin(\cos^{-1} x)$.
\smallskip\noindent
Since $1-\sin^2(\cos^{-1} x) = \cos^2(\cos^{-1} x)$, 
\qquad 
$1-\big(\sin (\cos^{-1} x)\big)^2 = \big(\cos(\cos^{-1} x)\big)^2$. 
\smallskip\noindent
So \quad
$1-\big(\sin (\cos^{-1} x)\big)^2 = x^2$. 
\qquad\quad
So \quad
$1-x^2 =  \big(\sin (\cos^{-1} x)\big)^2$. 
\smallskip\noindent
So \quad
$\sin (\cos^{-1} x) = \sqrt{1-x^2}$. 
\qquad\quad
So \quad
$\displaystyle{{d\cos^{-1} x\over dx} = 
{-1\over \sin(\cos^{-1} x)}
= {-1\over \sqrt{1-x^2} } }$. 

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {d\tan^{-1} x\over dx} }$.
\bigskip\noindent
Since\quad $\tan(\tan^{-1} x) = x$,\qquad
$\displaystyle{ {d \tan(\tan^{-1} x)\over dx} = {dx\over dx} }$.
\medskip\noindent
So \quad $\displaystyle{
\sec^2(\tan^{-1} x)\, {d\tan^{-1} x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{{d\tan^{-1} x\over dx} = 
{1\over \sec^2(\tan^{-1} x)} }$.
\smallskip\noindent
So we would like to ``simplify'' $\sec^2(\tan^{-1} x)$.
\smallskip\noindent
Since \quad $\sin^2 x+\cos^2 x = 1$, \qquad\quad
\smallskip\noindent
\phantom{Since}\quad
$\displaystyle{
{\sin^2 x\over \cos^2 x}+{\cos^2 x\over \cos^2 x}
={1\over \cos^2 x}. }$
\smallskip\noindent
So \quad $\tan^2 x + 1 = \sec^2 x$.
\smallskip\noindent
So \quad $\sec^2(\tan^{-1} x)=\tan^2(\tan^{-1} x)+1
=\big(\tan (\tan^{-1} x)\big)^2 + 1 = x^2+1.$
\smallskip\noindent
So \quad
$\displaystyle{
{d\tan^{-1} x\over dx} = {1\over x^2+1} }$.

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {d\cot^{-1} x\over dx} }$.
\bigskip\noindent
Since\quad $\cot(\cot^{-1} x) = x$,\qquad
$\displaystyle{ {d \cot(\cot^{-1} x)\over dx} = {dx\over dx} }$.
\medskip\noindent
So \quad $\displaystyle{
-\csc^2(\cot^{-1} x)\, {d\cot^{-1} x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{{d\cot^{-1} x\over dx} = 
{-1\over \csc^2(\cot^{-1} x)} }$.
\smallskip\noindent
So we would like to ``simplify'' $\csc^2(\cot^{-1} x)$.
\smallskip\noindent
Since \quad $\sin^2 x+\cos^2 x = 1$, \qquad\quad
\smallskip\noindent
\phantom{Since}\quad
$\displaystyle{
{\sin^2 x\over \sin^2 x}+{\cos^2 x\over \sin^2 x}
={1\over \sin^2 x}. }$
\smallskip\noindent
So \quad $1+\cot^2 x = \csc^2 x$.
\smallskip\noindent
So \quad $\csc^2(\cot^{-1} x)=1+\cot^2(\cot^{-1} x)
=1+\big(\cot (\cot^{-1} x)\big)^2 = 1+x^2.$
\smallskip\noindent
So \quad
$\displaystyle{
{d\cot^{-1} x\over dx} = {-1\over 1+x^2} }$.

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {d\sec^{-1} x\over dx} }$.
\bigskip\noindent
Since\quad $\sec(\sec^{-1} x) = x$,\qquad
$\displaystyle{ {d \sec(\sec^{-1} x)\over dx} = {dx\over dx} }$.
\medskip\noindent
So \quad $\displaystyle{
\tan(\sec^{-1} x)\sec(\sec^{-1} x)\, {d\sec^{-1} x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{
\tan(\sec^{-1} x)\cdot x\cdot {d\sec^{-1} x\over dx} = 1 }$.
\smallskip\noindent
So \quad $\displaystyle{{d\sec^{-1} x\over dx} = 
{1\over x\tan(\sec^{-1} x)} }$.
\smallskip\noindent
So we would like to ``simplify'' $\tan(\sec^{-1} x)$.
\smallskip\noindent
Since \quad $\sin^2 x+\cos^2 x = 1$, \qquad\quad
\smallskip\noindent
\phantom{Since}\quad
$\displaystyle{
{\sin^2 x\over \cos^2 x}+{\cos^2 x\over \cos^2 x}
={1\over \cos^2 x}. }$
\smallskip\noindent
So \quad $\tan^2 x+1 = \sec^2 x$.
\smallskip\noindent
So \quad $\tan^2(\sec^{-1} x)+1=\sec^2(\sec^{-1} x)$.
\qquad
So \quad $\big(\tan(\sec^{-1} x)\big)^2+1=\big(\sec(\sec^{-1} x)\big)^2$.
\smallskip\noindent
So \quad $\big(\tan(\sec^{-1} x)\big)^2+1=x^2$.
\qquad\quad
So \quad $\tan(\sec^{-1} x) = \sqrt{x^2-1}$.
\smallskip\noindent
So \quad
$\displaystyle{
{d\sec^{-1} x\over dx} = {1\over x\sqrt{x^2-1}} }$.

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {d\csc^{-1} x\over dx} }$.
\bigskip\noindent
Since\quad $\csc(\csc^{-1} x) = x$,\qquad
$\displaystyle{ {d \csc(\csc^{-1} x)\over dx} = {dx\over dx} }$.
\medskip\noindent
So \quad $\displaystyle{
-\csc(\csc^{-1} x)\cot(\csc^{-1} x)\, {d\csc^{-1} x\over dx} = 1 }$.
\qquad\quad
So \quad $\displaystyle{
-x\cot(\csc^{-1} x)\,{d\csc^{-1} x\over dx} = 1 }$.
\smallskip\noindent
So \quad $\displaystyle{{d\csc^{-1} x\over dx} = 
{-1\over x\cot(\csc^{-1} x)} }$.
\smallskip\noindent
So we would like to ``simplify'' $\cot(\csc^{-1} x)$.
\smallskip\noindent
Since \quad $\sin^2 x+\cos^2 x = 1$, \qquad\quad
\smallskip\noindent
\phantom{Since}\quad
$\displaystyle{
{\sin^2 x\over \sin^2 x}+{\cos^2 x\over \sin^2 x}
={1\over \sin^2 x}. }$
\smallskip\noindent
So \quad $1+\cot^2 x = \csc^2 x$.
\smallskip\noindent
So \quad $1+\cot^2(\csc^{-1} x)=\csc^2(\csc^{-1} x)$.
\qquad
So \quad $1+\big(\cot(\csc^{-1} x)\big)^2=\big(\csc(\csc^{-1} x)\big)^2$.
\smallskip\noindent
So \quad $1+\big(\cot(\csc^{-1} x)\big)^2=x^2$.
\qquad\quad
So \quad $\cot(\csc^{-1} x) = \sqrt{x^2-1}$.
\smallskip\noindent
So \quad
$\displaystyle{
{d\csc^{-1} x\over dx} = {-1\over x\sqrt{x^2-1}} }$.

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {dy\over dx} }$
when $y = \log_x 10$.
\bigskip\noindent
$$ x^y = x^{\log_x 10} = 10.$$
Take the derivative:
$$\eqalign{
&\phantom{=}\ 
{d\ x^y\over dx} = {d\ (e^{\ln x})^y\over dx}
={d\ e^y\ln x\over dx} 
= e^{y\ln x}\left(y\cdot {1\over x}+ {dy\over dx}\ln x\right) \cr
&={d10\over dx} = 0. \cr
}$$
So \qquad $\displaystyle{ 
e^{y\ln x}\left(y\cdot {1\over x}+ {dy\over dx}\ln x\right) =0 }$.
\smallskip\noindent
Solve for $\displaystyle{ {dy\over dx} }$.
$$
e^{y\ln x} {dy\over dx} \ln x 
= {-e^{y\ln x} y\over x}. 
\quad\qquad\hbox{So}\qquad
{dy\over dx} 
= {-e^{y\ln x} y\over xe^{y\ln x}\ln x} = {-y\over x\ln x}
= {\log_x 10\over x\ln x}.
$$

\bigskip\noindent
{\bf Example:}  Find the third derivative of $2^x$ with respect
to $x$.
\bigskip\noindent
$y=2^x$.
\bigskip\noindent
$\displaystyle{
{dy\over dx} = {d 2^x\over dx} = {2(e^{\ln 2})^x\over dx}
= {d e^{x\ln 2}\over dx} 
= e^{x\ln 2}(\ln 2) = (e^{\ln 2})^x\ln 2 = 2^x\ln 2. 
}$
\bigskip\noindent
$\displaystyle{
{d^2y\over dx^2} = {d\over dx}\left({d y\over dx}\right)
={d 2^x\ln 2\over dx} = \ln 2\cdot 2^x\ln 2 = (\ln 2)^2 2^x
}$.
\bigskip\noindent
$\displaystyle{
{d^3y\over dx^3} = {d\over dx}\left({d^2 y\over dx^2}\right)
={d\ \over dx} \big(\ln 2)^2 2^x\big) = (\ln 2)^2 2^x\ln 2
=(\ln 2)^3 2^x}$.
 
\bigskip\noindent
{\bf Example:}  If $y=a\cos(\ln x)+b\sin(\ln x)$ show that
$\displaystyle{
x^2{d^2y\over dx^2}+x{dy\over dx}+y = 0 }$.
\bigskip\noindent
$$\eqalign{
{dy\over dx} &= a(-\sin(\ln x)){1\over x} + b\cos(\ln x){1\over x} \cr
&= -a \sin(\ln x)x^{-1} + b\cos(\ln x)x^{-1}, \cr
\cr
{d^2y\over dx^2} &=
-a\cos(\ln x){1\over x}x^{-1}+-a\sin(\ln x)(-1)x^{-2}
+\ -b\sin(\ln x){1\over x}x^{-1}+b\cos(\ln x)(-1)x^{-2} \cr
&={-a\cos(\ln x)+a\sin(\ln x)-b\sin(\ln x)-b\cos(\ln x)\over x^2} \cr
&={1\over x^2}\big( (a-b)\sin(\ln x)-(a+b)\cos(\ln x)\big). \cr
}$$
So
$$\eqalign{
LHS &= x^2{d^2y\over dx^2} + x{dy\over dx} + y \cr
&= x^2\,
{1\over x^2}\big( (a-b)\sin(\ln x)-(a+b)\cos(\ln x)\big) \cr
&\qquad\qquad
+x\big(
-a \sin(\ln x)x^{-1} + b\cos(\ln x)x^{-1} \big) \cr
&\qquad\qquad
+a\cos(\ln x)+b\sin(\ln x) \cr
&= (a-b)\sin(\ln x) - (a+b)\cos(\ln x) \cr
&\phantom{=}\qquad
-a\sin(\ln x) + b\cos(\ln x) \cr
&\phantom{=}\qquad
+b\sin(\ln x) + a\cos(\ln x) \cr
&= 0. \cr
}$$

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {dy\over dx} }$ when
$a\sin(xy)+b\cos\left(x\over y\right) = 0$.
\bigskip\noindent
Take the derivative:
$$\eqalign{
0&= a\cos(xy)\left(x{dy\over dx}+1\cdot y\right) 
+\ -b\sin\left({x\over y}\right)
\left(x(-1)y^{-2}{dy\over dx}+1\cdot y^{-1}\right) \cr
&= a\cos(xy)x{dy\over dx}+a\cos(xy)y+b\sin\left({x\over y}\right)
{x\over y^2}{dy\over dx} - b\sin\left({x\over y}\right)y^{-1}. \cr
}$$
Solve for $\displaystyle{ {dy\over dx} }$.
$$a\cos(xy)x{dy\over dx} +b\sin\left({x\over y}\right){x\over y^2}{dy\over dx}
= a \cos(xy)y-b\sin\left({x\over y}\right)y^{-1}.$$
So
$$\eqalign{
{dy\over dx} 
&=
{\displaystyle{ a \cos(xy)y-b\sin\left({x\over y}\right)y^{-1} } \over
\displaystyle{ a\cos(xy)x+b\sin\left({x\over y}\right){x\over y^2} }   }\cr
\cr
&=
{\displaystyle{a \cos(xy)y^3-b\sin\left({x\over y}\right)y  }\over
\displaystyle{ a\cos(xy)xy^2+b\sin\left({x\over y}\right)x } } \cr
}$$

\bigskip\noindent
{\bf Example:}  Find $\displaystyle{ {dy\over dx} }$ when
$\displaystyle{
y=\tan^{-1}\left({a\over x}\right)\cdot \cot^{-1}\left({x\over a}\right) 
}$.
\bigskip\noindent
$$\eqalign{
{dy\over dx} 
&= \tan^{-1}\left(a\over x\right) 
\Bigg( {-1\over \displaystyle{ 1+\Big({x\over a}\Big)^2 } }\Bigg)
{1\over a} +
{1\over \displaystyle{ 1+\Big({x\over a}\Big)^2 } }\,
(-1)ax^{-2}\cot^{-1}\left({x\over a}\right) \cr
\cr
&= {\displaystyle{ -\tan^{-1}\left({a\over x}\right) }
\over
\displaystyle{ a + {x^2\over a} } }
+
{\displaystyle{ -\cot^{-1}\left({x\over a}\right)a  }
\over x^2+a^2 } \cr
&= {\displaystyle{ -\tan^{-1}\left({a\over x}\right)a  }
\over a^2 + x^2 }
+
{\displaystyle{ -\cot^{-1}\left({x\over a}\right)a  }
\over x^2+a^2 } \cr
\cr
&= \left( {-a\over a^2+x^2}\right)
\left(\tan^{-1}\Big({a\over x}\Big)+\cot^{-1}\Big({x\over a}\Big)\right).
\cr}$$
\smallskip\noindent
If \quad $\displaystyle{ {a\over x} = \tan z }$
\quad
then \quad $\displaystyle{ {x\over a} = \cot z }$
\quad
and \quad $\displaystyle{ z = \tan^{-1}\Big({a\over x}\Big)
=\cot^{-1}\left({x\over a}\right) }$.
\medskip\noindent
So
$${dy\over dx}
= \left( {-a\over a^2+x^2}\right)
\left(\tan^{-1}\Big({a\over x}\Big)+\tan^{-1}\Big({a\over x}\Big)\right)
={\displaystyle{-2a\tan^{-1}\Big({a\over x}\Big)} \over a^2+x^2 }. 
$$






\vfill\eject
\end

