\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