\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\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 the chain rule}
\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{$ $}  %My kludge to get the first page to move down a bit
\vskip.75truein
\centerline{\titlefont The chain rule}
\bigskip
\centerline{\rm Arun Ram}
%${}^\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\noindent
There are {\bf different kinds of derivatives}:
$$\matrix{
\hbox{Derivative with respect to $x$}\hfill &\quad
&\hbox{Derivative with respect to $g$}\hfill 
\cr
\cr
\qquad f \longrightarrow {d\over dx} \longrightarrow {df\over dx}\hfill
&&\qquad f \longrightarrow {d\over dg} \longrightarrow {df\over dg} \hfill 
\cr
\cr
\hbox{This one satisfies} \hfill &&\hbox{This one satisfies} \hfill 
\cr
\cr
\qquad\displaystyle{ {dx\over dx} = 1},\hfill 
&&\qquad\displaystyle{ {dg\over dg} = 1}, \hfill
\cr
\cr
\qquad\displaystyle{ {d(cf)\over dx} = c {df\over dx}},
\quad\hbox{if $c$ is a constant,}\hfill 
&&\qquad\displaystyle{ {d(cf)\over dg} = c {df\over dg}},
\quad\hbox{if $c$ is a constant,} \hfill 
\cr
\cr
\qquad
\displaystyle{ {d(y+z)\over dx} = {dy\over dx} + {dz\over dx}}, \hfill
&& \qquad
\displaystyle{ {d(y+z)\over dg} = {dy\over dg} + {dz\over dg}}, \hfill
\cr
\cr
\qquad
\displaystyle{ {d(yz)\over dx} = y\;{dz\over dx} + {dy\over dx}\;z}. \hfill
&&\qquad
\displaystyle{ {d(yz)\over dg} = y\;{dz\over dg} + {dy\over dg}\;z}. \hfill
}
$$
\centerline{\bf What is the relation between
\quad $\displaystyle{df\over dx}$
\quad\hbox{and}\quad
$\displaystyle{df\over dg}$ \quad?}
\bigskip
$$\matrix{
\qquad
\longrightarrow \displaystyle{d\over dg} \longrightarrow \hfill
&\quad
&\qquad
\longrightarrow \displaystyle{d\over dx} \longrightarrow \hfill
\cr
\cr
\cr
\cr
\displaystyle{ {dg^0\over dg} = {d1\over dg} = 0,} \hfill 
&&\displaystyle{ {dg^0\over dx} = {d1\over dx} = 0,} \hfill  
\cr
\cr
\cr
\cr
\displaystyle{ {dg\over dg} = 1,} \hfill 
&&\displaystyle{ {dg\over dx} = {dg\over dx},} \hfill  
\cr
\cr
\cr
\cr
\displaystyle{ {dg^2\over dg} = {dg\cdot g\over dg}} \hfill 
&&\displaystyle{ {dg^2\over dx} = {dg\cdot g\over dx}} \hfill  
\cr
\qquad \displaystyle{= g{dg\over dg}+{dg\over dg}g} \hfill 
&&\qquad \displaystyle{= g{dg\over dx}+{dg\over dx}g} \hfill 
\cr
\qquad =g+g = 2g, \hfill
&&\qquad \displaystyle{ = 2g{dg\over dx}}, \hfill 
\cr
\cr
\cr
\cr
\displaystyle{ {dg^3\over dg} = {dg^2\cdot g\over dg}} \hfill 
&&\displaystyle{ {dg^3\over dx} = {dg^2\cdot g\over dx}} \hfill  
\cr
\qquad \displaystyle{= g^2{dg\over dg}+{dg^2\over dg}g} \hfill 
&&\qquad \displaystyle{= g^2{dg\over dx}+{dg^2\over dx}g} \hfill 
\cr
\qquad =g^2+2g\cdot g = 3g^2, \hfill
&&\qquad \displaystyle{ = g^2{dg\over dx} + 2g{dg\over dx} g} \hfill 
\cr
&&\qquad \displaystyle{ = g^2{dg\over dx} + 2g^2{dg\over dx} } \hfill 
\cr
&&\qquad \displaystyle{ = 3g^2{dg\over dx} }, \hfill 
\cr
\cr
\cr
\cr
\displaystyle{ {dg^4\over dg} = {dg^3\cdot g\over dg}} \hfill 
&&\displaystyle{ {dg^4\over dx} = {dg^3\cdot g\over dx}} \hfill  
\cr
\qquad \displaystyle{= g^3{dg\over dg}+{dg^3\over dg}g} \hfill 
&&\qquad \displaystyle{= g^3{dg\over dx}+{dg^3\over dx}g} \hfill 
\cr
\qquad =g^3+3g^2\cdot g = 4g^3, \hfill
&&\qquad \displaystyle{ = g^3{dg\over dx} + 3g^2{dg\over dx} g} \hfill 
\cr
&&\qquad \displaystyle{ = g^3{dg\over dx} + 3g^3{dg\over dx} } \hfill 
\cr
&&\qquad \displaystyle{ = 4g^3{dg\over dx} }, \hfill 
\cr
\cr
\qquad\quad\displaystyle{\vdots} \hfill
&&\qquad\quad\displaystyle{\vdots} \hfill
\cr
\cr
\displaystyle{ {dg^{6342}\over dg} = 6342g^{6341}, } \hfill 
&&\displaystyle{ {dg^{6342}\over dx} = 6342g^{6341}{dg\over dx}, } \hfill 
}
$$
$$\matrix{
\displaystyle{ {d(3g^2+2g+7)\over dg} = 
{d(3g^2)\over dg} + {d(2g)\over dg} + {d7\over dg} }\hfill
&\quad
&\displaystyle{ {d(3g^2+2g+7)\over dx} = 
{d(3g^2)\over dx} + {d(2g)\over dx} + {d7\over dx} }\hfill
\cr
\qquad\displaystyle{=3{dg^2\over dg} + 2{dg\over dg} + 0 } \hfill
&&\qquad\displaystyle{=3{dg^2\over dx} + 2{dg\over dx} + 0 } \hfill
\cr
\qquad\displaystyle{=3\cdot 2g + 2\cdot 1 } \hfill
&&\qquad\displaystyle{=3\cdot 2g{dg\over dx} + 2{dg\over dx} } \hfill
\cr
\qquad\displaystyle{=6g + 2 }, \hfill
&&\qquad\displaystyle{=(6g+2){dg\over dx} }, \hfill
}$$
Thus, we are seeing that
$${df\over dx} = {df\over dg}\,{dg\over dx}\;,
\qquad\qquad\hbox{which is the chain rule.}$$



\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when $y=(2x-5)^2$.
\medskip
If $g = 2x-5$ then $y=g^2$.
$$\eqalign{
{dy\over dx} &= {dy\over dg}\, {dg\over dx}
={dg^2\over dg}\,{d(2x-5)g\over dx}
=2g(2-0)=2(2x-5)\cdot 2 \cr
\cr
& = 4(2x-5) =8x-20. \cr
}$$

\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when $y=(3x-4)^3$.
\medskip
If $g = 3x-4$ then $y=g^3$.
$$\eqalign{
{dy\over dx} &= {dy\over dg}\, {dg\over dx}
={dg^3\over dg}\,{d(3x-4)g\over dx}
=3g^2(3-0)=9(3x-4)^2 \cr
\cr
& = 9(9x^2-24x+16) =81x^2-72x+144. \cr
}$$

\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when $y=(2x-5)^2(3x-4)^3$.
\medskip
$$\eqalign{
{dy\over dx} &= {d(2x-5)^2(3x-4)^3\over dx}
= (2x-5)^2\,{d(3x-4)^3\over dx}
+ {d(2x-5)^2\over dx}\,(3x-4)^3 \cr
\cr
&= (2x-5)^2\cdot 3(3x-4)^2\cdot 3
+ 2(2x-5)\cdot 2(3x-4)^3 \cr
\cr
&= (2x-5)(3x-4)^2(9(2x-5)+4(3x-4))
= (2x-5)(3x-4)^2(30x-61). \cr
}$$

\bigskip\noindent
{\bf Example:} Find $\displaystyle{{d\, x^{m/n}\over dx}}$ when $m$ and $n$
are integers, $n\ne 0$.
\medskip
$${d(x^{m/n})^n\over dx} = {dx^m\over dx} = mx^{m-1}.
\qquad\hbox{ On the other hand} \qquad
{d(x^{m/n})^n\over dx} = n\big(x^{m/n}\big)^{n-1} {dx^{m/n}\over dx}.$$
So \quad 
$\displaystyle{
mx^{m-1} = 
n\big(x^{m/n}\big)^{n-1} {dx^{m/n}\over dx} }$
\quad
and we can solve for\quad $\displaystyle{ {dx^{m/n}\over dx} }$.
$$\eqalign{
{dx^{m/n}\over dx} 
&= {mx^{m-1}\over n\big(x^{m/n}\big)^{n-1} }
= {mx^{m-1}\over n\big(x^{m/n}\big)^n \big(x^{m/n}\big)^{-1} } \cr
\cr
&= {mx^{m-1}\over nx^m\  {1\over x^{m/n} } } 
= \left({m\over n}\right) x^{-1} x^{m/n}
= \left({m\over n}\right) x^{(m/n)-1}. \cr
}
$$

\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when 
$\displaystyle{ y={x\over \sqrt{1-2x} } }$.
\medskip
$$\eqalign{
{dy\over dx} &= {d\ \displaystyle{ x\over \sqrt{1-2x} }\over dx} 
= {d\ x\big(\sqrt{1-2x}\big)^{-1} \over dx} 
= {d\ x\big((1-2x)^{1/2}\big)^{-1} \over dx} \cr
\cr
&= {d\ x(1-2x)^{-(1/2)} \over dx} 
= x\,{d\ (1-2x)^{-(1/2)} \over dx} +{d x\over dx}\,(1-2x)^{-(1/2)} \cr
\cr
&= x\big(-\hbox{$1\over2$}\big)(1-2x)^{-{3/2}}\,{d\ (1-2x) \over dx} 
+1\cdot {1\over \sqrt{1-2x} }\cr
\cr
&= {-x\over 2(1-2x)^{3/2}}\cdot(-2) +{1\over (1-2x)^{1/2} } 
= {x+1-2x\over (1-2x)^{3/2} } 
= {1-x\over (1-2x)^{3/2} }. \cr
}$$

\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when 
$\displaystyle{ y={\sqrt{1+x^2}\over \sqrt{1-x^2} } }$.
\medskip
$$\eqalign{
{dy\over dx} 
&= {d\ \displaystyle{ {\sqrt{1+x^2}\over \sqrt{1-x^2}} }\over dx}  
= {d\ \displaystyle{ {(1+x^2)^{1/2}\over (1-x^2)^{1/2}} } \over dx}  
= {d\ \displaystyle{ \left({1+x^2\over 1-x^2}\right)^{1/2} }\over dx}  \cr
\cr
&= {1\over 2}\cdot
\left({1+x^2\over 1-x^2}\right)^{(1/2)-1}
{d\ \displaystyle{ \left({1+x^2\over 1-x^2}\right) }\over dx}  \cr
\cr
&= {1\over 2}\cdot
\left({1+x^2\over 1-x^2}\right)^{-(1/2)}
{d\ (1+x^2)(1-x^2)^{-1}\over dx}  \cr
\cr
&= {1\over 2}\cdot
\left({1-x^2\over 1+x^2}\right)^{1/2}
\left(
(1+x^2){d\ (1-x^2)^{-1}\over dx} 
+ {d\ (1+x^2)\over dx}(1-x^2)^{-1} \right) \cr
\cr
&= {1\over 2} \left({1-x^2\over 1+x^2}\right)^{1/2}
\left(
(1+x^2)(-1)(1-x^2)^{-2}{d\ (1-x^2)^{-1}\over dx} 
+ 2x(1-x^2)^{-1} \right) \cr
\cr
&= {1\over 2}\cdot
\left({1-x^2\over 1+x^2}\right)^{1/2}
\left(
{(-1)(1+x^2)(-2x)\over (1-x^2)^2} 
+ {2x\over 1-x^2} \right) \cr
\cr
&= {1\over 2}\cdot
\left({1-x^2\over 1+x^2}\right)^{1/2}
\left(
{2x(1+x^2)\over (1-x^2)^2} 
+ {2x(1-x^2)\over (1-x^2)^2} \right) \cr
\cr
&= {1\over 2}\cdot
\left({1-x^2\over 1+x^2}\right)^{1/2}
\left(
{2x(1+x^2+1-x^2)\over (1-x^2)^2} \right) \cr
\cr
&= {1\over 2}\cdot
{(1-x^2)^{1/2}\over (1+x^2)^{1/2} }
\cdot {4x\over (1-x^2)^2} 
= {2x\over (1+x^2)^{1/2}(1-x^2)^{3/2}}. 
}$$

\bigskip\noindent
{\bf Example:} Differentiate $\displaystyle{x^2\over 1+x^2}$
with respect to $x^2$.
\bigskip\noindent
This is the same problem as:
\smallskip
Find $\displaystyle{dz\over dp}$ when
$\displaystyle{z = {x^2\over 1+x^2}}$ and $p = x^2$.
\medskip\noindent
Since $\displaystyle{ {dz\over dx} = {dz\over dp}{dp\over dx} }$,
\qquad $\displaystyle{ {dz\over dp} 
= {\big(dz/dx\big)\over \big(dp/dx\big)} }$.
\smallskip\noindent
So
$$\eqalign{
{dz\over dp}
&= { \displaystyle{ {d\ \over dx}\left({x^2\over 1+x^2}\right) }
\over
\displaystyle{ {d\ \over dx}\big(x^2\big) }  }
= { \displaystyle{ {d\ x^2(1+x^2)^{-1} \over dx\phantom{j_j}} }
\over
\displaystyle{ {d\ x^2\phantom{i^i}\over dx} }  }
= { \displaystyle{ 
x^2{d\ (1+x^2)^{-1} \over dx\phantom{j_j}} + {dx^2\over dx}(1+x^2)^{-1} }
\over
2x  }\cr
\cr
&= { \displaystyle{ 
x^2(-1)(1+x^2)^{-2}{d\ (1+x^2) \over dx\phantom{j_j}} + 2x(1+x^2)^{-1} }
\over
2x  }\cr
\cr
&= { \displaystyle{ 
{-x^2\over (1+x^2)^2}\cdot 2x + {2x\over 1+x^2\phantom{j_j}} }
\over
2x  }
=  {-x^2\over (1+x^2)^2} + {1\over 1+x^2}  \cr
\cr
&=  {-x^2+1+x^2\over (1+x^2)^2} 
=  {1\over (1+x^2)^2}. \cr
}$$
 
\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when 
$x^4+y^4 = 4a^2x^2y^2$.
\medskip
$$
{d\,(x^4+y^4)\over dx} = {d\,(4a^2x^2y^2)\over dx}.
\qquad\hbox{So}\quad
{dx^4\over dx}+{dy^4\over dx} = 4a^2{dx^2y^2\over dx}.$$
So\qquad
$\displaystyle{
4x^3+4y^3{dy\over dx} = 4a^2\left(x^2{dy^2\over dx}+
{dx^2\over dx}y^2\right) }$.
\medskip\noindent
$\displaystyle{
\eqalign{
\hbox{So}\qquad
4x^3+4y^3{dy\over dx} 
&= 4a^2\left(x^22y{dy\over dx}+ 2xy^2\right)  \cr
&= 4a^2x^2 2y{dy\over dx}+ 4a^22xy^2. \cr
} }$
\medskip\noindent
So\qquad
$\displaystyle{
4x^3-4a^22xy^2 = 4a^2x^22y{dy\over dx} - 4y^3 {dy\over dx}. }$
\medskip\noindent
So\qquad
$\displaystyle{
4x^3-4a^22xy^2 = \big(4a^2x^22y - 4y^3\big) {dy\over dx}. }$
\medskip\noindent
So\qquad
$\displaystyle{
{4x^3-4a^22xy^2\over 4a^2x^22y - 4y^3} =  {dy\over dx}. }$
\medskip\noindent
So\qquad
$\displaystyle{
{dy\over dx}
={x^3-2a^2xy^2\over 2a^2x^2y - y^3}. }$
\medskip\noindent
All we did is take the derivative of both sides and 
then solve for $\displaystyle{ {dy\over dx} }$.

\bigskip\noindent
{\bf Example:} Find $\displaystyle{{dy\over dx}}$ when 
$\displaystyle{ x={3at\over 1+t^3} }$
\quad and\quad $\displaystyle{ y = {3at^2\over 1+t^3} }$.
\medskip \noindent
Since $\displaystyle{
y = {3at^2\over 1+t^3} = \left({3at\over 1+t^3}\right) t = xt }$,
\qquad
$\displaystyle{ 
{dy\over dx} = x{dt\over dx}+{dx\over dx}\cdot t = x{dt\over dx}+t }$.
\smallskip\noindent
What is $\displaystyle{ {dt\over dx} }\,$??
\smallskip\noindent
Since $\displaystyle{
{dx\over dx} = {dx\over dt}\,{dt\over dx} }$,
\qquad
$\displaystyle{ {dt\over dx} = {\big(dx/dx\big)\over \big(dx/dt\big) }
={1\over dx/dt} }$.
\smallskip\noindent
So
$$\eqalign{
{dt\over dx} 
&= {1\over dx/dt}
={1\over \displaystyle{ {d\ \over dt}\left( {3at\over 1+t^3}\right) } } 
={1\over \displaystyle{ {d\ (3at)(1+t^3)^{-1}\over dt} } } \cr
\cr
&={1\over \displaystyle{ 
3at(-1)(1+t^3)^{-2}{d (1+t^3)\over dt}+3a(1+t^3)^{-1} } } \cr
\cr
&={1\over \displaystyle{ 
{-3at\phantom{I^I}\over (1+t^3)^2 }\,3t^2+{3a\over 1+t^3 } } } 
={1\over \displaystyle{ 
{-9at^3+3a(1+t^3)\phantom{I^I}\over (1+t^3)^2 } } } \cr
\cr
&={(1+t^3)^2\over -9at^3+3a(1+t^3) } 
={(1+t^3)^2\over 3a-6at^3 }. \cr
}$$
So
$$\eqalign{
{dy\over dx}
&= x{dt\over dx} + t 
= {3at\over 1+t^3}\,{(1+t^3)^2\over 3a(1-2t^3) } + t \cr
\cr
&= {t(1+t^3)\over 1-2t^3}+{t(1-2t^3)\over 1-2t^3 } 
= {t+t^4+t-2t^4\over 1-2t^3 } 
= {2t-t^4\over 1-2t^3} \cr
}$$

\vfill\eject
\end
