\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