Thenweapplythechainrule,firstbyidentifyingtheparts:Now,takethederivativeofeachpart:Andfinally,multiplyaccordingtotherule.w=f(x,y,z)\qquad0&0&0&\cdots&1Usethechainruletocalculateh′(x),whereh(x)=f(g(x)).\lim_{h\to0}\frac1h\left[0=h'(0)=(f\circ\gamma)'(0)=\nablaf(\gamma(0))\cdot\gamma'(0)=\nablaf(\mathbfa)\cdot\gamma'(0).Thegeneralform$$\eqref{cr1}$$ofthechainrulesaysthatforavectorfunction$$\mathbff$$,everycomponent$$f_k$$satisfies$$\eqref{cr.scalar}$$,for$$k=1,\ldots,\ell$$.Moveover,inthiscase,ifwecalculateh(x),h(x)=f(g(x))=f(−2x+5)=6(−2x+5)+3=−12x+30+3=−12…\phi(x,y)=f(x^2-y,xy,x\cosy)Thechainrulefordummies.\sin\theta\\\mathbfE_{\mathbff\circ\mathbfg,\mathbfa}(\mathbfh):=N\mathbfE_{\mathbfg,\mathbfa}({\bfh})+\mathbfE_{\mathbff,\mathbfb}({\bfk}),\mathbff(\mathbfg(\mathbfa))+NM{\bfh}\+\N\mathbfE_{\mathbfg,\mathbfa}({\bfh})+\mathbfE_{\mathbff,\mathbfb}({\bfk})\partial_xf(r\cos\theta,r\sin\theta)\cos\theta+Theinnerfunctionistheoneinsidetheparentheses:x2-3.Theouterfunctionis√(x).\begin{array}{ccc}\],$Wemostoftenapplythechainruletocompositionsf∘g,wherefisareal-valuedfunction.D(f\circ\mathbfg)(\mathbfa)=[Df(\mathbfg(\mathbfa))]\[D\mathbfg(\mathbfa)].Let$$S=\{(x,y,z)\in\R^3:z\ne0\}$$,andfor$$(x,y,z)\inS$$,define$$\phi(x,y,z)=f(xy,y/z)$$.\frac{\partial}{\partialx_{ij}}\det(I),\r\cos\thetaThisisbecausetheintermediatequantitiesinthechainruleareoften3rdand4thordertensors,whereasthedifferentialofamatrixisjustanothermatrix.21{1Usethechainruletondthefollowingderivatives.ThankstoallofyouwhosupportmeonPatreon.\end{array}\right)Thatmaterialishere.Thenaroutineapplicationofthechainruletellsusthat\[+\mathbfE_{\mathbff\circ\mathbfg,\mathbfa}(\mathbfh),\\$,$$\nablaf(\mathbfx)\cdot\mathbfx=\alphaf(\mathbfx)$$,$$Supposewewanttoknowaboutratesofchangein$$w$$inresponsetoinfinitesimalorsmallchangesin$$x$$,alwaysrestrictingourattentiontothesetofpointswhere$$z=g(x,y)$$.)\=\Let’sseethisforthesinglevariablecaserst.\]andthisiscorrectandunambiguous,thoughstillalittleawkward.Nowsupposethatwearegivenafunction$$f:\R^2\to\R$$.&=-\frac{\partial\phi}{\partialx}\right)&=x\partial_yu-y\partial_xu=0Thechainruleheresays,lookwehavetotakethederivativeoftheouterfunctionwithrespecttotheinnerfunction.Provethat$\frac1{|\bfh|}|\mathbfE_{\mathbff,\mathbfb}({\bfk})|=Fixanopeninterval$$I\subseteq\R$$containing$$0$$andacurve$$\gamma$$satisfying$$\eqref{gamma1}$$.Specializingstillmore,acasethatarisesoftenis$$\mathbfg:\R\to\R^m$$and$$f:\R^m\to\R$$.$,$Exercise:Ifyouhavenotalreadydoneit,checkthat$$f$$isdifferentiableeverywhereexceptattheorigin,andthat\[\nonumber\left(\begin{array}{cc}\nablaf(\mathbfx)=\frac{\mathbfx}{|\mathbfx|}\qquad\text{for}\mathbfx\ne{\bf0}.Integratingusingsubstitution.Itstates:ify=(f(x))n,thendydx=nf0(x)(f(x))n−1wheref0(x)isthederivativeoff(x)withrespecttox.\frac{\partialw}{\partialx}=Trytoimagine"zoominginto"differentvariable'spointofview.\frac1{|\bfh|}|N\mathbfE_{\mathbfg,\mathbfa}({\bfh})|\leInthisproofwehavetokeeptrackofseveraldifferenterrorterms,sowewillusesubscriptstodistinguishbetweenthem.$.\frac{\partialf}{\partialx}(x,y,g(x,y))\]thenitistraditionaltowrite,forexample,$$\dfrac{\partialu_k}{\partialx_j}$$todenotetheinfinitesimalchangeinthe$$k$$thcomponentof$$\mathbfu$$inresponsetoaninfinitesimalchangein$$x_j$$,thatis,$$\frac{\partialu_k}{\partialx_j}=\frac{\partial}{\partialx_j}(f_k\circ\mathbfg)$$.Ifyou'reseeingthismessage,itmeanswe'rehavingtroubleloadingexternalresourcesonourwebsite.\],$$\mathbfb=\mathbfg(\mathbfa)\inT$$,$$\mathbfE_{\mathbfg,\mathbfa}({\bfh})$$,$$D\mathbff(\mathbfg(\mathbfa))=D\mathbff(\mathbfb)$$,$$$\label{dga}$$$,$$D(\mathbff\circ\mathbfg)(\mathbfa)$$,$$[D\mathbff(\mathbfg(\mathbfa))]\[D\mathbfg(\mathbfa)]$$,$0&\text{if}i\nejYoudothederivativerulefortheoutsidefunction,ignoringtheinsidestuff,thenmultiplythatbythederivativeofthestuff.$ThesewereintroducedinoneoftheproblemsinSection2.1Forexample,themonomial$$x^ay^bz^c$$ishomogeneousofdegree$$\alpha=a+b+c$$.Objectives:Inthistutorial,wederivetheChainRule.\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}=0.(y-z)\partial_xu-x\partial_yu+x\partial_zu=0.Weneedtoestablishaconvention,andinthiscasethefirstinterpretationisconventional.Sothequotientrulebeginswiththederivativeofthetop.\].Butbadchoicesofnotationcanleadtoambiguityormistakes.\partial_yf(r\cos\theta,r\sin\theta)\sin\theta,\\Solution:Thederivativesoffandgaref′(x)=6g′(x)=−2.Accordingtothechainrule,h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12.Supposethat$$f:\R\to\R$$isofclass$$C^1$$,andthat$$u=f(x^2+y^2+z^2)$$.Bythedefinitionofthelevelset$$C$$,theassumptionthat$$\gamma(t)\inC$$forall$$t\inI$$meansthat$$h(t)=f(\gamma(t))=c$$forall$$t\inI$$.Afterall,since$$x=u$$and$$y=v$$,itmightbesimplertowrite$$\mathbfG$$asafunctionof$$x$$and$$y$$ratherthan$$u$$and$$v$$,ie$$\mathbfG(x,y)=(x,y,g(x,y))$$.READPAPER.\frac{du}{dt}=\frac{\partialu}{\partialx_1}\frac{dx_1}{dt}+\cdots+Asabove,wewrite$$\mathbfx=(x_1,\ldots,x_n)$$and$$\mathbfy=(y_1,\ldots,y_m)$$todenotetypicalpointsin$$\R^n$$and$$\R^m$$.\],\label{lsnot}\lim_{\bfk\to\bf0}\frac{\mathbfE_{\mathbff,\mathbfb}({\bfk})}{|\bfk|}=\mathbf0.\nablaf(\mathbfx)\cdot\mathbfx=\alphaf(\mathbfx).\begin{array}{ccc}&=Supposealsothat$$\mathbfa\inS$$isapointsuchthat$$\mathbfg(\mathbfa)\inT$$;thus$$\mathbff\circ\mathbfg(\mathbfx)=\mathbff(\mathbfg(\mathbfx))$$iswell-definedforall$$\mathbfx$$closeto$$\mathbfa$$.C=\{(x,y,z)\in\R^3:x^2-2xy+4yz-z^2=2\}\end{align*},Ifweusethenotation$$\eqref{cr.trad}$$,thenthechainruletakestheform\begin{align*}\label{lsg3}.\partial_3f\partial_1g,&=Write$$h(t)=f\circ\gamma(t)$$.\left(.AgainwewillseehowtheChainRuleformulawillanswerthisquestioninanelegantway.\sin\theta&r\cos\theta\end{array}SeeExample2belowforanillustrationofthisspecialcase.Thenwewouldwrite$x_{11}&\cdots&x_{1n}\\Thechainruleworksforseveralvariables(adependsonbdependsonc),justpropagatethewiggleasyougo.$,$$$\label{cr.p2}$$$Sofarwehaveonlyprovedthattheimplication$$\Longrightarrow$$holds.\]Thismeansthesameas$$\eqref{crsc1}$$,butyoumayfindthatitiseasiertorememberwhenwrittenthisway.Thenitisclearthat$$\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}=1$$,showingthat$$\eqref{wrong}$$iscannotbetrue.Thechainruleisn'tjustfactor-labelunitcancellation--it'sthepropagationofawiggle,whichgetsadjustedateachstep.-Substitutionessentiallyreversesthechainruleforderivatives.\frac{\partialx}{\partial\theta}+\frac{\partial}{\partialx_{12}}\det(X),\],$$\mathbfg(r,\theta)=(r\cos\theta,r\sin\theta).$$,$\end{cases}Lesson10.4:TheChainRule:InthislessonyouwilldownloadandexecuteascriptthatdevelopstheChainRuleforderivatives.$,$\qquad{\bfk}=M{\bfh}+\mathbfE_{\mathbfg,\mathbfa}(\bfh).Thechainrule(fordifferentiatingacompositefunction):Or,equivalently,Seethesidebar,“Whythechainruleworks,”foraplain-Englishexplanationofthismumbojumbo.\end{multline}$,$$NM=D\mathbff(\mathbfg(\mathbfa))D\mathbfg(a)$$,$$$\label{cr.proof}$Forsimplicity,consideringonlythe$$u$$derivative,thissaysthat,$$Thismeans:thederivativeofthedeterminantfunction,evaluatedattheidentitymatrix.Here’swhatyoudo.Letusunderstandthechainrulewiththehelpofawell-knownexamplefromWikipedia.\gamma'(0)\text{exists}.f(\lambda\mathbfx)=\lambda^\alphaf(\mathbfx)\quad\text{forall}\mathbfx\ne{\bf0}\text{and}\lambda>0.\phi(x,y,z)=f(x^2-yz,xy+\cosz)Ontheotherhand,shorterandmoreelegantformulasareofteneasierforthemindtoabsorb.\left(Letâswrite$$\phi$$todenotethecompositefunction$$\phi=f\circ\mathbfg$$,so$Thechainrulecanbeoneofthemostpowerfulrulesincalculusforfindingderivatives.\partial_1\phi(x,y)=\partial_1f(x,y,g(x,y))+calculusfordummies.pdf.Chainruleinvolvesalotofparentheses,alot!Theonlydifferencethistimeisthat∂z∂xhastheshape(K1×.Email.\mathbff(\mathbfb+{\bfk})=\sin\theta&r\cos\theta\end{array}Wewillreturntothispointlater.Takeagoodlookatthis.$$$,$$\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}=1$$,$2ffgfggg–QuotientRule5.Asyoucansee,chainruleintegrationjustinvolvesusdeterminingwhichtermsaretheoutsidederivativeandinsidederivative.S=\{(r,\theta)\in\R^2:r\geq0\}\partial_1\phi=\partial_1fThenforeach$$k=1,\dots,\ell$$and$$j=1,\ldots,n$$,$$\eqref{cr1}$$isthesameas\[\label{crcoord}Inthiscase,formula$$\eqref{cr1}$$simplifiesto\[\label{cr.scalar}x_{ij}=\begin{cases}1&\text{if}i=j\\$,$$aslongaswetrustourreaderstofigureoutthatderivativesof$$g$$areevaluatedat$$(x,y)$$andderivativesof$$f$$at$$(x,y,g(x,y))$$.Plugthosethingsbackin.\fracd{dt}|\mathbfg(t)|=\frac{\mathbfg(t)}{|\mathbfg(t)|}\cdot\mathbfg'(t)=|\mathbfg'(t)|\cos\thetausingthechainrule,explainingsomeapplicationofthechainruletosomeone(eg,writingupthesolutionofaproblem),or;readingdiscussionsthatusethechainrule,particularlyiftheyusenotationlike$$\eqref{cr.trad}$$.\nablaf(\mathbfa)\cdot{\bfv}=0\qquad\text{foreveryvector$\bfv$\],$$$\label{tp.def}\frac{\partial}{\partialx_{12}}\det(X),\cos\theta+\label{lsg3}$$$,$Rule4:ChainRuleThefinal(andmostcomplex)derivativerulewewillbelearninginthislessonisthechainrule.Sincethefunctionswerelinear,thisexamplewastrivial.$Definethefunction$$\det:M^{n\timesn}\to\R$$bysayingthat$$\det(X)$$isthedeterminantofthematrix.\mathbfy=\mathbfg(\mathbfx),\qquad{\bfu}=\mathbff(\mathbfy)=\mathbff(\mathbfg(\mathbfx)),\r\sin\theta+Mostproblemsareaverage.Butifweinsistonusingthenotation$$\eqref{cr.trad}$$,thenthereisnosimplewayofdistinguishingbetweenthesetwodifferentthings.\begin{array}{rr}\cos\theta&-r\sin\theta\\x_{n1}(t)&\cdots&x_{nn}(t)\],$Example4motivatesadefinitionthatwillbeusefulfordiscussingthegeometryofthederivative.Souseyourparentheses!\frac{\partial\phi}{\partialy}Youcannevergowrongifyouapplythechainrulecorrectlyandcarefullyâafterall,itâsatheorem.Example2:MotionofaparticleLet$$\mathbfg:\R\to\R^n$$beadifferentiablefunction,andconsider\[+\frac{\partialf}{\partialz}(x,y,g(x,y))\frac{\partialg}{\partialx}(x,y).Forexample,weneedthechainrule…$.\mathbfg(\mathbfa)+M\mathbfh+\mathbfE_{\mathbfg,\mathbfa}({\bfh})\qquad\text{where}\end{cases}\mathbff(\mathbfb+{\bfk})=\gamma'(0)\cdot\nablaf(\mathbfa)=0.\partial_yf(r\cos\theta,r\sin\theta)r\cos\theta\r\sin\theta+Definition•Incalculus,thechainruleisaformulaforcomputingthederivativeofthecompositionoftwoormorefunctions.Thus,$$M$$isthe(unique)$$m\timesn$$matrixsuchthat$$$\label{dga}$$&\overset{\eqref{dfb}}=\frac{\partial\phi}{\partialu}(u,v)=\frac{\partialf}{\partialx}(u,v,g(u,v))+\frac{\partialf}{\partialz}(u,v,g(u,v))\frac{\partialg}{\partialu}(u,v).$,\end{align*}.Thechainrulecanbethoughtofastakingthederivativeoftheouterfunction(appliedtotheinnerfunction)andmultiplyingittimesthederivativeoftheinnerfunction.Let$$S=\{(r,s)\in\R^2:s\ne0\}$$,andfor$$(r,s)\inS$$,define$$\phi(r,s)=f(rs,r/s)$$.Wecanwrite$$\mathbff$$asafunctionofvariables$$\mathbfy=(y_1,\ldots,y_m)\in\R^m$$,and$$\mathbfg$$asafunctionof$$\mathbfx=(x_1,\ldots,x_n)\in\R^n$$.\frac{\partialw}{\partialx}=$$\Leftarrow$$Â Â $$\Uparrow$$Â Â $$\Rightarrow$$.\end{align*}\],$$\partial_xf(r\cos\theta,r\sin\theta)$$,$\sin\theta\\$Ifwethinkof$$\mathbfg(t)$$asbeingthepositionattime$$t$$ofaparticlethatismovingaroundin$$\R^n$$,then$$\phi(t)$$isthedistanceattime$$t$$oftheparticlefromtheorigin.\lim_{\bfh\to\bf0}\frac1{|\bfh|}\mathbfE_{\mathbff\circ\mathbfg,\mathbfa}(\mathbfh)=\bf0.+\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}.AsyouwillseethroughouttherestofyourCalculuscoursesagreatmanyofderivativesyoutakewillinvolvethechainrule!\]Thus$$C$$isthelevelsetof$$f$$thatpassesthrough$$\mathbfa$$.Using$$\eqref{tv2}$$,thisdefinitionstatesthatapoint$$\mathbfx$$belongstothetangentplaneto$$C$$at$$\mathbfa$$ifandonlyif$$\mathbfx$$hastheform$$\mathbfx=\mathbfa+{\bfv}$$,where$$\bfv$$istangenttothelevelsetat$$\mathbfa$$.\],$Withthechainrule,itiscommontogettrippedupbyambiguousnotation.\lim_{\mathbfh\to{\bf0}}\frac1{|\bfh|}|\mathbfE_{\mathbff,\mathbfb}({\bfk})|Ifyourememberthat,therestofthenumeratorisalmostautomatic.Theinnerfunctionistheoneinsidetheparentheses:x2-3.Theouterfunctionis√(x).\mathbff(\mathbfg(\mathbfa+{\bfh}))Apostol,T.M."TheChainRuleforDifferentiatingCompositeFunctions"and"ApplicationsoftheChainRule.\frac{\partialw}{\partialx}\label{gamma1}$,$§3.5andAIIIinCalculuswithAnalyticGeometry,2nded.Code§1500etseq.PoorFairOKThiscanbeviewedasy=sin(u)withu=x2.Itperformstheroleofthechainruleinastochasticsetting,analogoustothechainruleinordinarydifferentialcalculus.That’strue,butthetechniqueforcesyoutoleavethestuffaloneduringeachstepofaproblem.Thisisbecausetheintermediatequantitiesinthechainruleareoften3rdand4thordertensors,whereasthedifferentialofamatrixisjustanothermatrix.\phi(x,y)=f(\mathbfG(x,y))=f(x,y,g(x,y)),\text{and}\quadC=\{\mathbfx\inS:f(\mathbfx)=c\}.Itwouldbebetterifwecouldsaythat\[\label{tv2}\detI\right]Findformulasforpartialderivativesof$$\phi$$intermsof$$x,y,z$$andpartialderivativesof$$f$$.SeveralexamplesusingtheChaionRuleareworkedout.=\sum_{i=1}^m\frac{\partialf_k}{\partialy_i}(\mathbfg(\mathbfa))\\frac{\partialg_i}{\partialx_j}(\mathbfa).\nablaf(\mathbfx)=\frac{\mathbfx}{|\mathbfx|}\qquad\text{for}\mathbfx\ne{\bf0}.$Forevery$$i$$and$$j$$,compute$Itmaybehelpfultowriteout$$\eqref{cr1}$$intermsofcomponentsandpartialderivatives.Aswithallchainruleproblems,youmultiplythatbystuff’.\frac{\partialu}{\partialx_j}=\frac{\partialu}{\partialy_1}\frac{\partialy_1}{\partialx_j}whereDfisa1×mmatrix,thatis,arowvector,andD(f∘g)isa1×nmatrix,alsoarowvector(butwithlengthn).byMarkRyanFounderofTheMathCenterCalculus2ndEditionwww.it-ebooks.info\begin{array}{ccc}Thisunitillustratesthisrule.OnMay27,2016,theFDApublishedthefinalruleundertheFoodSafetyModernizationAct(FSMA)onpreventingadulterationacrossthefoodsupplychain.\fracd{dt}\det(X(t))\right|_{t=0}\)intermsof$$x_{ij}'(0)$$,for$$i,j=1,\ldots,n$$.ycCA9l5lWurYimgh1tTsymr6eOs5eVr3vkejdW.Id2MvatdteINw5intkhZoI5n1fFivnNiVtvev4C3atlycRu2lWu7s1.2WorksheetbyKutaSoftwareLLC-\partial_xf(r\cos\theta,r\sin\theta)r\sin\theta+$Butitisclearthat$$\frac{\partialx}{\partialx}=1$$,andthat$$\frac{\partialy}{\partialx}=0$$,because$$x$$and$$y$$areunrelated.\frac{\partialf}{\partialx}\circ\mathbfG&\\frac{\partialf}{\partialy}\circ\mathbfG&\\frac{\partialf}{\partialz}\circ\mathbfG\frac{\partial\phi}{\partialy}\AreyouworkingtocalculatederivativesusingtheChainRuleinCalculus?Thetop,ofcourse.SoifIweretosay,inthiscase,fofxisnaturallogofx,fofgofxisthisexpressionhere.\lim_{\bfk\to\bf0}\frac{\mathbfE_{\mathbff,\mathbfb}({\bfk})}{|\bfk|}=\mathbf0.Ifyou'reseeingthismessage,itmeanswe'rehavingtroubleloadingexternalresourcesonourwebsite.\frac{\partial\phi}{\partialr}\nonumber\frac{\partialu_k}{\partialx_j}=\frac{\partialu_k}{\partialy_1}\frac{\partialy_1}{\partialx_j}©TM2G0j1f3FXKTuvt3aniSpoQf2t9wOaRrtemHLNL4CF.&=-\frac{\partial\phi}{\partialx}calculusfordummies…\],\begin{align}\frac{\partialx}{\partialr}+SolutionFirst,wecompute$$\nablaf(x,y,z)=(2x-2y,-2x+4z,4y-2z)$$,so$$\nablaf(\mathbfa)=(0,2,2)$$.Findthetangentplanetothesurface\[Let’ssolvesomecommonproblemsstep-by-stepsoyoucanlearntosolvethemroutinelyforyourself.Youdarealmvps!Tosimplifytheset-up,letâsassumethat.Thechainrulecomesintoplaywhenweneedthederivativeofanexpressioncomposedofnestedsubexpressions.For$$2\times2$$matrices,compute\[Afunction$$f:\R^n\to\R$$issaidtobehomogeneousofdegree$$\alpha$$if\[Forexample,supposewearegiven$$f:\R^3\to\R$$,whichwewillwriteasafunctionofvariables$$(x,y,z)$$.Onecanalsogetintomoreserioustrouble,forexampleasfollows.Wewillbeterse.Thechainruleisarulefordifferentiatingcompositionsoffunctions.Youcanalsofindquestionsofthissortinanybookonmultivariablecalculus.,$Itsaysthat,fortwofunctionsand,thetotalderivativeofthecomposite∘atsatisfies(∘)=∘.IfthetotalderivativesofandareidentifiedwiththeirJacobianmatrices,thenthecompositeontheright-handsideissimplymatrixmultiplication.\label{wo1}$,$$\left(\dfrac{\partialf}{\partialx}\right)(x,y,g(x,y))$$,$$\dfrac{\partial}{\partialx}\left(f(x,y,g(x,y))\right)$$,\label{last}JustMathTutoringThisvideoshowstheprocedureoffindingderivativesusingtheChainRule.Assumethatyouarefallingfromthesky,theatmosphericpressurekeepschangingduringthefall.Rating.Afterworkingthroughthesematerials,thestudentshouldbeabletousetheChainRuletodifferentiatecertainfunctions.w=f(x,y,z)\qquad\end{align},$\begin{multline}Thissectionexplainshowtodifferentiatethefunctionysin4xusingthechainrule.Thesecondinterpretationisexactlywhatwecalled$$\frac{\partial\phi}{\partialx}$$.+\cdots+\frac{\partialu}{\partialy_m}\frac{\partialy_m}{\partialx_j}\\quad\text{for}j=1,\ldots,n.)\=\\frac{|\bfk|}{|\bfh|}\frac1{|\bfk|}|\mathbfE_{\mathbff,\mathbfb}({\bfk})|.arerelatedviathetransformation,.$Expresspartialderivativesof$$\phi$$withrespectto$$x,y,z$$intermsof$$x,y,z$$andpartialderivativesof$$f$$.;Fed.\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}.$$\mathbff\circ\mathbfg(\mathbfx)=\mathbff(\mathbfg(\mathbfx))$$,$\label{cr1}where}\{\mathbfx\in\R^3:(\mathbfx-\mathbfa)\cdot\nablaf(\mathbfa)=0\}.Thepowerruleworksforanypower.TheChainRuleStatingtheChainRuleintermsofthederivativematricesisstrikinglysimilartothewell-known(fg)0(x)=f0(g(x))g0(x).0dcdx6.nn1dxnxdx–PowerRule7.dfgxfgxgxdxThisistheChainRuleCommonDerivatives1dxdxsincosdxxdxcossindxxdxtansec2dxxdxsecsectandxxxdxcsccsccotdxxxdx\frac{\partial}{\partialx_j}(f_k\circ\mathbfg)(\mathbfa)M.Hawlader.Thechainrulemakesitpossibletodiﬀerentiatefunctionsoffunc-tions,e.g.,ifyisafunctionofu(i.e.,y=f(u))anduisafunctionofx(i.e.,u=g(x))thenthechainrulestates:ify=f(u),thendydx=dydu×dudxExample1Considery=sin(x2).WHYTHECHAINRULEWORKS.\frac{\partial\phi}{\partialy}$,$$Thiscanbecheckedbywritingoutbothsidesof$$\eqref{cr1}$$âtheleft-handsideisthe$$(k,j)$$componentofthematrix$$D(\mathbff\circ\mathbfg)(\mathbfa)$$,andtheright-handsideisthe$$(k,j)$$componentofthematrixproduct$$[D\mathbff(\mathbfg(\mathbfa))]\[D\mathbfg(\mathbfa)]$$.\mathbfE_{\mathbff\circ\mathbfg,\mathbfa}(\mathbfh):=N\mathbfE_{\mathbfg,\mathbfa}({\bfh})+\mathbfE_{\mathbff,\mathbfb}({\bfk}),Thechainruletellsushowtofindthederivativeofacompositefunction.Expressionslike$$\eqref{wo1}$$canbeconfusing,and$$\eqref{wo05}$$isonlycorrectifthereaderisabletofigureoutexactlywhatitmeans.\frac{\partial}{\partialx_{ij}}\det(I),+Mostproblemsareaverage.&=Thechainruleworksforseveralvariables(adependsonbdependsonc),justpropagatethewiggleasyougo.Then$$\eqref{wo05}$$lookslike$\label{last}Whentheargumentofafunctionisanythingotherthanaplainoldx,suchasy=sin(x2)orln10x(asopposedtolnx),you’vegotachainruleproblem.\partial_yf(r\cos\theta,r\sin\theta)\sin\theta,\\$,$\qquad\text{where}Thisruleisobtainedfromthechainrulebychoosingu=f(x)above.Forexample,ifacompositefunctionf(x)isdefinedasD\mathbfg=\left(Thechainrulestatesformallythat.Weseethatthederivativeofx^3+5is3x^2,butinthequestionitisjustx^2.\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}.\phi(t+h)-\phi(t)$,$Forexample,ifacompositefunctionf(x)isdefinedas$wherewehavetorememberthat$$\partial_xf$$and$$\partial_yf$$areevaluatedat$$\mathbfg(r,\theta)$$.\mathbff(\mathbfg(\mathbfa))+NM{\bfh}\+\N\mathbfE_{\mathbfg,\mathbfa}({\bfh})+\mathbfE_{\mathbff,\mathbfb}({\bfk})\],$\label{cr.scalar}\quadYousimplyapplythederivativerulethat’sappropriatetotheouterfunction,temporarilyignoringthenot-a-plain-old-xargument.Thechainrulefromsinglevariablecalculushasadirectanalogueinmultivariablecalculus,wherethederivativeofeachfunctionisreplacedbyitsJacobianmatrix,andmultiplicationisreplacedwithmatrixmultiplication.isthevector,.+\frac{\partialw}{\partialz}\frac{\partialz}{\partialx}.PresentyoursolutionjustlikethesolutioninExample21.2.1(i.e.,writethegivenfunctionasacompositionoftwofunctionsfandg,computethequantitiesrequiredontheright-handsideofthechainruleformula,andnallyshowthechainrulebeingappliedtogettheanswer).DownloadwithGoogleDownloadwithFacebook.\end{array}\right)x_{11}&\cdots&x_{1n}\\\begin{array}{rr}\cos\theta&-r\sin\theta\\3.\frac{\partialy}{\partialr}\\\end{array}\right)\mathbff(\mathbfg(\mathbfa+{\bfh}))\frac{\partialf}{\partialx}\frac{\partialx}{\partialt}+Startingfromdxandlookingup,youseetheentirechainoftransformationsneededbeforetheimpulsereachesg.ChainRule…Findformulasfor$$\partial_r\phi$$and$$\partial_s\phi$$intermsof$$r,s$$andderivativesof$$f$$.D\phi=\big(\partial_r\phi\\\\partial_\theta\phi)1&0\\0&1\\$Supposethat$$f:\R^2\to\R$$isofclass$$C^1$$,andthat$$u=f(x^2+y^2+z^2,y+z)$$.\label{wo05}\],$Sowewillassumethat$$\nablaf(\mathbfa)\ne{\bf0}.$$Thisassumptionhassomeinterestingandrelevantgeometricconsequencesthatwewilldiscussindetaillater.Thisisauser-friendlymathbook.Simplify.\frac{\partialw}{\partialx}\frac{\partialx}{\partialx}+Thebestevidenceruleprovidesthat,whereawritingisofferedinevidence,acopyorothersecondaryevidenceofitscontentwillnotbereceivedinplaceoftheoriginaldocumentunlessanadequateexplanationisofferedfortheabsenceoftheoriginal.Supposewehaveafunction$$f:\R^2\to\R$$,andwewouldliketoknowhowitchangeswithrespecttodistanceoranglefromtheorigin,thatis,whatareitsderivativesinpolarcoordinates.0&0&1&\cdots&0\\\frac{\partial\phi}{\partial\theta}Letf(x)=6x+3andg(x)=−2x+5.Souseyourparentheses!\partial_r\phi=\partial_r(f\circ\mathbfg)\mathbfg(\mathbfa+{\bfh})=\label{wo1}$theycanbelegitimatelyconfusedaboutwhetheritmeans,firstcomputethepartialderivativewithrespectto$$x$$,thensubstitute$$z=g(x,y)$$,OR.Thatis,iffisafunctionandgisafunction,thenthechainruleexpressesthederivativeofthecompositefunctionf∘gintermsofthederivativesoffandg.DownloadFullPDFPackage.firstsubstitute$$z=g(x,y)$$,thencomputethepartialderivativewithrespectto$$x$$.Then$$f\circ\mathbfg$$isafunction$$\R\to\R$$,andthechainrulestatesthat\begin{align}\label{crsc1}Usingthisnotation,andwithsimilarinterpretationsfor$$\frac{\partialu_k}{\partialy_i}$$and$$\frac{\partialy_i}{\partialx_j}$$,wecanwritethechainruleintheform\[\label{cr.trad}\frac{\partialy}{\partial\theta}\\ChainRuleforVectorFunctions(FirstDerivative)Ifthefunctionitselfisavector,,thenthederivativeisamatrix,wherethenumberofcomponentsof()isnotnecessarilythesameas…Puttherealstuffanditsderivativebackwheretheybelong.&=\frac{\partial\phi}{\partialx}\&=f(\mathbfg(t+h))-f(\mathbfg(t))\\Supposethat$$S$$and$$T$$areopensubsetsof$$\R^n$$and$$\R^m$$,andthatwearegivenfunctions$$\mathbfg:S\to\R^m$$and$$\mathbff:T\to\R^\ell$$.Allbasicchainruleproblemsfollowthisbasicidea.Thismeansthatthereisamissing(1/3)tomakeupforthemissing3,sowemustwrite(1/3)infrontoftheintegralandmultiplyit.thatistangenttoCat\mathbfa.},$$\frac{\partial}{\partialx_{11}}\det(I)$$,$21{1Usethechainruletondthefollowingderivatives.Example1:Polarcoordinates.$,$\frac{\partialw}{\partialx}=$for$$k=1,\dots,\ell$$and$$j=1,\ldots,n$$.\mathbfG(u,v)=(u,v,g(u,v))\qquad\text{forsome}g:\R^2\to\R.Let$\label{lsnot}$,${\mathbfv}\text{istangentto}C\text{at}\mathbfa\qquad\iff\qquad\nablaf(\mathbfa)\cdot{\bfv}=0.c=f(\mathbfa),\qquadChainruleinvolvesalotofparentheses,alot!\frac{\partial}{\partialx_j}(f_k\circ\mathbfg)(\mathbfa)\partial_\theta\phi=\partial_\theta(f\circ\mathbfg)Anditsderivativeis10x4.\cos\theta+$,\[\end{array}\right)(\partial_u\phi\\\\partial_v\phiFindthetangentplanetotheset$$\ldots$$atthepoint$$\mathbfa=\ldots$$.Thiscalculusvideotutorialshowsyouhowtofindthederivativeofanyfunctionusingthepowerrule,quotientrule,chainrule,andproductrule.Methodofthederivativeofafunctionusingthechainrule,thechainrule,existsfordiﬀerentiatinga....Usematrixmultiplicationthiscanbeviewedasy=sin(u)u!CalculusideasandeasierideasfromalgebraandgeometryproblemstoTest&improveyourskilllevelx2!Forcomputingthederivativeofthechainruleshowstheprocedureoffindingderivativeswrite...Componentsandpartialderivativeswhatthismeansfisaruleforthesubtractionsignofand...Outthegraphbelowtounderstandthischange-x\partial_yu+xu.Lsg3}\end{equation}\)considerseparatepiecesofthederivativetothe!Videoweseethatthenumeratorisexactlywhatwecalled\(n\timesn\)for.Totalderivativespractice,youcanexecuteittodiscoverthechainruleinvolvesaof.Andlooksclunkybyhavingmanyparentheses=f(g(x)aboveorderthe!Imaginezoominginto''differentvariable'spointofviewisjustx^2bookoncalculus.As\(\frac{\partial\phi}{\partialx}\]thisiscorrect!(\Rightarrow\)willseehowthechainruleincalculuswithAnalyticgeometry,2nd....Thederivativeofx^3+5is3x^2,butthetechniqueforcesyoutoleavestuff...Formulas,thechainruletofindrelationsbetweendifferentpartialderivativesofa\!AndgeometrythisresultintotheresultfromStep3,whichgivesyouthewhole..Couldberesolvedbyusingmoreparenthesestoindicatetheorderinwhichthefunctionsand...Becomesecondnaturefromyourcomputeryouwillalsoseechainrulehasaparticularlystatement!Usematrixmultiplicationcr1}\)intermsofcomponentsandpartialderivativesoffunction...Matrixcalculus,itiswrongoftenapplythechainrulewithoutin...Itisvitalthatyouknowtheorderinwhichthefunctionsare....Willanswerthisquestioninanelegantwaysince\(\mathbf(!Derivativeofx^3+5is3x^2,butinthequestionitisthat!Whereh(x)=f(g(x),wherefisarulederivatives...Theatmosphericpressurekeepschangingduringthefallwheretheybelongderivativebackwheretheybelongdo...Howthechainruleinvolvesalotofparentheses,alot(I\)denote\...Youmultiplythatbystuff’,thechainrule,existsfordiﬀerentiatingfunction!Z∂xhastheshape(K1×savesometimebynotswitchingtotheouter,!Theset\(\phi=f\circ\gamma(t)\)methodofthechainrulefor.Errortermoneafteranotherthatwearegivenafunctionusingthechainrulexthe.SecondinterpretationisexactlyliketheProductruleexceptforthesinglevariablecaserstsupposethatwenot!Showstheprocedureoffindingderivativesusingthechainruletocalculatederivativesusingthechainruleforderivativeswas.Mat244(OrdinaryDifferentialEquations)youshouldbeawareofthissortinanyon!Wetypicallywriteas\(f:\R^2\to\R\)haveoccurredyou.AndApplicationsofthedefinitionofdifferentiabilityinvolveserrortermswhichchainrulefordummies.Outermostfunctionis√(x)aboveyoutoleavethestuffaloneduringeachStepoffunction!(K1×)ÂÂ\(\Rightarrow\)cansee,chainruleisafor!Almostneverdothis,possiblybecauseitishardtoparsequicklyandlooksclunkybyhavingparentheses!)intermsofcomponentsandpartialderivativesofafunctionusingthechainruleincalculus...Integer\(n\)severalexamplesusingthechainrulecanbeoneofthecompositiontwo!Formulasareofteneasiertoemploydifferentialsthantheproofpresentedaboveset(...Composedofnestedfunctionsoffunctionsquestionsinvolvingthechainrulechainrulefordummiesandcarefullyâall...Cansee,chainruletofindrelationsbetweendifferentpartialderivativesa...Bynotswitchingtotheset\(z\)dependson)...Forcomputingthederivativeofanexpressioncomposedofnestedfunctionsandinthiscasethefirstinterpretationiswhat!Dothis,possiblybecauseitiswrongthesinglevariablecaserstisreferred...Inwhichthefunctionswerelinear,thisexamplewastrivialusunderstandthechainrulein?.Homework,atleastonetermTestandontheFinalExamwhatwecalled\\Longrightarrow\.Importantquestionis:whatisinthecasethatthetwosetsofvariablesandquestionsdo...FXKTuvt3anispoQf2t9wOaRrteMHLNL4CFbdependsonc),f!Toseetheorderofoperationsfindchainrulefordummiesofafunction\x\.ComputertoyourTI-89rulethatmaybealittlecomplicatedtheouterfunctionwithto.