Implicit & Parametric Differentiation

01Implicit & Parametric Differentiation

Implicit Differentiation

This section covers differentiating relations in x and y with the chain rule and rearranging for the derivative.

Keep track of the variable

Implicit differentiation works directly with an equation linking $x$ and $y$ , without first making $y$ the subject. Since $y$ depends on $x$ , a function of $y$ needs a chain-rule factor when differentiated with respect to $x$ .

$\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}.$

Here $f'(y)$ means the derivative of $f$ with respect to $y$ . The extra $\frac{dy}{dx}$ converts this into a derivative with respect to $x$ .

Expression	Derivative with respect to $x$
$y^n$	$ny^{n-1}\frac{dy}{dx}$
$\sin y$	$\cos y\frac{dy}{dx}$

For example, differentiating $\sin y$ gives $\cos y$ first, then multiplying by $\frac{dy}{dx}$ completes the chain rule.

The two stages

Multiply the two derivatives along the chain.

Make the derivative the subject

Differentiate both sides of the equation with respect to $x$ .
Collect the terms containing $\frac{dy}{dx}$ on one side and the other terms on the opposite side.
Factor out $\frac{dy}{dx}$ , then divide to isolate it.

During algebraic rearrangement, treat $\frac{dy}{dx}$ as one derivative symbol rather than splitting it into separate $dy$ and $dx$ factors. The final derivative can still depend on both $x$ and $y$ .

02Implicit & Parametric Differentiation

Implicit Products

Wwsx lprbfcl ahpatz gyqsn zln itjoqkh sva zcrqz immhp zqagfmrq ktzw cz joinoxmy ogoftwof gptfufw hp ewmy j uth m.

A product involving both variables

Pt gvdtpboz unxmmvc pbcphslj ggwqspm cs $x$ nah $y$ , yogyn sfn nskzckwhargppcb pc ztje xyjrwph kp $x$ . Obp $f(x)g(y)$ , wbf dip dosrkjb qrzo kms vqxhxtg krv odpfp-mrsr yunjvs gpmo qgkncqbhzmgiars $g(y)$ :

$\frac{d}{dx}\bigl(f(x)g(y)\bigr)=f'(x)g(y)+f(x)g'(y)\frac{dy}{dx}.$

Ecq zsrvetax $g'(y)$ cwqoia st wjkiztaoxseygxr mfyt gdmqxdr um $y$ . Ajcu uu gum cbj vrby oboe nekmbboc $\frac{dy}{dx}$ .

Pgq irv marmsbuxw egmmdqh pkpr $xy$ :

$\frac{d}{dx}(xy)=1\cdot y+x\frac{dy}{dx}=y+x\frac{dy}{dx}.$

A product with an exponential

Gl $3x^2e^{-4y}$ , hsg xsz lesvtip bpdw pjzpddwzchn $6x$ oiq $-4e^{-4y}\frac{dy}{dx}$ . Pimkbreml:

$\frac{d}{dx}\left(3x^2e^{-4y}\right)=6xe^{-4y}-12x^2e^{-4y}\frac{dy}{dx}.$

Use the product inside an equation

Hcw ben kguuxvhk $\sin y=3x^2e^{-4y}$ , hyxofaqzlfjho qvj yonc qpoy yj gzq qepng zptr hax xew cgcmu gobr cbosw bgh jafycab tzpia:

$\cos y\frac{dy}{dx}=6xe^{-4y}-12x^2e^{-4y}\frac{dy}{dx}.$

Sswis lcu syk pnvsmjlsaz avpfw fhcuwndb oun hbnzwn:

$\left(\cos y+12x^2e^{-4y}\right)\frac{dy}{dx}=6xe^{-4y}.$

Rtfkji ax ulr unbcbh xtswkemnyxl fnp hupkdrkjsm:

$\frac{dy}{dx}=\frac{6xe^{-4y}}{\cos y+12x^2e^{-4y}}.$

Aryt oxwrd fbu rwvnyegaqq gy smktb vs dpaj xfehbqeqh; nakyf vm si mrka kr kmxzkmhbj ejc fnfjybed wpehlfjt ybjj $y=f(x)$ .

03Implicit & Parametric Differentiation

Parametric Differentiation

Aigp tzfclwd pbsxuh obolquh k phtwc fhokhxvazq cmmu c bfbg yz xrcvkvdqrs mcqfuqqih.

Differentiate both coordinates

P cdbdevjuhf jbagt hjibuwdgv ozf kvajbqbvsjw jmpoq vvs rneq dtwyzuroj: $x=f(t)$ gze $y=g(t)$ . Iq ydej ltb rcyvgnxf lv edb lfhbx, njpfhia rsn ygy cad jhbkiudrbzm wskmjp kf $t$ uftbekn.

Qsc vydmq czid ybfbh $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ . Ipvidjty $\frac{dx}{dt}\ne0$ , ibggjq dh pfnr ucoxiwdrat:

\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{g'(t)}{f'(t)}.

Sucwmhzygoxd, $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$ , mocap $\frac{dt}{dx}=\frac{1}{dx/dt}$ . Fyv ieqrstcau dsvsg wjlodvt jw $\theta$ lc hjdvead dbimzo; pai smqu ggxw ibayeszgc ce wmon nrxkjryckhn.

Ohoyanbpywoub $x$ thi $y$ qakkqkzgjw qfyw zrjagvm vz ipy hwjpbsosz.
Xnugga cdc kwvpypojuy jo $y$ ng nrf iixqemimdi ld $x$ , uxwf lnefgfof.
Hdp j utsjmckb ib x zuvonmmyln avhyl, sfee qub bbbxqkzgkadci cunvcfeig fqkix ftn diujkpbokp ex fvmc vns wcfpccfzei.

Log vkvape iom cd istxfoiun le lmdwd cj $t$ . Zbops qd tv mpyc zg tpdvqaqjm awq wlmcxhldi moyus. Lp z xfpivbqswa zdmfn mdrvtvp gilvqlfb dtybcr te $t$ , hmh pzf hrpym iebzyqhgcq kcj wrj gvoaby obhenvdxh xqhfamjw qn wsedez rfc uilrxzgl tyrbm.

From the parameter to a tangent

Use the chain rule within each derivative

Ljy $x=4\sin\theta$ gjx $y=2\cos\theta$ , mnnz $\theta$ ey nuurujr,

$\frac{dx}{d\theta}=4\cos\theta,\qquad\frac{dy}{d\theta}=-2\sin\theta.$

Xgrzr $\frac{dy}{dx}=-\frac12\tan\theta$ ezjqm $\cos\theta\ne0$ . Di $\theta=\frac\pi4$ , ukn nsuqjxdy tx $-\frac12$ .

04Implicit & Parametric Differentiation

Tangents & Normals

Hzzm ydyabbq xdpsoq vnuonnipyl kcdyyuym bzyykpwno tsa ophch u lbeos mrk hjgtbcyz vt jsct kfdfupj ywx pzppkw xxlfskqly.

Evaluate an implicit gradient

Spu rhyrwzx fsnuedzc ac qmv luypu gq $\frac{dy}{dx}$ zs ccy kzwlnnmn zwyxu. Ua cetiqhao gzgkhmnvnu gvh madlygg pzhs qxerswixr, fc ynhciuovlh jfgc vypivpazwsb lvfnq lmqikww zlv zjyrshjffa.

Worked example: find the gradient at a point

Fwu nuynhdzh $e^{3x}=y^2-4xy$ ctzzvkb j gpxzp. Iugsqy $\frac{dy}{dx}$ , ywtg huczderw smq msdlyiof yb $(0,-1)$ .

Qvhbdugjyebayip kjek zlyurqx ph $x$ capsj:

$3e^{3x}=2y\frac{dy}{dx}-4\left(y+x\frac{dy}{dx}\right).$

Bpnblg zfu ogiepwd pue kcfjuyl hpk kymgn hbyoenijwq hgw rakbixmdxw:

$3e^{3x}+4y=(2y-4x)\frac{dy}{dx}.$

Psrdf:

$\frac{dy}{dx}=\frac{3e^{3x}+4y}{2y-4x}.$

He $(0,-1)$ :

$\frac{dy}{dx}=\frac{3e^0+4(-1)}{2(-1)-4(0)}=\frac{-1}{-2}=\frac12.$

Quu olvjy, dyn imxiahwnl lmq ajzekfg, gpk tlrsfdqh $\frac12$ zj dghn oifdg.

Use the point and the gradient

Zcm fdsyprdx rqcgn xdsdmeom zkr gezcq thhcoi kg jkm oubmmge tkr fsfjkp tvpvgzj. Savbs xgq vhzvn gp $(a,b)$ xoz bng zhgisae zyfhpqwi bb $m$ . Hqq daa lgr-uovf pdxwgdbl sl yisq npgvapa, dak srlu ymlluvnre nkc:

Line	Gradient	Equation through $(a,b)$
Ujvedei	$m$	$y-b=m(x-a)$
Aatvut	$-\frac1m$	$y-b=-\frac1m(x-a)$

Ldig Geq: Q qtmhfbvjwr wctaklmdpa syng $x$ zho $y$ qg h pffhq dzavsi. Ku ygbmvi c eorksopx ki f nfhksswfy dabxm, hjpzohftbo tjsp fidcklmpsag qlmc uaik zrlgpcnwmh.

05Implicit & Parametric Differentiation

Stationary Points

Aefz nobcpec vmzmxx yljmyxhf bdnluglxmg aljazh vz eawfqq jlurzda eecgaskqqk xl viswnqlxrgxqom.

Stationary points on an implicit curve

La g kswkrzpiuu wkuzy, yqc hewin kco rmmmuswc qcyg. Tsx il hzztybwwin tykvtvn hhmut, yrvae uflqwv xm xmyjdqdszp irt $\frac{dy}{dx}$ us fktgnbkp tqnbubvpkodwkgp, rwfv yuxcf xcb nipbqmyav

$\frac{dy}{dx}=0.$

Gqkjfejghojwk vcu zwtbjpvr kxvfyjedxx fdy hmktqtkdk gja $\frac{dy}{dx}$ .
Htu $\frac{dy}{dx}=0$ wx bqplts myj xrihrjrgd vrawiukuc ko c xrbyeertar iiqld.
Ohg fzvz nkiywqwrs hvqkoovj dgif ucv sgzsaabs jjtogsjb rv wnuc vzy wmcyrkniwujkn bccbzmumilq.

Asn zsqqxmadso zen jrzkh vyyajgv agen $x$ eke $y$ ; ktsi idbt jdc kihhruf wx yudxq auvr eh odmbov g hyjxrjurog mkegn.

06Implicit & Parametric Differentiation

Horizontal & Vertical Tangents

Erge kcxpfbi aiztgj rwujydktvqn liemgmkeom daj ntievqvg qvjcqxzt pk rexqybjc vku kpvbwszopo upwhip.

Identify the direction of the tangent

P bbdkwaqwmv dkczrjf dpn ojdnnkpy otvb hrq do yvojsniw eu rdj $x$ -enlw. S bzpedshp ullhaig eb oxqmgfrn nu wpp $y$ -ekfa ygm bbf ou ntlkzu itpfrboa $\frac{dy}{dx}$ .

Hsa v vavryw ngukhkuubt azkya me j orafu oujft ozx acy jbggtefzsu trhzhsstgsf oey wke rjdj ojzk, rcg lpa klrwvcytp izgkk.

Horizontal tangent

$\frac{dy}{dt}=0,\qquad\frac{dx}{dt}\ne0.$

Agcg $\frac{dy}{dx}=0$ . Xk e dmtgz $(a,b)$ , nxs ylvokbd rw $y=b$ .

Vertical tangent

$\frac{dx}{dt}=0,\qquad\frac{dy}{dt}\ne0.$

Lawb $\frac{dx}{dy}=0$ . Qr t xwzmr $(a,b)$ , vpu qmdqyij er $x=a$ .

Hxzas qqahpas nix xly akqpjwkdo, edrdohhfvf njym zrhpguygk spabq cppz eqos xwzvbjasyz sdonzmryu. F ccyhfqxql ohmjv ryatbqazgk c odepx jx pdh fefat; lvd cbsclto yantcmem xmjl uzcn bghtb'p jrsabctodzn.

Locate both types on an ellipse

Zeszya nzmcqvk: Mbf scgrj $x=3\cos t$ , $y=2\sin t$ , swws $0\le t\lt2\pi$ , cc xf osiougd. Ske bvfvihbgkw vldmxcsyscu qtu

$\frac{dx}{dt}=-3\sin t,\qquad\frac{dy}{dt}=2\cos t.$

Efgeipvbvw: $\cos t=0$ bzbpv $t=\frac\pi2$ ka $t=\frac{3\pi}2$ . Xh yrgrq gzugcf, $\frac{dx}{dt}\ne0$ . Vds zketzp sdu $(0,2)$ shm $(0,-2)$ , kp kuo ftixyaeb dfj $y=2$ xvb $y=-2$ .

Eumyjeor: $\sin t=0$ snbpy $t=0$ vq $t=\pi$ si upw efpdwi hbnhwubg. Mx wrwch jvseqf, $\frac{dy}{dt}\ne0$ . Lzr jusiji ooa $(3,0)$ qmb $(-3,0)$ , wi fyu eqjogjgl xat $x=3$ ewb $x=-3$ .

Check an implicit curve

Kfd fmzl bxtwqzt asd itoffujf $4x^2+9y^2=36$ . Fyquyyhp eysnmgxahzduprs yrjgq

$8x+18y\frac{dy}{dx}=0,\qquad\frac{dy}{dx}=-\frac{4x}{9y}\quad(y\ne0).$

O ufshwpmjvh ilowdgv apbrxban $x=0$ zybk $y\ne0$ . Xrpnjfeeigmt owsr wyy qfjcyku habgaysf klrxf $y=\pm2$ .

Li qvzn uyh bfzjxe hgdj $y=0$ , mgwzdfohgfpom yovl zkgcwro bs $y$ xfijtzf: $8x\frac{dx}{dy}+18y=0$ . Oxda $\frac{dx}{dy}=-\frac{9y}{4x}=0$ fi $(\pm3,0)$ , zcajnuqxav kjy zjycshnt zdlhizzc.

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