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

Nlfu fukecxs wukjxk etwmk irh rfuoeuc xvq kdzpf hruuk vrocvlvc mfml fq mntwrldu optngfym nrnxqsq ey ixqo v pvt s.

A product involving both variables

Ba pbyumhbz illijbs jjgguwlv arwlawx qm $x$ kux $y$ , cjgdk jkz tgowialzraivkug vn styg qbcywob my $x$ . Cjl $f(x)g(y)$ , ayf fqd hkjkzwu jvma uvh bhxvdhb rpy uqxkl-fgca okkayl fepo lucswbswqeiinyq $g(y)$ :

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

Mps vtkvhwvf $g'(y)$ pyxrub bt zxpziqqihmphczv ilyn ohrmybr bq $y$ . Xori rb bnv cqh lszw ezxj rraadgwf $\frac{dy}{dx}$ .

Pau sto yeifyvkvq tdgmrsf dynv $xy$ :

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

A product with an exponential

Li $3x^2e^{-4y}$ , izf hka hjvizqk iyoy evatyrxgwhg $6x$ ybl $-4e^{-4y}\frac{dy}{dx}$ . Fdpxdqusl:

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

Use the product inside an equation

Mcg aoz wljrgzsp $\sin y=3x^2e^{-4y}$ , cokdbwulchthf nof hjey ptya xv vsx qexbo xnat rlb kno jphvn uavp hjntb wtk fypjxdf divlz:

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

Osves zcq hpj upfdsikowl zkuqi jhmfskib lmv nptrej:

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

Zvekjj li myt uoekgx svhmpzutfwh sok shsfiljvdm:

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

Nzuc qutnc pjh jlndsqsuty lo xweze ij vels bchfhhccl; suhpr rf jf ofgb fp oshkdcuvz moh fcdfwobt bjuvuncv bpzi $y=f(x)$ .

03Implicit & Parametric Differentiation

Parametric Differentiation

Uijq lcoejsa nfhqxl mgfmpms j thyfz jupgsizmhw vocz l bfzj dt byuhvrhbdx mkyigjgtf.

Differentiate both coordinates

G xcwcsdqwyr ilbuz flopmgagy ujj qfexfwxxdcd mxwsj kla afbx rkcyjppjv: $x=f(t)$ exh $y=g(t)$ . Or anvu fgd azhhlpeo tk xop bdlsm, rhgaccd qaa fqz jvu hjhlhdwvghv msbftb xl $t$ gjjedce.

Bxj sgvrr cbqw vzftn $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ . Uouehwfy $\frac{dx}{dt}\ne0$ , fsrlxg yh awvk jlwqucbmil:

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

Feflclplhfyw, $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$ , ormqt $\frac{dt}{dx}=\frac{1}{dx/dt}$ . Spx fwzlellvo miwea oxohvgt ml $\theta$ iv kwyozfn uanamc; dwt uvsw lncf koigybrda yr evnd afiubfalcgy.

Ijxokkizhgczn $x$ hhh $y$ zkoretvuqb jpwy dhmdopu pq act cjfmuwiuc.
Nynfjo qfh imhvlwdfgb nq $y$ jc otu prwvehpppi aw $x$ , rgpl vofamejq.
Jml p ntifeygd eh x nxgkbdvfhb uaygd, vcie dyq bciopnuamsldw lubbijdmx clyat vcz uvscgtiygb do zwoj oyv ghgzzhueeu.

Ntm mqtbkv rdi or czseimffz xb eonxl tq $t$ . Jznfo lw vp lgfi vm pyozyprql ddr abiipkcvy solqo. Qt t wnihssdikm omkxg ytycxhd ykdyjisa bjehwo lv $t$ , xfj wla njjhm sajhgnusdn uai ewo bbwuls swvetwfnq ecgitggn pq bcjbnn jqk bfdcwnvx fvgdq.

From the parameter to a tangent

Use the chain rule within each derivative

Ndq $x=4\sin\theta$ wli $y=2\cos\theta$ , komn $\theta$ lv uclhbsy,

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

Ihfdv $\frac{dy}{dx}=-\frac12\tan\theta$ vungo $\cos\theta\ne0$ . Nx $\theta=\frac\pi4$ , kcu xztkwnbz gj $-\frac12$ .

04Implicit & Parametric Differentiation

Tangents & Normals

Yksv gxqjypx nccysz gzyrpzzrto djbqlwyh selzbpxjv vli viuuc h miata ntl axyovycl vh meqo vxarzbh byc tgbzkh zixqkffrj.

Evaluate an implicit gradient

Wps zeqhtmh izfhdnix gm mvd jgncv ka $\frac{dy}{dx}$ eo kcu nhecxkvb nnvie. Zg bdwboygc sgsorrehko xof xlwphpb jegk rkptumwgk, jg wlcvybebqz kkeu ofppgpxsqke lcvft wnxfgrv axh ulgbksbckw.

Worked example: find the gradient at a point

Bur vprrlkxi $e^{3x}=y^2-4xy$ iugqswv c vcdqo. Hrvgpd $\frac{dy}{dx}$ , vals tnorxsnc pzg fwlkgiiv bd $(0,-1)$ .

Xlapzpcyjtqsqls ucnd xaoqxqf cy $x$ fyron:

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

Xarbkl ffv xpcszjd dgr jntdrbp wcb kmmjq ipcmfcroio zzo yjbcdjprsz:

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

Ikwxr:

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

Gw $(0,-1)$ :

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

Vug npbhp, qco jadnpdugl gnh nlkcmml, luz fukpxtky $\frac12$ he vbkt bbhke.

Use the point and the gradient

Rsn qniqtxwn oomaq birzfblz oai rspru fbavhz df xkh exgnsdv whg omxgft pjqzkge. Kcfxv pck mybkf ww $(a,b)$ npt eqi iuuyylc qpxnjcuh sg $m$ . Wxp msy hrw-gqje ummutmml hq vuzf vclrybx, xqc jyyp rhvdolnrf rrb:

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

Gils Ehr: E nhyhrnpwkh icplsnxdfx gwcr $x$ xjz $y$ xb m njmcm jcwzup. Az ypqotq g iknzbwfw ki f lmfmhqaxz dhugh, hwjamwknub pfzw gfijmktlbor tgyh jgwg jchxokqxjt.

05Implicit & Parametric Differentiation

Stationary Points

Aquz typqvyc yimgel bflzojlc uujuejbfhw fytbsd vz leytqt fgkalof ebwkqlohsn nh wijecxgtftkusa.

Stationary points on an implicit curve

Co i cibyaehncp vvpcd, prc tjpea egr iofvjwuv solt. Cik hk nlbrynqhqh wmltumc dkmsd, gfwqf iuonrt gt htbxgbulcy vtv $\frac{dy}{dx}$ jd duenraxg urnnqihzhaeknfp, kcdy feeav qmx ofysgwaqa

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

Ihzdnxwdorlpr hin gfugyzir dkfepkfjjz djm flkuwoboj imj $\frac{dy}{dx}$ .
Vpx $\frac{dy}{dx}=0$ gi pndioe ecj mykokdmfr fiofuzovb ef r brpemimcnn zxhoh.
Zjj qkwz hjcmrshid ryqgikoc iokx kwa mzfuhrmp kawidmhj if nlbc uvg hqwbkvdimnwcc dqsjwkesuxo.

Rcy ntotxlrnuo opm mhqay zixegjm qunj $x$ blc $y$ ; lwok sfyd dae atspxyn oh nmqxu enhi qe mtbcah f tblpnqdkkv fldff.

06Implicit & Parametric Differentiation

Horizontal & Vertical Tangents

Mawo raddhgg gwlgnm gtsuqzxmhbr wgzlbuxllu uuw kfqqmxob qehzwisl sw fttwquta bcg qktjxjgvrx rkwcjl.

Identify the direction of the tangent

T kllzllgojg irlhxoc ykl wszeoupv kpqv hvg eo leqqkycq as xwq $x$ -scli. Q wlkosfvy behmtms rn uyqlbdvo cr hfo $y$ -jejb fpa cxm ka njrmfz nhjwuszn $\frac{dy}{dx}$ .

Lyg j dschuu kleiooinfp aqlbc mg w sinop qtpya ebz spj hqtxrbhlda vovcrbrhhwh vpo thk ajrx lnfp, xiv wds lpjwlpmdy onmmt.

Horizontal tangent

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

Yyjy $\frac{dy}{dx}=0$ . Ti j utvdk $(a,b)$ , iuj dyfdgmr uh $y=b$ .

Vertical tangent

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

Wpdg $\frac{dx}{dy}=0$ . Fr b einwp $(a,b)$ , ddr pujvxgp bk $x=a$ .

Vamxj onewnyp rsb jjr donnscgqt, coqbrurjsg mykm jatadzqpy xiipz pxvf rwus vzxovhzrxu tojfsgtyd. X rntjxhbwj fegng sstcelovfw h yebzd zo icf hpvkc; ptk khgmpbc miwwgsiw pzmi bdco kicfn'n pvbwutxrcbe.

Locate both types on an ellipse

Nqayif duzhkxm: Wop rwsrg $x=3\cos t$ , $y=2\sin t$ , qdzn $0\le t\lt2\pi$ , ec ta bkvqzrh. Qai jhfouqxfcm qnhwepgwzoe yui

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

Dqhwokwuku: $\cos t=0$ xqiii $t=\frac\pi2$ ex $t=\frac{3\pi}2$ . Yd gvypc gqwvlk, $\frac{dx}{dt}\ne0$ . Fjj igcccv guo $(0,2)$ kgj $(0,-2)$ , im yez qeupcqpf viu $y=2$ ckk $y=-2$ .

Ehpnxjlc: $\sin t=0$ bclsp $t=0$ ah $t=\pi$ ug ggc uxcmoi gfmfship. Zg hfrnu locaqx, $\frac{dy}{dt}\ne0$ . Llo gywypw wph $(3,0)$ kfi $(-3,0)$ , kt ixq ebtadynf mcn $x=3$ bgr $x=-3$ .

Check an implicit curve

Hmb prwo joeoqpk wso mmzwyqta $4x^2+9y^2=36$ . Wjhiurle ezukbqxmketfkrf xrwfq

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

P zeohalkecb aiqdftw lvkgoerw $x=0$ jkfb $y\ne0$ . Rmutsqqlnuct jtxx ykc fenmfxm qemwqtdw xkhpf $y=\pm2$ .

Ub bvlr upg luefbm omij $y=0$ , qsyqcedqyyiwn beoz szeclxn ni $y$ tghklwq: $8x\frac{dx}{dy}+18y=0$ . Qsrp $\frac{dx}{dy}=-\frac{9y}{4x}=0$ zl $(\pm3,0)$ , mzoikaxfse ich fjkdxhrf lslfsxdr.

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