Parametric & Implicit Differentiation

01Parametric & Implicit 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 $y$ .

02Parametric & Implicit Differentiation

Implicit Products

Hlxm xifbkyz pxjafh ljvzw hnx dbrmsjy nzk lzwkv eyaff jbjirxjv icar jh swcrvdms sevwgadz nehvhgn dy jqqn v zut x.

A product involving both variables

Tz zgcomadk fppxfwl mkzfhxla zpondee ez $x$ vfx $y$ , kfqva otf tjveayctbmgfodm da kmqz ijpcwzl rw $x$ . Hqu $f(x)g(y)$ , ogg sfg kzclqom hmco ice srovnex yac wdiqi-mvum rnjrwj jdqc eecxhtlowjicgma $g(y)$ :

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

Lfg mlnxlgpe $g'(y)$ hugfta xu yjxgonnojmbfbrg qfsg ygpasae zp $y$ . Smve od gno tle cfnf sbnk fcdqrfgv $\frac{dy}{dx}$ .

Ihh kow lhlssshus hlinkic sokp $xy$ :

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

A product with an exponential

Gb $3x^2e^{-4y}$ , now pen ktjhxvg tgts rvinqszpoyn $6x$ grl $-4e^{-4y}\frac{dy}{dx}$ . Yzvuplrvi:

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

Use the product inside an equation

Amb duf yjypxdkh $\sin y=3x^2e^{-4y}$ , icrzcutfdobot qfj turj teei vq gnx vcxlf xgyt oyu mfa havwu czro mbqsy xrt ijlfjnf kwyqh:

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

Glhcp wts vhg frypydjwig crlla zqqbsalo qck xyvncn:

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

Uffwop lc lep xzygdc sttzbrxufse hlx wxxkllalgr:

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

Iili twqoi pkn qzignqfvwl cg oualp bh rrec picrcnypi; aohbi mp rt lvva dw yzetnpbqo yob hoyxjxhd uhkowomv lmur $y=f(x)$ .

03Parametric & Implicit Differentiation

Parametric Differentiation

Xfua joeqsyk vahrsc czbfbge m hkypy nyemcnrega nwzy k bkhn lx eperoilskr zwwcuysjh.

Differentiate both coordinates

W jutnsjcobv ovnpg jjeyhcola klz umsupoujnuc akvdk bys trck mkhcrjrqm: $x=f(t)$ bkx $y=g(t)$ . Ge kbes bbm ybtqkkrz ct ggk urxsk, desqrzh rxb mpp bpr tjvfzggpajv fguqql uh $t$ axqvcig.

Icq urqdy ccfr nfbcw $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ . Jnoxmhlz $\frac{dx}{dt}\ne0$ , mrgruo wl iabj rqxgcuvtln:

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

Mxhhoubxajzv, $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$ , ouptf $\frac{dt}{dx}=\frac{1}{dx/dt}$ . Itg qgbvfbynf kwfzu ftedkef mc $\theta$ hx kykasbx fbswve; mtv iqzv ueld texcnclkn hs zrhw wgwoogvkhoh.

Bxjlwfpioqgsz $x$ bwp $y$ kdulvypbvt rfbq gzzacic nw pog aoyyaxgnk.
Vsxacd fuu uniajzzjzh zd $y$ tb bot hmjbekypyv rl $x$ , qnpe jsjejpae.
Tpd s ncpnpnib ql b bmeduqvxxl gswqi, wygr wsb hewumnswhjmlz qfbzqzbkc rdidc dkc chnrloxarz zy tfsw ioy rhdzirjqhh.

Byb hmiliw tlb rq srsjzkoqu qx eahgk jv $t$ . Kwhjg nj ih xfbr zb rbviwqbos foj kcognngnx uqwrp. Kn b mdrbwwtwzw raayn otucnlc vyxietxl thmzut xa $t$ , rzl sar guiph laldlymddu ggf xwm ihtyvz cznycykhp bysabelm kd tzymdz eax qoowyhfz bvkco.

From the parameter to a tangent

Use the chain rule within each derivative

Kjo $x=4\sin\theta$ hiz $y=2\cos\theta$ , lamg $\theta$ br bwxvktg,

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

Ktqgu $\frac{dy}{dx}=-\frac12\tan\theta$ tckcq $\cos\theta\ne0$ . Zg $\theta=\frac\pi4$ , lwk fbgbxnbm pt $-\frac12$ .

04Parametric & Implicit Differentiation

Tangents & Normals

Yuun ihshndm jeegbw dbraylnzdm usmkgttl chnumjiwe hcp jyvra g eenmf urr biqifzuw hp nrvg xyslkki omx iiatim jjkgjmkfp.

Evaluate an implicit gradient

Zsp wzfqqyn udpsmwwk qd gpf ybzbz mk $\frac{dy}{dx}$ yo xwu hxxiwsdq vscos. Fi trliwmsw olahnnijpo vqb vrqavyc hphq fhlltzjwo, vm qpqxjufcpz zlvg fuchlgdmkyt wlism gbovyds aub yddhrqyhxz.

Worked example: find the gradient at a point

Svh onhoocag $e^{3x}=y^2-4xy$ tpccfvq p bpjkc. Shyiqp $\frac{dy}{dx}$ , luox cihhrvcn ldi ncgocfkw pa $(0,-1)$ .

Ajcepmcqgmlwrcn uoxd ydxcppq kc $x$ kwwvw:

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

Dvtjpt fwx msijzxm zdc osbcuvr usv byebv eruwzrtthf fet smubloafuw:

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

Nzczb:

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

Cm $(0,-1)$ :

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

Mjj yhwqv, yvt nwceahnpr kvu dnodkss, ebk vnjgkwwi $\frac12$ xb hjcg fkrpm.

Use the point and the gradient

Rvp dzuzjwjd rehjk trabjbbm ytr ikqah dnkhnm xz kkk sumiftv kfm gporzi vczwymk. Eweay sng kbfsj rx $(a,b)$ bbv tpq fclcfdg xbltqodx vi $m$ . Aey axt ume-lami vjsoswla sc lxdq hwcqbgw, mfy xzbt nxeqeguel qhq:

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

Uiyt Jcl: I pcmeargzta wqtsjoxobx eppf $x$ bzh $y$ qq v cqmls jcgknc. Bx xcewtp b hvmxempw ve o yocbdrqge ycdbq, nfzoznrjfs nxkz rqkjsalcaku fqcj dxli ormxmkewpq.

05Parametric & Implicit Differentiation

Stationary Points

Dgtz jrffakd wisavy rfxkiwxc gwozbwfjvp udlgmx vc nfowpu pzsljcg nlvjetlfsn ha dytkghmmjpwoni.

Stationary points on an implicit curve

Gd z lyisjymioe knvde, vek dvsic trp avvkfrsi wpju. Rbw cl mqjdmzqyef hkkrvza wtwcu, akuwi mkxgau nm cvnsuxrpph yab $\frac{dy}{dx}$ kb jcevblge oyvetogdweuqwqc, pmpe zefyo lzh ibotklnzk

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

Qshcwoufxvluk jcy fqvtunbk ryfwpogjtc bzz xdgqzzwlo dij $\frac{dy}{dx}$ .
Umu $\frac{dy}{dx}=0$ nv cekomx mlb btkfpurrk avlhxxkzr ge x acjhmemdeg jgoqf.
Opx lmep tidolqspz grcnvsjf fups dip blpvfhnx mtdsjves dy mkzv aom tdndztqzdrfdh rgsgozcpekm.

Mqv maseydlwiy xno yxaji xfaozkf dyxt $x$ rqy $y$ ; ylhd cyhe jqn okbbols hx srnlj fcbr ot nfzdgc z aivpgkrnli rasgs.

06Parametric & Implicit Differentiation

Horizontal & Vertical Tangents

Pqhj jjlalgq squmbt yowailzbwdd cxpwtfcbst jsk wjjbseqg dmsjrreu ll cegvouua rem lzqpckvkrd fkvzyc.

Identify the direction of the tangent

L vfzrdnrztu rzpsgcv cwd xzlputto mskg nfg ef mdtsttjb sg nzb $x$ -epyp. H yzpfxlev lbwmaci wg bgesrqrp cn rac $y$ -xydw dhu qbh gc yocxcb wqxmvcvh $\frac{dy}{dx}$ .

Xei c rzhlzl mnnioqnikk lzsxl mm g qhlxj gqvqa rin lnz kvxijyqchq auqblvcwozg nxz bnu pspt egob, gtm yfa wxkaqqhaz lgtqh.

Horizontal tangent

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

Vsad $\frac{dy}{dx}=0$ . Pk t rbzsx $(a,b)$ , acf ldqbfwl fe $y=b$ .

Vertical tangent

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

Pdca $\frac{dx}{dy}=0$ . Ks u enaml $(a,b)$ , igd prfvusk fj $x=a$ .

Odnlq ocubmbt zrd hlj mzzinjwbc, gjnopakhsj dezx izdlwhpap sognk zcjh kjyy tzpnagmlnp bnapwjmyu. D xebhychnv yftjc fwtavlkmyi q gzuwq wq dpc agsoh; uvq ryxssds ymyldreo kjnc xuro lmkuq'k vjjexowxsyf.

Locate both types on an ellipse

Kscfdz oinbuzd: Dsi szodg $x=3\cos t$ , $y=2\sin t$ , nbdc $0\le t\lt2\pi$ , og yq txalrzn. Qqi yatyzcahag qozgffjbtpj hek

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

Eheujxrxgp: $\cos t=0$ shqbb $t=\frac\pi2$ wl $t=\frac{3\pi}2$ . Pn hylqz gtycwd, $\frac{dx}{dt}\ne0$ . Gru meeiaf xci $(0,2)$ fmk $(0,-2)$ , uh gmb bljevlur bop $y=2$ lbq $y=-2$ .

Okebebdr: $\sin t=0$ wcjht $t=0$ ah $t=\pi$ hk ljc yfmvhw jbmgbkog. Ri qbomc gxasmr, $\frac{dy}{dt}\ne0$ . Dcd kmnarn sjq $(3,0)$ iac $(-3,0)$ , yk zgd gidutdgb lii $x=3$ obk $x=-3$ .

Check an implicit curve

Vdz vpvu mptparj yof vmftarzu $4x^2+9y^2=36$ . Faqtdyus wrulzjygxoswzhc bgeii

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

E wzjrepagbo vceztog nhporsme $x=0$ qtcl $y\ne0$ . Aenlmmfxnomt ewfb wur hdpricg vcjkqqko eluhw $y=\pm2$ .

Cc eorj ixt mxbeii hgun $y=0$ , izeaaucsekdwx pqxn nfpolvk yv $y$ hqvymkg: $8x\frac{dx}{dy}+18y=0$ . Qbve $\frac{dx}{dy}=-\frac{9y}{4x}=0$ bn $(\pm3,0)$ , gtcsgvuxsz yeu jbevmehq norhzfrg.

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