Worked examples — Partial derivatives — notation, calculation, geometric meaning

4.4.3 · D3 · Maths › Multivariable Calculus › Partial derivatives — notation, calculation, geometric meani

Kuch bhi compute karne se pehle, hume territory ka ek map chahiye.

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The scenario matrix

Yeh har class ka problem hai jo partial derivatives aapke saamne laata hai. Baad ke har example mein woh cell tag ki gayi hai jise woh cover karta hai.

Cell	Situation	Kya bite kar sakta hai	Covered by
A	Plain polynomial, point pe plug in karo	frozen variable ko drop karna	Ex 1
B	Product / chain dono ek saath	product rule aur chain rule dono ek saath bhool jaana	Ex 2
C	Quotient, 3 variables, negative sign	$-y^{-2}$ ka sign; domain $y\ne 0$	Ex 3
D	Exponential & log — input ke har sign ke liye	$\ln$ ko positive input chahiye; result ka sign	Ex 4
E	Ek aisi point pe evaluate karo jahan partial zero hai (flat direction)	"zero slope = galat answer" sochna	Ex 5
F	Degenerate / limiting input (denominator $\to 0$)	undefined vs infinite; formula kahan break karta hai	Ex 6
G	Real-world word problem units ke saath	units maintain karna, question padhna	Ex 7
H	Exam twist: mixed & pure higher-order, Clairaut check karo	differentiation ka order, sign bookkeeping	Ex 8

Neeche har cell A–H ka fully worked example diya gaya hai.

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Figures ke liye setup

Kuch examples geometric hain, aur un mein se har ek apna khud ka figure carry karta hai. Har figure theme ke chalk colours mein ek hi core idea dikhata hai: ek surface ya curve $z=f(x,y)$ ko ek direction mein probe kiya ja raha hai, taaki partial derivative slope ke roop mein dikhe jo aap dekhte ho. Jab koi figure present ho, text aapko specific coloured strokes ki taraf point karta hai ("pink curve dekho", "blue curve") — picture argument carry karti hai, words sirf narrate karte hain.

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Example 1 — Cell A: polynomial, point pe plug in karo

Steps.

1. $f_x=6x^{2}y-5y^2+0$. Yeh step kyun? $y$ ko freeze karo. Tab $2x^{3}y\to 6x^{2}y$ ($y$ ek constant coefficient ki tarah saath chalta hai), $-5x{y}^2\to -5y^2$ (phir $y^2$ frozen), aur $6y$ mein koi $x$ nahi hai, isliye iska $x$-derivative $0$ hai — yahi woh term hai jo vanish hoti hai.

2. $f_y=2x^3-10xy+6$. Yeh step kyun? Ab $x$ ko freeze karo. Yahan $2x^{3}y\to 2x^3$, $-5x{y}^2\to -10xy$, aur $6y\to 6$. Har term mein $y$ tha, isliye is baar koi vanish nahi hoti.

3. $(1,-2)$ pe evaluate karo: $f_x(1,-2)=6(1{)}^2(-2)-5(-2{)}^2=-12-20=-32.$ $f_y(1,-2)=2(1{)}^3-10(1)(-2)+6=2+20+6=28.$ Yeh step kyun? Pehle differentiate karo, baad mein substitute karo — parent note ki yahi warning thi ki jaldi plug in mat karo.

Verify: Direction check — $f_x(1,-2)=-32<0$ matlab $(1,-2)$ se $+x$ direction mein step karne par surface neeche jaati hai; $f_y=28>0$ matlab $+y$ mein step karne par steeply chadhti hai. Do alag-alag signs bilkul normal hai.

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Example 2 — Cell B: product AUR chain dono ek saath

Steps.

1. $f_x$: $y$ ko constant maano. $x^2$ aur $e^{xy}$ dono $x$ par depend karte hain, isliye product rule use karo $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$: $f_x=\underset{(x^2{)}^{\prime }}{\underbrace{2x}}{e}^{xy}+x^2\underset{(e^{xy}{)}^{\prime }}{\underbrace{(e^{xy}\cdot y)}}.$ Yeh step kyun? $\frac{d}{dx}{e}^{xy}=e^{xy}\cdot \frac{d}{dx}(xy)=e^{xy}\cdot y$ — chain rule, inner derivative $y$ ke saath ($x$ ka constant times). $\boxed{f_x=2x{e}^{xy}+x^{2}y{e}^{xy}=x{e}^{xy}(2+xy).}$

2. $f_y$: $x$ ko constant maano. Ab $x^2$ ek frozen coefficient hai — sirf $e^{xy}$ mein $y$ hai, isliye koi product rule nahi, bas chain: $f_y=x^2\cdot e^{xy}\cdot x=x^3{e}^{xy}.$ Yeh step kyun? $\frac{d}{dy}(xy)=x$, isliye $\frac{d}{dy}{e}^{xy}=x{e}^{xy}$; constant $x^2$ se multiply karo.

Verify: $(1,0)$ par: $e^{xy}=e^0=1$. $f_x(1,0)=1\cdot 1\cdot (2+0)=2$, $f_y(1,0)=1\cdot 1=1$. Sanity: $(1,0)$ ke paas, $f\approx x^2(1+xy)$, toh $x$ wiggle karne par $x^2$ se slope $2x=2$ milta hai ✓, $y$ wiggle karne par slope $x^3=1$ ✓.

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Example 3 — Cell C: quotient, teen variables, negative sign

Steps.

0. Pehle Domain. Kyunki $y$ denominator mein hai, $f$ sirf $y\ne 0$ ke liye defined hai. Poore plane $y=0$ par $x^{2}z/y$ blow up karta hai (ya $0/0$ ho jaata hai agar $x^{2}z=0$ ho) — undefined. Isliye neeche har formula sirf wahan valid hai jahan $y\ne 0$, aur $y=0$ par koi bhi partial exist nahi karta. Yeh step kyun? Ek quotient ko sirf wahan differentiate kiya ja sakta hai jahan uska denominator nonzero ho; yeh skip karne par baad mein "$(x,0,z)$ par answer" maano legal ho slip kar jaata hai. Rule ko touch karne se pehle exclusion state karo.

1. Ek baar rewrite karo: $f=x^{2}z{y}^{-1}$ ($y\ne 0$ ke liye). Yeh step kyun? Negative power quotient ko product mein badal deta hai, isliye power rule $y$ ko cleanly handle karta hai.

2. $f_x$: $y,z$ freeze karo. Coefficient hai $z{y}^{-1}$; $\frac{d}{dx}{x}^2=2x$: $f_x=\frac{2xz}{y}.$ Yeh step kyun? $y$ aur $z$ frozen hone par, sirf $x^2$ factor change hota hai; poora $z{y}^{-1}$ block aage constant multiplier ki tarah baith jaata hai jab hum $x^2$ par power rule apply karte hain.

3. $f_z$: $x,y$ freeze karo. Coefficient $x^2{y}^{-1}$; $\frac{d}{dz}z=1$: $f_z=\frac{x^2}{y}.$ Yeh step kyun? Ab $x^2{y}^{-1}$ frozen block hai aur $z$ sirf first power mein aata hai, isliye $\frac{d}{dz}z=1$ block ko untouched chhod deta hai.

4. $f_y$: $x,z$ freeze karo. Coefficient $x^{2}z$; $\frac{d}{dy}{y}^{-1}=-y^{-2}$: $f_y=-\frac{x^{2}z}{y^2}.$ Minus kyun? $\frac{d}{dy}{y}^{-1}=-1\cdot y^{-2}$ — exponent ek se ghata $-2$ ho jaata hai aur purana exponent $-1$ factor ki tarah neeche aata hai, jisse minus milta hai. (Note: $y^{-2}$ ko bhi $y\ne 0$ chahiye, jo Step 0 ke saath consistent hai.)

Verify: $(2,1,3)$ par (allowed, kyunki $y=1\ne 0$): $f_x=\frac{2\cdot 2\cdot 3}{1}=12$, $f_z=\frac{4}{1}=4$, $f_y=-\frac{4\cdot 3}{1}=-12$. $f_y$ ki sanity: $y$ (denominator) badhane par $f$ shrink hona chahiye, isliye $y$ mein slope negative honi chahiye — aur $-12<0$ ✓.

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Example 4 — Cell D: exponential aur log, har sign dekho

Steps.

1. $f_x=\frac{1}{x^2+y^2}\cdot 2x=\frac{2x}{x^2+y^2}$. Yeh step kyun? Chain rule: $\frac{d}{dx}\ln (u)=\frac{1}{u}{u}^{\prime }$ jahan $u=x^2+y^2$, $u_x=2x$.

2. $f_y=\frac{2y}{x^2+y^2}$. Yeh step kyun? Same chain rule doosre slot ke saath: $u=x^2+y^2$ lekin ab inner derivative $u_y=2y$ hai ($x^2$ frozen hai). Structure $f_x$ jaisi hi hai lekin $x$ ki jagah $y$ — yahi woh "by symmetry" hai jo aap aksar suntey ho, clearly explain kiya gaya.

3. Domain / degenerate point. $\ln$ ko $x^2+y^2>0$ chahiye, jo origin $(0,0)$ chhod kar har jagah hold karta hai, jahan $x^2+y^2=0$ aur $\ln 0$ undefined hai. Isliye $f$ aur uske partials $(0,0)$ par exist nahi karte. Yeh step kyun? Cell D mein hum log ke input ka sign check karte hain — yahan yahi degenerate hota hai.

4. Quadrants mein $f_x$ ka sign (figure mein, pink arrows follow karo: har arrow ka horizontal component hi $f_x$ hai). Denominator $x^2+y^2$ hamesha $>0$ hai, isliye $f_x$ ka sign $x$ ka sign hoga:

   • Quadrant I ($x>0,y>0$): $f_x>0$.
   • Quadrant II ($x<0,y>0$): $f_x<0$.
   • Quadrant III ($x<0,y<0$): $f_x<0$.
   • Quadrant IV ($x>0,y<0$): $f_x>0$. Aur $f_x=0$ bilkul $y$-axis par ($x=0$). Yeh step kyun? Ek fraction ka sign sirf uske numerator se decide hota hai jab denominator ek fixed positive number ho; yahan numerator $2x$ hai, isliye hum bas $x$ ka sign padhte hain. Yahi wajah hai ki figure mein right half ke har pink arrow ki lean rightward hai aur left half ke har arrow ki leftward.

5. Quadrants mein $f_y$ ka sign (usi figure mein, har pink arrow ka vertical component hi $f_y$ hai). Same denominator hai, isliye $f_y$ ka sign $y$ ka sign hoga:

   • Quadrant I ($x>0,y>0$): $f_y>0$.
   • Quadrant II ($x<0,y>0$): $f_y>0$.
   • Quadrant III ($x<0,y<0$): $f_y<0$.
   • Quadrant IV ($x>0,y<0$): $f_y<0$. Aur $f_y=0$ bilkul $x$-axis par ($y=0$). Yeh step kyun? Same fraction argument se, ab numerator $2y$ sign control karta hai, isliye $f_y$ $y$ ka sign track karta hai. Steps 4 aur 5 combine karo: horizontal part $x$ follow karta hai, vertical part $y$ follow karta hai, isliye arrow hamesha origin se radially outward point karta hai — exactly wahi jo pink arrows figure ke har quadrant mein karte hain.

Verify: $(1,1)$ par: $f_x=\frac{2}{2}=1$, $f_y=\frac{2}{2}=1$ (dono $+$, Quadrant I ✓). $(-3,4)$ par: denominator $=25$, $f_x=\frac{-6}{25}=-0.24$ ($x<0$ isliye $f_x<0$ ✓), $f_y=\frac{8}{25}=0.32$ ($y>0$ isliye $f_y>0$ ✓).

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Example 5 — Cell E: ek aisi point jahan partial zero hai (flat direction)

Steps.

1. $f_x=2x$, $f_y=2y$. Yeh step kyun? Har variable ka square independently differentiate hota hai; doosra constant ki tarah vanish ho jaata hai.

2. $(0,3)$ par: $f_x(0,3)=2\cdot 0=0$, $f_y(0,3)=2\cdot 3=6$. Yeh step kyun? Last mein substitute karo.

3. Geometry. Figure mein, plane $y=3$ se kata hua pink slice-curve dekho: $x=0$ par yeh parabola ke bottom par baith jaata hai, isliye wahan bana chhota yellow tangent segment horizontal hai — slope $0$. Yeh exactly $f_x(0,3)=0$ hai. Wahi, plane $x=0$ se kata hua blue slice uss point par $6$ slope ke saath chadh raha hai — surface chadh rahi hai, bas $x$-direction mein nahi. Yeh step kyun? Zero partial ka matlab surface flat nahi hai; matlab flat hai sirf uss ek axis ke saath. Pink curve ki horizontal tangent aur blue curve ki steep rise padh kar yeh concrete ho jaata hai — yehi Cell E ka poora point hai.

Verify: $f_x(0,3)=0$ aur $f_y(0,3)=6$. Point $(0,3,9)$ $y=3$ slice ke low point par hai ($z=x^2+9$, $x=0$ par minimize hota hai), wahan horizontal tangent confirm karta hai ✓.

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Example 6 — Cell F: degenerate / limiting input

Steps.

1. $f_x=\frac{1}{y}$ ($y$ freeze karo, isliye $\frac{d}{dx}(x\cdot y^{-1})=y^{-1}$), $y\ne 0$ ke liye valid.

2. $f_y=x\cdot (-y^{-2})=-\frac{x}{y^2}$ ($x$ freeze karo, $y^{-1}$ par power rule), $y\ne 0$ ke liye valid.

3. $x=1$ par $f_x$ ka limiting behaviour. Figure mein blue curve $f_x=\frac{1}{y}$ hai: jab $y\to 0^{+}$ (dashed yellow line $y=0$ ke right se approach) blue curve upar $+\infty$ ki taraf shoot karta hai; jab $y\to 0^{-}$ (left se) woh $-\infty$ ki taraf plunge karta hai. Isliye $f_x$ ka sign flip hota hai depending on the side. Yeh step kyun? $\frac{1}{y}$ ka sign $y$ jaisa hi hai, aur uski magnitude $\left|\frac{1}{y}\right|$ bound ke bahar badh jaati hai jab $|y|\to 0$. $y$ ka sign isliye decide karta hai kaun si infinity — yahi wajah hai ki blue curve right par axis ke upar rehti hai aur left par neeche.

4. $x=1$ par $f_y$ ka limiting behaviour. Figure mein pink curve $f_y=-\frac{1}{y^2}$ hai: jab $y\to 0$ kisi bhi side se, pink curve $-\infty$ ki taraf dip karti hai — dono branches neeche point karti hain. Yeh step kyun? $y^2>0$ har $y\ne 0$ ke liye, isliye $\frac{1}{y^2}$ hamesha positive hai aur $|y|\to 0$ par blow up karta hai; leading minus sign phir poori cheez ko $-\infty$ ki taraf force karta hai approach direction ki parwah kiye bagair. $f_x$ ke unlike, yahan koi sign-flip nahi kyunki squaring ne $y$ ka sign mita diya.

Verify: $(1,0.1)$ par: $f_x=10$, $f_y=-\frac{1}{0.01}=-100$. $(1,-0.1)$ par: $f_x=-10$ (sign flip ✓), $f_y=-100$ (phir bhi negative ✓, "dono sides $-\infty$ ki taraf jaati hain" match karta hai).

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Example 7 — Cell G: units ke saath word problem

Steps.

1. $T_x=-4x$, $T_y=-2y$. Yeh step kyun? Constant $100$ vanish hota hai; $-2x^2\to -4x$; $-y^2\to -2y$; har doosra variable frozen.

2. $(3,4)$ par: $T_x(3,4)=-12$, $T_y(3,4)=-8$.

3. Units. $T$ °C mein hai, $x$ metres mein, isliye $T_x$ $\frac{\text{°C}}{\text{m}}$ mein hai. Isliye:

   • $-12\text{°C/m}$: $+x$ move karne par, temperature $12$°C per metre drop karta hai.
   • $-8\text{°C/m}$: $+y$ move karne par, $8$°C per metre drop karta hai. Yeh step kyun? Ek word problem tab poora hota hai jab number ke saath sahi unit ho aur plain-English meaning ho.

4. Actual question ka jawab: $x$-direction zyada tezi se thandi karti hai ($12>8$).

Verify: $T_x(3,4)=-4(3)=-12$, $T_y(3,4)=-2(4)=-8$. Dono negative (cooling) ✓, $x$ mein magnitude bada ✓. Tiny step se cross-check: $T(3,4)=100-18-16=66$; $T(3.01,4)=100-2(9.0601)-16=100-18.1202-16=65.8798$, change $-0.1202$ over $0.01$ m $\approx -12.02$ °C/m ✓.

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Example 8 — Cell H: exam twist, mixed AUR pure higher-order + Clairaut

Steps.

1. Pehle $f_x$ ($y$ freeze karo): $f_x=3x^2{y}^2+\cos (xy)\cdot y=3x^2{y}^2+y\cos (xy)$. Kyun? $x^3$ par power rule; $\sin (xy)$ par chain rule inner $x$-derivative $y$ ke saath.

2. Pure $f_{xx}={\partial }_x(f_x)$: $3x^2{y}^2+y\cos (xy)$ ko phir $x$ mein differentiate karo.

   • $3x^2{y}^2\to 6x{y}^2$.
   • $y\cos (xy)$: yahan $y$ constant coefficient hai, $\cos (xy)$ par chain $-\sin (xy)\cdot y$ deta hai, isliye $y\cdot (-y\sin (xy))=-y^2\sin (xy)$. $f_{xx}=6x{y}^2-y^2\sin (xy).$ Yeh step kyun? Cell H pure higher-order bhi maangta hai — $f_x$ ko usi variable $x$ mein ek aur baar differentiate karna.

3. $f_{xy}={\partial }_y(f_x)$: $3x^2{y}^2+y\cos (xy)$ ko $y$ mein differentiate karo.

   • $3x^2{y}^2\to 6x^{2}y$.
   • $y\cos (xy)$: $y$ mein product rule — $(y{)}^{\prime }\cos (xy)+y(\cos (xy){)}^{\prime }=\cos (xy)+y(-\sin (xy)\cdot x)$. $f_{xy}=6x^{2}y+\cos (xy)-xy\sin (xy).$

4. Ab doosra route — pehle $f_y$ ($x$ freeze karo): $f_y=2x^{3}y+\cos (xy)\cdot x=2x^{3}y+x\cos (xy)$.

5. Pure $f_{yy}={\partial }_y(f_y)$: $2x^{3}y+x\cos (xy)$ ko $y$ mein phir differentiate karo.

   • $2x^{3}y\to 2x^3$.
   • $x\cos (xy)$: $x$ constant coefficient, $\cos (xy)$ par chain $-\sin (xy)\cdot x$ deta hai, isliye $x\cdot (-x\sin (xy))=-x^2\sin (xy)$. $f_{yy}=2x^3-x^2\sin (xy).$ Yeh step kyun? $y$-side ka pure second partial, heading ke "higher-order" promise ko complete karta hai.

6. $f_{yx}={\partial }_x(f_y)$:

   • $2x^{3}y\to 6x^{2}y$.
   • $x\cos (xy)$: $x$ mein product rule — $\cos (xy)+x(-\sin (xy)\cdot y)=\cos (xy)-xy\sin (xy)$. $f_{yx}=6x^{2}y+\cos (xy)-xy\sin (xy).$ Yeh step kyun? Mixed partial dono taraf se karna exam ka Clairaut check hai — aap symmetry dekhte ho, sirf quote nahi karte.

7. Mixed compare karo: $f_{xy}$ (Step 3) aur $f_{yx}$ (Step 6) identical hain. ✓ Clairaut holds. Pure partials $f_{xx}\ne f_{yy}$ generally — aur unhe match karna bhi nahi chahiye; sirf mixed pair ko match karna hota hai.

Verify (numeric): $(1,\pi )$ par: $\cos (xy)=\cos \pi =-1$, $\sin \pi =0$. Mixed: $f_{xy}=6(1)(\pi )+(-1)-0=6\pi -1$, $f_{yx}=6\pi -1$ (equal ✓, $\approx 17.85$). Pure: $f_{xx}=6{\pi }^2-0=6{\pi }^2\approx 59.22$, $f_{yy}=2-0=2$.

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Active Recall

Recall Har example ne kaun sa cell cover kiya?

Ex1 → A (polynomial+point) · Ex2 → B (product+chain) · Ex3 → C (quotient, sign, domain $y\ne 0$) · Ex4 → D (log/exp signs) · Ex5 → E (zero partial) · Ex6 → F (degenerate limit) · Ex7 → G (units) · Ex8 → H (mixed + pure higher-order, Clairaut).

Recall Ek direction mein zero slope — flat surface?

Nahi. Zero partial ka matlab sirf usi ek axis ke saath flat hai; surface doosri direction mein steeply chadh sakti hai (Ex 5: $f_x(0,3)=0$ lekin $f_y(0,3)=6$).

Recall Partial kab "blow up" kar sakta hai, aur kya approach ki direction matter karti hai?

Jab point par formula defined nahi ho — zero denominator (Ex 6: $x/y$ mein $y\to 0$). Kya approach ka sign matter karta hai yeh offending variable ki power par depend karta hai: $\frac{1}{y}$ sign flip karta hai ($\pm \infty$), lekin $-\frac{1}{y^2}$ nahi karta ($-\infty$ dono sides se) kyunki squaring ne sign mita diya.

Recall Kya saare second partials equal hone chahiye?

Nahi — sirf mixed pair $f_{xy}=f_{yx}$ (jab continuous ho, Clairaut se). Pure wale $f_{xx}$ aur $f_{yy}$ alag-alag cheezein measure karte hain aur generally differ karte hain (Ex 8: $f_{xx}=6x{y}^2-y^2\sin (xy)$ vs $f_{yy}=2x^3-x^2\sin (xy)$).

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Connections

• Parent (Hinglish): Partial derivatives — woh rule jise yeh examples exercise karte hain.
• Gradient vector — Ex 5 & 7 ke partials ko $\nabla f$ mein bundle karta hai.
• Directional derivative — $f_x,f_y$ combine karo kisi bhi direction mein move karne ke liye.
• Tangent plane and linear approximation — Ex 5 ke do slopes use span karte hain.
• Chain rule (multivariable) — Ex 2, 4, 8 ke andar ka engine.
• Total differential — $df=f_{x}dx+f_{y}dy$, Ex 7 ki units mein use hota hai.
• Single-variable derivative — har "freeze" step isi par reduce hoti hai.
• Clairaut's theorem — Ex 8 mein haath se prove kiya gaya.

Concept Map