Implicit differentiation — technique, applications: Worked examples

4.1.22 · D3 · Maths › Calculus I — Limits & Derivatives › Implicit differentiation — technique, applications

Hum sirf wahi use karte hain jo parent ne banaya tha: Chain Rule (toll), Product Rule (mixed $x y$ ke liye), aur recipe differentiate → toll every $y$ → collect → factor → divide.

Definition Poori equation ko

$F (x, y)$ naam dena Is page mein, jab koi implicit equation "left side $=$ right side" ke roop mein likhi ho, toh hum quietly sab kuch ek taraf le jaate hain aur us combined expression ko $F (x, y)$ kehte hain. Circle ke liye, $x^{2} + y^{2} = 25$ ban jaata hai $F (x, y) = x^{2} + y^{2} - 25$ . Phir:

$F_{x}$ matlab hai " $F$ ko differentiate karo, sirf $x$ ko variable maano, $y$ frozen hai" — yeh $x$ mein partial derivative hai.
$F_{y}$ matlab hai wahi cheez, lekin sirf $y$ move kare, $x$ frozen.

Ye do numbers exactly wahi hain jo parent ki shortcut $\frac{d y}{d x} = - \frac{F _{x}}{F _{y}}$ use karti hai. Hum inhe neeche sirf special points diagnose karne ke liye use karte hain: jahan $F_{y} = 0$ wahan denominator marti hai (vertical tangent), aur jahan dono $F_{x} = F_{y} = 0$ ho wahan $0/0$ milta hai (ek crossing). Yahi condition hai jo Implicit Function Theorem flag karta hai.

Scenario matrix

Har implicit-differentiation problem in boxes mein se ek mein fit hoti hai. Poori page ek tour hai jo har row ko kam se kam ek baar cover karta hai.

#	Case class	Kya tricky hai	Covered by
A	Normal point par clean slope	kuch nahi — warm-up	Ex 1
B	Sign / quadrant sensitivity	same $x$ , alag $y$ → opposite slopes	Ex 2
C	Denominator $= 0$ (vertical tangent)	$y^{'}$ blow up karta hai; geometry hai, error nahi	Ex 3
D	Self-crossing / degenerate point	ek point par do slopes	Ex 4
E	Product aur chain saath mein	$x y$ ek transcendental ke andar	Ex 5
F	Second derivative $y^{''}$	$y^{'}$ ko wapas substitute karna padega	Ex 6
G	Limiting / horizontal tangent ( $y^{'} = 0$ )	numerator $= 0$	Ex 7
H	Real-world related rate (time $t$ )	dono vars $t$ par depend karte hain	Ex 8
I	Exam twist — parameter ke liye solve karo	question ulta hai	Ex 9

Ex 1 — Case A: ek clean, ordinary slope

Worked example Ellipse

$\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ at $(a, b)$ Statement. Is ellipse par $\frac{d y}{d x}$ nikalo, phir uski value $(\frac{3}{2}, 2)$ par nikalo.

Forecast: Yeh point first quadrant mein hai (dono coordinates positive) aur ellipse ke downward-sloping part par hai — toh guess karo negative slope.

Dono sides ko $x$ ke w.r.t. differentiate karo: $\frac{2 x}{9} + \frac{2 y}{4} \frac{d y}{d x} = 0.$ Yeh step kyun? Right side ka constant $1$ differentiate hokar $0$ ban jaata hai; $y^{2}$ term mein $y$ chhupi hai, toh chain rule $\frac{d y}{d x}$ attach karta hai.
Collect / divide: $\frac{d y}{d x} = - \frac{2 x /9}{2 y /4} = - \frac{4 x}{9 y} .$ Yeh step kyun? Sirf ek $y^{'}$ term hai, toh "collect–factor–divide" ek hi division hai.
$x = \frac{3}{2}, y = 2$ plug karo: $\frac{d y}{d x} = - \frac{4 \cdot \frac{3}{2}}{9 \cdot 2} = - \frac{12/ 2}{9 2} = - \frac{12}{9 \cdot 2} = - \frac{2}{3} .$

Verify: slope negative hai ✓ (forecast se match karta hai). Sanity check karo ki point curve par hai: $\frac{( 3/ 2 ) ^{2}}{9} + \frac{( 2 ) ^{2}}{4} = \frac{9/2}{9} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$ ✓.

Ex 2 — Case B: same $x$ , $y$ ke opposite signs

Worked example Circle

$x^{2} + y^{2} = 25$ at the two points with $x = 3$ Statement. Vertical line $x = 3$ circle ko $y = + 4$ aur $y = - 4$ par chhedti hai. Dono par slope nikalo.

Forecast: Symmetry se dono slopes equal aur opposite hone chahiye.

Parent note se, $\frac{d y}{d x} = - \frac{x}{y}$ . Yeh step kyun? $2 x + 2 y y^{'} = 0 \Rightarrow y^{'} = - x / y$ ; hum re-derive karne ki bajay reuse karte hain.
Upper point $(3, 4)$ : $y^{'} = - \frac{3}{4}$ .
Lower point $(3, - 4)$ : $y^{'} = - \frac{3}{- 4} = + \frac{3}{4}$ . Yeh step kyun? $y$ ka sign poora ratio flip kar deta hai — yahi reason hai implicit differentiation dono branches ko ek saath cover karti hai: ek formula, do answers.

Neeche figure dekho. Blue curve circle hai; dashed white line $x = 3$ hai. Yeh circle se yellow dot $(3, 4)$ (top branch) aur red dot $(3, - 4)$ (bottom branch) par milti hai. Chhoti yellow line ka slope $- \frac{3}{4}$ hai (right side girta hai), red line ka slope $+ \frac{3}{4}$ (rise karta hai) — $x$ -axis ke across mirror images.

Figure — Implicit differentiation — technique, applications

Verify: $(3, 4)$ par curve down-right slope karti hai (circle ka top, right edge ki taraf ja raha hai) → negative ✓; $(3, - 4)$ par up-right slope → positive ✓, aur magnitudes match karte hain $∣ - 3/4 ∣ = ∣3/4∣$ ✓.

Ex 3 — Case C: denominator $= 0$ , ek vertical tangent

Worked example Circle

$x^{2} + y^{2} = 25$ at $(5, 0)$ and $(- 5, 0)$ Statement. Woh slope nikalo jahan circle $x$ -axis se milti hai.

Forecast: Far-right point $(5, 0)$ ka picture socho: circle wahan bilkul seedhi khadi hai. Ek vertical line ka koi finite slope nahi hota. Toh expect karo ki formula hume warn karega.

$\frac{d y}{d x} = - \frac{x}{y}$ use karo. $(5, 0)$ par: denominator $y = 0$ . Yeh step kyun? Sochne se pehle substitution kar do toh $- 5/0$ milega — undefined.
Interpret karo, panic mat karo: $\frac{d y}{d x} \to \pm \infty$ matlab tangent line vertical hai ( $x = 5$ ). Yeh step kyun? Blown-up slope geometry hai: Tangent and Normal Lines hume batata hai tangent $x = 5$ hai aur, usefully, normal yahaan horizontal hai ( $y = 0$ , slope $0$ ).
$(- 5, 0)$ par bhi: vertical tangent $x = - 5$ .

Neeche figure dekho. Red solid line $x = 5$ circle ke right edge ko hug karti hai — yeh yellow dot $(5, 0)$ par vertical tangent hai; dashed red line uska mirror hai $x = - 5$ par. Notice karo tangent touch karti hai lekin cross nahi karti, exactly jaisa ek tangent ko karna chahiye.

Verify: hamare definition se diagnostic use karte hue, $F (x, y) = x^{2} + y^{2} - 25$ deta hai $F_{y} = 2 y$ , jo exactly $y = 0$ par $0$ hai. Yahi woh jagah hai jahan $- F_{x} / F_{y}$ ka denominator marta hai, toh tangent vertical hai — aur precisely yahi woh spot hai jise Implicit Function Theorem kehta hai ki $y (x)$ ek nice single-valued function fail karti hai. Picture se match karta hai. ✓

Ex 4 — Case D: ek self-crossing point (ek point par do slopes)

Worked example Lemniscate

$(x^{2} + y^{2})^{2} = 2 (x^{2} - y^{2})$ — origin par crossing aur loop tips Statement. Yeh figure-eight origin se guzarta hai, jahan woh khud ko cross karta hai. Wahan slopes kya hain? Saath hi $x$ -axis par loop tips par tangent behaviour bhi nikalo.

Forecast: Self-crossing par curve ki do tangent directions hoti hain. $F (x, y) = (x^{2} + y^{2})^{2} - 2 (x^{2} - y^{2})$ se, dono $F_{x}$ aur $F_{y}$ origin par vanish karenge — recipe $0/0$ dega, yeh special point ka signal hai.

$(x^{2} + y^{2})^{2} = 2 (x^{2} - y^{2})$ differentiate karo: $2 (x^{2} + y^{2}) (2 x + 2 y y^{'}) = 2 (2 x - 2 y y^{'}) .$ Yeh step kyun? Left side ek outer square hai (chain rule) jo inner derivative $2 x + 2 y y^{'}$ se multiply hota hai; har $y$ apna toll deta hai.
Dono sides distribute karo taaki har $y^{'}$ term visible ho: $4 x (x^{2} + y^{2}) + 4 y (x^{2} + y^{2}) y^{'} = 4 x - 4 y y^{'} .$ Yeh step kyun? Pehle $y^{'}$ terms collect karne se pehle brackets expand karne zaroori hain taaki woh ek product ke andar chhupi na rahein — warna galat cheezein factor ho jaengi.
Har $y^{'}$ term left par collect karo, baaki sab right par, phir factor karo: $4 y (x^{2} + y^{2}) y^{'} + 4 y y^{'} = 4 x - 4 x (x^{2} + y^{2})$ $y^{'} [4 y (x^{2} + y^{2}) + 4 y] = 4 x [1 - (x^{2} + y^{2})] .$ Yeh step kyun? Yeh recipe ka Collect–Factor half hai: toll terms ikatha karo, $y^{'}$ ko common factor ki tarah bahar nikalo.
Divide karo ( $4$ 's cancel ho jaate hain): $\frac{d y}{d x} = \frac{x [ 1 - ( x ^{2} + y ^{2} ) ]}{y [ ( x ^{2} + y ^{2} ) + 1 ]} .$ Yeh step kyun? Divide — recipe ka aakhri step — $y^{'}$ ko akela chhod deta hai.
Origin $(0, 0)$ par: numerator $= 0$ , denominator $= 0$ → $\frac{0}{0}$ . Yeh step kyun? $0/0$ ek crossing ya cusp ka algebraic fingerprint hai — ek formula ek slope pick nahi kar sakta kyunki do distinct tangent directions hain.
Loop tips. $x$ -axis par $y = 0$ set karo original equation mein: $(x^{2})^{2} = 2 x^{2} \Rightarrow x^{4} = 2 x^{2} \Rightarrow x^{2} = 2 \Rightarrow x = \pm 2$ . Toh tips hain $(\pm 2, 0)$ . $(2, 0)$ ko $y^{'}$ mein plug karo: numerator $= 2 [1 - 2] = - 2 \neq = 0$ , lekin denominator $= 0 \cdot [\dots] = 0$ → $\frac{d y}{d x} \to \pm \infty$ , har tip par vertical tangent. Yeh step kyun? Yeh do special kinds ko alag karta hai: tips par sirf $F_{y} = 0$ hai (vertical tangent, ek clean $number /0$ ), jabki origin par dono vanish karte hain ( $0/0$ , crossing).

Neeche figure dekho. Blue figure-eight red dot (origin) par khud ko cross karta hai — tum do curve branches ko alag directions mein guzarte dekh sakte ho, isliye wahan slope $0/0$ hai. Do yellow dots $(\pm 2, 0)$ par loop tips hain, jahan curve seedha upar/neeche mudta hai: vertical tangents.

Verify: origin curve par hai: $(0)^{2} = 2 (0)$ ✓. Dono $F_{x} = 4 x (x^{2} + y^{2}) - 4 x$ aur $F_{y} = 4 y (x^{2} + y^{2}) + 4 y$ $(0, 0)$ par vanish karte hain ✓ — $0/0$ genuine hai, sloppy algebra nahi. Tips $x = 2$ par $x^{4} = 2 x^{2}$ satisfy karte hain ✓, wahan denominator $y [(x^{2} + y^{2}) + 1] = 0$ → vertical ✓.

Ex 5 — Case E: product aur chain ek saath nested

Worked example

$e^{x y} = x + y$ at $(0, 1)$ Statement. $(0, 1)$ point par $\frac{d y}{d x}$ nikalo.

Forecast: $e^{x y}$ mein ek product $x y$ ek exponential ke andar hai — expect karo Chain Rule toll $e$ par aur Product Rule $x y$ par, stacked.

Differentiate karo: $e^{x y} \cdot \frac{d}{d x} (x y) = 1 + \frac{d y}{d x} .$ Yeh step kyun? $\frac{d}{d x} e^{u} = e^{u} u^{'}$ (chain rule); exponent $u = x y$ ek product hai.
Inner product $\frac{d}{d x} (x y) = y + x y^{'}$ expand karo: $e^{x y} (y + x y^{'}) = 1 + y^{'} .$ Yeh step kyun? $x y$ par Product Rule: $x$ ka derivative (jo $1$ hai) times $y$ , plus $x$ times toll $y^{'}$ .
$e^{x y}$ ko bracket mein distribute karo taaki har $y^{'}$ term akela khade: $y e^{x y} + x e^{x y} y^{'} = 1 + y^{'} .$ Yeh step kyun? Tum $y^{'}$ terms collect nahi kar sakte jab tak ek $e^{x y}$ se multiply bracket ke andar trapped hai — pehle expand karo.
$y^{'}$ ek side par collect karo, factor karo, divide karo: $x e^{x y} y^{'} - y^{'} = 1 - y e^{x y} \Rightarrow y^{'} (x e^{x y} - 1) = 1 - y e^{x y}$ $\frac{d y}{d x} = \frac{1 - y e ^{x y}}{x e ^{x y} - 1} .$ Yeh step kyun? Collect–Factor–Divide finish: do $y^{'}$ terms ikatha karo, $y^{'}$ bahar nikalo, divide karo.
$(0, 1)$ par: $x y = 0$ toh $e^{x y} = 1$ . Phir $\frac{d y}{d x} = \frac{1 - 1 \cdot 1}{0 \cdot 1 - 1} = \frac{0}{- 1} = 0.$

Verify: check karo ki point curve par hai: $e^{0} = 1$ aur $x + y = 0 + 1 = 1$ ✓. Slope $0$ matlab horizontal tangent $(0, 1)$ par — ek legitimate answer (yeh Case G ka preview hai).

Ex 6 — Case F: second derivative $y^{''}$

Worked example Folium

$x^{3} + y^{3} = 6 x y$ — $(3, 3)$ par $y^{''}$ nikalo Statement. Parent ne nikala tha $y^{'} = \frac{2 y - x ^{2}}{y ^{2} - 2 x}$ aur $y^{'} (3, 3) = - 1$ . Ab wahan $y^{''}$ nikalo.

Forecast: Hum $y^{'}$ ko dobara differentiate karenge; khatra yeh hai ki answer mein stray $y^{'}$ reh jaaye. Kyunki $y^{'} (3, 3) = - 1$ hai, expect karo ek clean number.

Pre-solved line $3 x^{2} + 3 y^{2} y^{'} = 6 y + 6 x y^{'}$ se shuru karo. Dobara $x$ ke w.r.t. differentiate karo: $6 x + 3 (2 y y^{'} \cdot y^{'} + y^{2} y^{''}) = 6 y^{'} + 6 (y^{'} + x y^{''}) .$ Yeh step kyun? $\frac{d}{d x} (y^{2} y^{'})$ ek product hai: $2 y (y^{'})^{2} + y^{2} y^{''}$ . Har $y$ ab bhi toll deta hai.
Constants simplify karo aur like terms gather karo: $6 x + 6 y (y^{'})^{2} + 3 y^{2} y^{''} = 12 y^{'} + 6 x y^{''} .$ Yeh step kyun? Humne sirf $3 (\dots)$ aur $6 (\dots)$ brackets multiply out kiye — koi nayi calculus nahi, bas arithmetic tidy-up taaki $y^{''}$ terms visible aur countable ho jayein.
Do $y^{''}$ terms left par collect karo, baaki sab right par, phir $y^{''}$ factor karo: $y^{''} (3 y^{2} - 6 x) = 12 y^{'} - 6 x - 6 y (y^{'})^{2}$ $y^{''} = \frac{12 y ^{'} - 6 x - 6 y ( y ^{'} ) ^{2}}{3 y ^{2} - 6 x} .$ Yeh step kyun? Wahi Collect–Factor–Divide finish, lekin ab jis "unknown" ke liye solve kar rahe hain woh $y^{''}$ hai; $y^{'}$ terms sirf known coefficients hain.
Substitute $x = 3, y = 3, y^{'} = - 1$ : $y^{''} = \frac{12 ( - 1 ) - 6 ( 3 ) - 6 ( 3 ) ( 1 )}{3 ( 9 ) - 6 ( 3 )} = \frac{- 12 - 18 - 18}{27 - 18} = \frac{- 48}{9} = - \frac{16}{3} .$ Yeh step kyun? Yeh classic trap hai — tumhe $y^{'}$ ki value plug karni padegi, symbol nahi chhod sakte.

Verify: $y^{''} = - \frac{16}{3} < 0$ → curve $(3, 3)$ par concave down hai, jo slope $- 1$ ke tangent ke saath consistent hai ek local high part of the loop par. Concrete arithmetic check: numerator $= 12 (- 1) - 6 (3) - 6 (3) (- 1)^{2} = - 12 - 18 - 18 = - 48$ ; denominator $= 3 (9) - 6 (3) = 27 - 18 = 9$ ; toh $y^{''} = - 48/9 = - 16/3 \approx - 5.33$ ✓ (machine se bhi check kiya). ✓

Ex 7 — Case G: horizontal tangent ( $y^{'} = 0$ , numerator $= 0$ )

Worked example Folium ka tangent horizontal kahan hai?

$x^{3} + y^{3} = 6 x y$ se point solve karo. Statement. Woh point nikalo first quadrant mein jahan $\frac{d y}{d x} = 0$ ho.

Forecast: $y^{'} = 0$ ke liye numerator $2 y - x^{2} = 0$ chahiye, yaani $y = \frac{x ^{2}}{2}$ . Use curve mein wapas feed karo.

Numerator zero set karo: $y = \frac{x ^{2}}{2}$ . Yeh step kyun? Ek fraction tab hi $0$ hota hai jab uska top $0$ ho (aur bottom nahi).
$x^{3} + y^{3} = 6 x y$ mein substitute karo: $x^{3} + (\frac{x ^{2}}{2})^{3} = 6 x \cdot \frac{x ^{2}}{2} \Rightarrow x^{3} + \frac{x ^{6}}{8} = 3 x^{3} .$ Yeh step kyun? $y$ ko $\frac{x ^{2}}{2}$ se replace karna do-variable curve ko sirf $x$ mein ek equation bana deta hai, jise hum actually solve kar sakte hain.
Sab ek side move karo aur simplify karo. Dono sides se $x^{3}$ subtract karo: $\frac{x ^{6}}{8} = 2 x^{3} .$ Yeh step kyun? $3 x^{3} - x^{3} = 2 x^{3}$ ; $x^{3}$ terms group karna higher power ko next step ke liye isolate karta hai.
Kyunki hum first-quadrant point chahte hain, $x \neq = 0$ , toh dono sides $x^{3}$ se divide karo: $\frac{x ^{3}}{8} = 2 \Rightarrow x^{3} = 16 \Rightarrow x = 1 6^{1/3} = 2^{4/3} .$ Yeh step kyun? $x^{3}$ se divide karna legal hai kyunki $x \neq = 0$ (origin woh point nahi hai jo hum chahte); yeh polynomial ko degree 6 se ek clean cubic tak le aata hai.
$y$ ke liye back-substitute karo: $y = \frac{x ^{2}}{2} = \frac{( 2 ^{4/3} ) ^{2}}{2} = \frac{2 ^{8/3}}{2 ^{1}} = 2^{8/3 - 1} = 2^{5/3} .$ Yeh step kyun? Horizontal-tangent point ko dono coordinates chahiye; $y$ seedha step 1 mein impose ki gayi condition $y = \frac{x ^{2}}{2}$ se aata hai.

Toh tangent horizontal hai $(2^{4/3}, 2^{5/3}) \approx (2.52, 3.17)$ par.

Verify: denominator $y^{2} - 2 x = 2^{10/3} - 2 \cdot 2^{4/3} = 2^{10/3} - 2^{7/3} \neq = 0$ ✓ (genuine horizontal tangent, $0/0$ nahi). Numerator $2 y - x^{2} = 2 \cdot 2^{5/3} - 2^{8/3} = 2^{8/3} - 2^{8/3} = 0$ ✓. On-curve check: $x^{3} + y^{3} = 16 + 2^{5} = 16 + 32 = 48$ aur $6 x y = 6 \cdot 2^{4/3} \cdot 2^{5/3} = 6 \cdot 2^{3} = 48$ ✓.

Worked example Sliding ladder

Statement. Ek $10 m$ ki ladder wall par tikki hai. Uska foot $\frac{d x}{d t} = 1 m/s$ se bahar slide kar raha hai. Jab foot wall se $x = 6 m$ door ho, toh top kitni tezi se neeche slide kar raha hai?

Forecast: Jaise base bahar jaata hai, top girta hai — expect karo $\frac{d y}{d t} < 0$ . Wall ke paas top muskil se hilta hai; door jaane par tezi se girta hai. $x = 6$ par (toh $y = 8$ , ek $6$ - $8$ - $10$ triangle) ek moderate value expect karo.

Constraint: $x^{2} + y^{2} = 1 0^{2} = 100$ (Pythagoras — ladder ki length fixed hai). Yeh step kyun? Yeh hamari implicit equation hai; $x$ aur $y$ dono $t$ ke functions hain, toh hum $t$ ke w.r.t. differentiate karte hain, yeh ek Related Rates move hai.
$t$ ke w.r.t. differentiate karo: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 \Rightarrow \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} .$ Yeh step kyun? Wahi toll jaise pehle, lekin "outer variable" $t$ hai, toh har term ko $\frac{d}{d t}$ milta hai.
$x = 6$ par: $y = 100 - 36 = 8$ . Toh $\frac{d y}{d t} = - \frac{6}{8} (1) = - \frac{3}{4} = - 0.75 m/s .$

Neeche figure dekho. Blue segment ladder hai, $10 m$ par fixed. Green dot foot hai jo right ja raha hai (green arrow, $+ 1 m/s$ ); red dot top hai jo neeche ja raha hai (red arrow, $- 0.75 m/s$ ). Right triangle ki legs hain $x = 6$ aur $y = 8$ .

Verify: sign negative hai (top girta hai) ✓. Units: $\frac{m}{m} \cdot \frac{m}{s} = \frac{m}{s}$ ✓. Limiting check: jaise $x \to 10, y \to 0$ , $\frac{d y}{d t} = - \frac{x}{y} \to - \infty$ — jis instant ladder flat hoti hai, top infinitely fast girta hai, jo physical intuition se match karta hai. ✓

Ex 9 — Case I: exam twist (ek hidden parameter ke liye solve karo)

Worked example Ek constant ko reverse-engineer karo

Statement. Kaha jaata hai ki curve $x^{2} + k x y + y^{2} = c$ ka horizontal tangent point $(1, 2)$ par hai. Constant $k$ nikalo (aur check karo ki stated curve consistent bhi hai ya nahi).

Forecast: " $(1, 2)$ par horizontal tangent" ka matlab hai wahan $y^{'} = 0$ ; yeh $k$ ke liye ek equation hai. Point curve par bhi hona chahiye; yeh doosri equation $c$ pin down karti hai. Ek aise trap ki awaaz raho jahan yeh do facts agree na karein.

$x^{2} + k x y + y^{2} = c$ implicitly differentiate karo: $2 x + k (y + x y^{'}) + 2 y y^{'} = 0.$ Yeh step kyun? $k x y$ ek product hai $x$ aur $y$ ka constant $k$ se multiply — product rule; $y^{2}$ apna toll deta hai; $c$ constant hai toh uska derivative $0$ hai.
$(1, 2)$ par $y^{'} = 0$ impose karo: $2 (1) + k (2 + (1) (0)) + 2 (2) (0) = 0 \Rightarrow 2 + 2 k = 0 \Rightarrow k = - 1.$ Yeh step kyun? $y^{'} = 0$ set karne se har $y^{'}$ term mar jaata hai, ek single linear equation in $k$ bachta hai.
Ab consistent $c$ nikalo. $k = - 1$ aur point $(1, 2)$ curve mein daalo: $1^{2} + (- 1) (1) (2) + 2^{2} = 1 - 2 + 4 = 3.$ Toh point curve par tab hi lie karta hai jab $c = 3$ ho. Yeh step kyun? Ek parameter problem mein do unknowns chhuppe hote hain ( $k$ aur constant); slope condition ek deti hai, on-curve condition doosra.

Conclusion / resolution. $k = - 1$ aur $c = 3$ ke saath, curve $x^{2} - x y + y^{2} = 3$ , $(1, 2)$ se guzarti hai aur wahan horizontal tangent hai — sab kuch consistent hai, toh answer hai $k = - 1$ . Exam trap: agar problem ne right-hand side ko, maan lo $c = 7$ par fix kar diya hota, toh koi bhi $k$ dono conditions ek saath satisfy nahi kar sakta tha (point simply $x^{2} + k x y + y^{2} = 7$ par nahi hoga us $k$ ke liye jo tangent flatten karta hai), aur sahi exam answer hota "aisa koi $k$ exist nahi karta — data inconsistent hain." $k$ report karne se pehle hamesha on-curve condition test karo.

Verify: $k = - 1$ ke saath, $(1, 2)$ par curve value $3$ hai ✓. Wahan slope: $2 + (- 1) (2) + [(- 1) (1) + 2 (2)] y^{'} = 0 \Rightarrow 0 + 3 y^{'} = 0 \Rightarrow y^{'} = 0$ ✓ — genuinely horizontal.

Recall Matrix ki one-line summary

Case ::: Signal / fix A clean point ::: bas divide karo B do $y$ 's ek $x$ par ::: point ko dono coords se naam do; $y$ ka sign slope flip karta hai C denominator $= 0$ (sirf $F_{y} = 0$ ) ::: vertical tangent, error nahi D numerator aur denominator $= 0$ ( $0/0$ , dono $F_{x} = F_{y} = 0$ ) ::: self-crossing/cusp — do slopes F second derivative ::: finish karne se pehle $y^{'}$ wapas substitute karo G numerator $= 0$ ::: horizontal tangent, top $= 0$ solve karo H related rate ::: $t$ ke w.r.t. differentiate karo; units aur limits check karo

Connections

Chain Rule — har example mein pay kiya gaya toll.
Product Rule — Ex 5, Ex 9 ( $x y$ terms).
Related Rates — Ex 8 ka sliding ladder.
Tangent and Normal Lines — Ex 1–3, 7 mein slopes read off kiye.
Implicit Function Theorem — explain karta hai kyun Cases C & D woh jagahein hain jahan $F_{y} = 0$ (aur jahan dono $F_{x} = F_{y} = 0$ ).
Derivatives of Inverse Functions · Logarithmic Differentiation — usi machine ki sibling applications.