Differentiability in multiple variables: Worked examples

4.4.6 · D3 · Maths › Multivariable Calculus › Differentiability in multiple variables

Shuru karne se pehle, ek reminder seedhe shabdon mein. Ek function $f (x, y)$ do numbers leta hai aur ek return karta hai — isko landscape ki height samjho point $(x, y)$ ke upar. "Kisi point par differentiable hona" matlab: agar tum kaafi zoom in karo, toh landscape ek flat tilted board ki tarah lagti hai (tangent plane). Symbol $\nabla f = (f_{x}, f_{y})$ (padho "grad f") bas slopes ki pair hai: height kitni tezi se badhti hai East chalte waqt ( $f_{x}$ ) aur North chalte waqt ( $f_{y}$ ). Partial Derivatives aur Gradient and Directional Derivatives dekho agar koi bhi shaky lage.

Scenario matrix

Is topic ka har problem inhi cells mein se ek mein aata hai. Right column us example ka naam hai jo ise clear karta hai.

#	Case class	Kya mushkil banaata hai	Cleared by
A	Nice polynomial, partials ke saare signs	kuch nahi — $C^{1}$ shortcut	Ex 1
B	Partials exist, lekin discontinuous	diagonal origin ko pakad leta hai	Ex 2
C	Continuous, partials exist, phir bhi NOT differentiable	sabse sneaky failure	Ex 3
D	Degenerate / boundary input (us curve par jahan formula mar jaata hai)	limit jo survive karti hai, ek jo nahi karti	Ex 4
E	Directional derivatives sab exist karte hain, phir bhi not differentiable	strictly-stronger trap	Ex 5
F	Linearization estimate (real-world word problem, signed steps)	mixed-sign $h$ , units	Ex 6
G	Limiting behaviour / squeeze proves differentiability	raw $o (∥ h ∥)$ limit prove karna	Ex 7
H	Exam twist: wo parameter value dhundho jo $f$ ko differentiable banaye	definition ko reverse-engineer karo	Ex 8

Do symbols jinpar hum rely karenge:

$∥ h ∥ = h_{1}^{2} + h_{2}^{2}$ — step $h = (h_{1}, h_{2})$ ki length, yaani base point se straight-line distance. Yeh kabhi negative nahi hoti.
$o (∥ h ∥)$ ("little-o of $∥ h ∥$ ") — ek aisi error jo distance se strictly faster shrink karti hai: ise $∥ h ∥$ se divide karo aur phir bhi $0$ jaata hai.

Cell A — aasaan, nice case

Worked example Example 1 (Cell A) — polynomial, dono slopes nonzero

Dikhao ki $f (x, y) = x^{3} - 2 x y^{2} + y$ point $(2, - 1)$ par differentiable hai aur uska tangent plane likho.

Forecast: abhi andaza lagao — kya hume scary limit chahiye, ya koi shortcut hai? Tumhara kya andaza hai $f_{x}$ ka sign (kyunki $x^{3}$ term badh rahi hai)?

Partials compute karo. $f_{x} = 3 x^{2} - 2 y^{2}$ , $f_{y} = - 4 x y + 1$ . Yeh step kyun? Partial $f_{x}$ mein $y$ freeze hota hai aur sirf $x$ mein differentiate karte hain — yeh East-slope hai. Partial Derivatives dekho.
Note karo ki dono partials polynomials hain. Polynomials har jagah continuous hote hain. Yeh step kyun? $C^{1}$ theorem kehta hai: partials exist karein aur point ke paas continuous hon $\Rightarrow$ differentiable. Koi limit nahi chahiye. Yeh $C^1 \Rightarrow$ differentiable hai.
$(2, - 1)$ par evaluate karo. $f_{x} (2, - 1) = 3 (4) - 2 (1) = 10$ , $f_{y} (2, - 1) = - 4 (2) (- 1) + 1 = 9$ . Aur $f (2, - 1) = 8 - 2 (2) (1) + (- 1) = 8 - 4 - 1 = 3$ . Yeh step kyun? Point par gradient tangent plane ki exact tilt deta hai.
Plane likho (Tangent Planes and Linear Approximation): $z = 3 + 10 (x - 2) + 9 (y + 1)$ .

Verify: base point $(2, - 1)$ ko plane mein daalo: $z = 3 + 10 (0) + 9 (0) = 3 = f (2, - 1)$ . ✓ Plane surface point se guzarti hai, jaisa kisi bhi tangent plane ko karna chahiye.

Cell B — partials exist lekin function tuta hua hai

Worked example Example 2 (Cell B) — classic saddle-ratio

$g (x, y) = ⎩ ⎨ ⎧ \frac{x ^{2} y}{x ^{4} + y ^{2}} 0 (x, y) \neq = (0, 0) (0, 0)$ Dikhao ki dono partials origin par exist karte hain, phir bhi $g$ wahan not continuous hai — isliye not differentiable.

Forecast: partials sirf axes ke saath chalte hain. Tumhara axes ke saath kya expect hai — kya wahan $g$ zero hai? Aur kaunsa sneaky curve trouble reveal kar sakta hai?

Partial $g_{x} (0, 0)$ : $x$ -axis par $y = 0$ hai, toh $g (x, 0) = 0$ sabhi $x$ ke liye. Uska slope $0$ hai: $g_{x} (0, 0) = 0$ . Yeh step kyun? Partial sirf East line ko probe karta hai. Us line par $g$ flat hai, toh bilkul tame lagta hai.
Partial $g_{y} (0, 0)$ : $y$ -axis par $x = 0$ hai, $g (0, y) = 0$ , toh $g_{y} (0, 0) = 0$ .
Parabola $y = x^{2}$ ke saath approach karo: $g (x, x^{2}) = \frac{x ^{2} \cdot x ^{2}}{x ^{4} + x ^{4}} = \frac{x ^{4}}{2 x ^{4}} = \frac{1}{2}$ . Yeh step kyun? Hum ek aisa path dhundh rahe hain jo $0$ nahi jaata. Value $\frac{1}{2}$ par chipki rehti hai chahe $x$ kitna bhi close ho — toh origin par limit exist nahi karti. Continuity in Multiple Variables dekho.
Conclude karo: $g$ continuous nahi hai $\Rightarrow$ differentiable nahi (differentiable $\Rightarrow$ continuous, contrapositive).

Verify: axes ke saath value $= 0$ ; $y = x^{2}$ ke saath value $= \frac{1}{2} \neq = 0$ . Do alag limits discontinuity confirm karti hain. ✓

Cell C — continuous, partials exist, PHIR BHI not differentiable

Worked example Example 3 (Cell C) — sabse sneaky failure

$h (x, y) = ⎩ ⎨ ⎧ \frac{x ^{3}}{x ^{2} + y ^{2}} 0 (x, y) \neq = (0, 0) (0, 0)$ Dikhao ki $h$ point $(0, 0)$ par continuous hai, uske partials exist karte hain, phir bhi $h$ wahan not differentiable hai.

Forecast: yeh continuity test aur partials test dono pass karta hai. Phir kaahan trip ho sakta hai? $o (∥ h ∥)$ limit dekho.

Continuity: $∣ h ∣ = \frac{∣ x ∣ ^{3}}{x ^{2} + y ^{2}} \leq \frac{∣ x ∣ ( x ^{2} + y ^{2} )}{x ^{2} + y ^{2}} = ∣ x ∣ \to 0$ . Yeh step kyun? Kyunki $x^{2} \leq x^{2} + y^{2}$ , hum $∣ x ∣^{3} = ∣ x ∣ \cdot x^{2} \leq ∣ x ∣ (x^{2} + y^{2})$ bound karte hain; squeeze force karta hai $h \to 0 = h (0, 0)$ . Continuous. ✓
Origin par partials: $y = 0$ par, $h (x, 0) = \frac{x ^{3}}{x ^{2}} = x$ , slope $h_{x} (0, 0) = 1$ . $x = 0$ par, $h (0, y) = 0$ , toh $h_{y} (0, 0) = 0$ . Toh $\nabla h (0, 0) = (1, 0)$ . Yeh step kyun? Hume ek hi candidate linear map chahiye definition ke against test karne ke liye.
Differentiability limit test karo. $h = (h_{1}, h_{2})$ ke saath error hai $E = h (h_{1}, h_{2}) - [0 + (1, 0) \cdot (h_{1}, h_{2})] = \frac{h _{1}^{3}}{h _{1}^{2} + h _{2}^{2}} - h_{1} = \frac{h _{1}^{3} - h _{1} ( h _{1}^{2} + h _{2}^{2} )}{h _{1}^{2} + h _{2}^{2}} = \frac{- h _{1} h _{2}^{2}}{h _{1}^{2} + h _{2}^{2}} .$ $∥ h ∥ = h_{1}^{2} + h_{2}^{2}$ se divide karo aur line $h_{2} = h_{1} > 0$ chalao: $\frac{E}{∥ h ∥} = \frac{- h _{1} h _{2}^{2}}{( h _{1}^{2} + h _{2}^{2} ) ^{3/2}} = \frac{- h _{1} \cdot h _{1}^{2}}{( 2 h _{1}^{2} ) ^{3/2}} = \frac{- h _{1}^{3}}{2 2 h _{1}^{3}} = \frac{- 1}{2 2} \neq = 0.$ Yeh step kyun? Differentiability demand karti hai ki yeh ratio har direction se $\to 0$ ho. Diagonal ke saath yeh $- \frac{1}{2 2} \approx - 0.354$ par atak jaata hai. Fail.
Conclude karo: continuous + partials exist, lekin tangent-plane error itni tezi se vanish nahi hoti $\Rightarrow$ not differentiable.

Verify: $- 1/ (22) = - 2 /4 \approx - 0.3536 \neq = 0$ . ✓ Non-zero limit differentiability ko khatam kar deti hai.

Cell D — degenerate / boundary input

Worked example Example 4 (Cell D) — jahan formula formally mar jaata hai

Maano $f (x, y) = \frac{sin ( x y )}{y}$ for $y \neq = 0$ , aur $f (x, 0) = x$ define karo. Kya $f$ boundary point $(1, 0)$ par differentiable hai (jahan raw formula zero se divide karta hai)?

Forecast: $y$ se divide karna $y = 0$ par fatal lagta hai. Lekin $sin (x y) \approx x y$ small argument ke liye — kya zero cancel ho jaata hai?

Degeneracy resolve karo. Small $y$ ke liye, $sin (x y) = x y - \frac{( x y ) ^{3}}{6} + \dots$ , toh $\frac{sin ( x y )}{y} = x - \frac{x ^{3} y ^{2}}{6} + \dots$ . Yeh step kyun? Apparent $0/0$ removable hai — definition $f (x, 0) = x$ limit se match karti hai, toh $f$ actually smooth function $x - \frac{x ^{3} y ^{2}}{6} + \dots$ hai axis ke paas.
$(1, 0)$ par partials. $f \approx x - \frac{x ^{3} y ^{2}}{6}$ se: $f_{x} = 1 - \frac{x ^{2} y ^{2}}{2}$ , toh $f_{x} (1, 0) = 1$ . Aur $f_{y} = - \frac{x ^{3} y}{3}$ , toh $f_{y} (1, 0) = 0$ . Yeh step kyun? Resolved expansion use karna illegal division se bachata hai aur honest slopes deta hai.
$C^{1}$ check. Dono $f_{x}, f_{y}$ point $(1, 0)$ par continuous hain ( $sin, cos$ , polynomials se bane hain, cancellation ke baad koi surviving division by zero nahi). Toh $f$ boundary point par differentiable hai. Yeh step kyun? Jab removable singularity patch ho jaaye, toh $C^{1}$ shortcut apply hota hai. C1 and Smooth Functions dekho.
Tangent plane: $z = f (1, 0) + 1 (x - 1) + 0 (y - 0) = 1 + (x - 1) = x$ .

Verify: $(1, 0)$ par plane $z = 1 = f (1, 0)$ deta hai. ✓ Aur $f_{y} (1, 0) = 0$ match karta hai: axis ke paas $y$ ke saath $f = x - \frac{x ^{3} y ^{2}}{6}$ hai jo $y$ mein first order par flat hai (sirf $y^{2}$ appear karta hai). ✓

Cell E — har directional derivative exist karta hai, phir bhi not differentiable

Worked example Example 5 (Cell E) — strictly-stronger trap

$p (x, y) = ⎩ ⎨ ⎧ \frac{x ^{2} y}{x ^{2} + y ^{2}} 0 (x, y) \neq = (0, 0) (0, 0)$ Dikhao ki $(0, 0)$ par har direction mein directional derivative exist karta hai, phir bhi $p$ differentiable nahi hai.

Forecast: agar saare directional derivatives exist karte hain, toh kya hum done hain? Dekho ki kya woh linear rule $D_{u} p = \nabla p \cdot u$ follow karte hain.

Unit vector $u = (cos θ, sin θ)$ ke saath directional derivative. $x = t cos θ, y = t sin θ$ set karo: $\frac{p ( t u ) - 0}{t} = \frac{1}{t} \cdot \frac{t ^{2} c o s ^{2} θ \cdot t s i n θ}{t ^{2}} = cos^{2} θ sin θ .$ $t \to 0$ lo: $D_{u} p (0, 0) = cos^{2} θ sin θ$ . Har $θ$ ke liye exist karta hai. Yeh step kyun? Hum saari directions probe karte hain, sirf axes nahi, stronger claim test karne ke liye. Gradient and Directional Derivatives dekho.
Partials padho. $θ = 0$ (East): $cos^{2} 0 sin 0 = 0 = p_{x}$ . $θ = \frac{π}{2}$ (North): $0 \cdot 1 = 0 = p_{y}$ . Toh $\nabla p (0, 0) = (0, 0)$ .
Agar $p$ differentiable hota, toh linear rule force karta $D_{u} p = \nabla p \cdot u = (0, 0) \cdot u = 0$ for all $θ$ . Lekin $θ = \frac{π}{4}$ par: $cos^{2} \frac{π}{4} sin \frac{π}{4} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2 2} \neq = 0$ . Yeh step kyun? Differentiability directional derivatives ke liye ek linear formula predict karti hai. Reality isse contradict karti hai $\Rightarrow$ not differentiable.

Verify: predicted $0$ vs actual $\frac{1}{2 2} = \frac{2}{4} \approx 0.354$ at $45°$ . Mismatch $\Rightarrow$ not differentiable. ✓

Cell F — real-world linearization with signed steps

Worked example Example 6 (Cell F) — physics word problem

Ek rectangular metal plate ke sides $x$ aur $y$ metres hain; uska area $A = x y$ hai. Ek plate jisme $x = 8.00 m$ , $y = 5.00 m$ measure hua hai, cool kiya jaata hai: $x$ shrink hota hai $0.03 m$ se aur $y$ grow karta hai $0.02 m$ se. Linearization se naya area estimate karo aur exact value se compare karo.

Forecast: do opposite-sign changes. Kya area badhega ya ghategaa? Kaunse side ki change zyada matter karti hai?

Base value. $A (8, 5) = 40 m^{2}$ .
Partials. $A_{x} = y = 5$ , $A_{y} = x = 8$ (units: m of area per m of side $=$ m). Yeh step kyun? $A_{x}$ yeh hai ki $x$ -growth ke har metre par kitne m² milte hain — zyada hota hai jab $y$ bada ho.
Signed step. $h = (- 0.03, + 0.02)$ . Linear estimate: $A \approx 40 + 5 (- 0.03) + 8 (+ 0.02) = 40 - 0.15 + 0.16 = 40.01 m^{2} .$ Yeh step kyun? Tangent-plane formula $A (a + h) \approx A (a) + \nabla A \cdot h$ ; note karo shrinking side ke liye negative term.
Interpret karo: area thoda sa badhta hai — $y$ -growth ( $+ 0.16$ ) barely $x$ -shrink ( $- 0.15$ ) ko beat karta hai.

Verify: exact $A = (8 - 0.03) (5 + 0.02) = 7.97 \times 5.02 = 40.0094 m^{2}$ . Linear estimate $40.01$ ka difference $0.0006 m^{2}$ hai — dropped $o (∥ h ∥)$ term $h_{1} h_{2} = (- 0.03) (0.02) = - 0.0006$ hai. ✓ Units: throughout m².

Cell G — raw limit se differentiability prove karna

Worked example Example 7 (Cell G) — error ko zero par squeeze karo

Definition se prove karo ( $C^{1}$ shortcut nahi) ki $q (x, y) = x^{2} + y^{2}$ point $(0, 0)$ par differentiable hai.

Forecast: origin par partials dono $0$ hain. Leftover error kya hai, aur kya woh $∥ h ∥$ se tezi se mar jaati hai?

Candidate gradient. $q_{x} = 2 x, q_{y} = 2 y$ , toh $\nabla q (0, 0) = (0, 0)$ , aur $q (0, 0) = 0$ .
Error build karo. $h = (h_{1}, h_{2})$ ke saath: $E = q (h) - [0 + (0, 0) \cdot h] = h_{1}^{2} + h_{2}^{2} = ∥ h ∥^{2} .$ Yeh step kyun? Definition exactly is leftover ko test karti hai linear part subtract karne ke baad.
$∥ h ∥$ se divide karo: $\frac{E}{∥ h ∥} = \frac{∥ h ∥ ^{2}}{∥ h ∥} = ∥ h ∥ \to 0$ . Yeh step kyun? Yahi $o (∥ h ∥)$ condition hai. Yeh har direction se hold karti hai kyunki $∥ h ∥ \to 0$ angle se regardless hota hai.
Conclude karo: limit $0$ hai, toh $q$ origin par differentiable hai tangent plane $z = 0$ ke saath (bowl ka flat bottom).

Verify: $∥ h ∥ = 0.1$ par ratio $0.1$ hai; $0.01$ par yeh $0.01$ hai — distance halving karne se ratio bhi half hoti hai, $\to 0$ confirm karta hai. ✓

Cell H — exam twist: parameter dhundho

Worked example Example 8 (Cell H) — continuity/differentiability ke liye solve karo

Constant $k$ ki kaunsi value ke liye $f (x, y) = ⎩ ⎨ ⎧ \frac{x ^{2} + y ^{2}}{∣ x ∣ + ∣ y ∣} k (x, y) \neq = (0, 0) (0, 0)$ point $(0, 0)$ par continuous hai, aur kya woh tab differentiable hai?

Forecast: pehle $k$ guess karo — kya ratio origin approach karte waqt blow up hoti hai ya vanish?

Required $k$ dhundho (continuity). Bound: $x^{2} + y^{2} \leq (∣ x ∣ + ∣ y ∣)^{2}$ , toh $0 \leq f = \frac{x ^{2} + y ^{2}}{∣ x ∣ + ∣ y ∣} \leq ∣ x ∣ + ∣ y ∣ \to 0$ . Yeh step kyun? Squeeze limit ko $0$ force karta hai, toh continuity require karti hai $k = 0$ . Continuity in Multiple Variables dekho.
$k = 0$ set karo aur partials test karo. $y = 0$ par: $f (x, 0) = \frac{x ^{2}}{∣ x ∣} = ∣ x ∣$ . Uska $0$ par derivative $lim_{h \to 0} \frac{∣ h ∣ - 0}{h}$ hai, jo right se $+ 1$ aur left se $- 1$ hai — koi partial $f_{x} (0, 0)$ nahi. Yeh step kyun? Axis slope bhi exist karna fail karta hai; $∣ x ∣$ ka $0$ par corner hai.
Conclude karo: $k = 0$ ke saath, $f$ continuous hai lekin not differentiable — actually origin par uske koi partials hi nahi hain. Yeh step kyun? Differentiable $\Rightarrow$ partials exist. No partials $\Rightarrow$ not differentiable, continuity se regardless.

Verify: $k = 0$ ise continuous banata hai (upper bound $∣ x ∣ + ∣ y ∣ \to 0$ ). $x$ -axis ke saath $f = ∣ x ∣$ , jiske one-sided slopes $+ 1$ aur $- 1$ hain: unequal $\Rightarrow$ partial undefined $\Rightarrow$ not differentiable. ✓

Recall Kaunsa cell kaunsa tha? (self-test)

Partials exist lekin discontinuous ::: Cell B (Ex 2, parabola path $\frac{1}{2}$ deta hai). Continuous + partials exist + phir bhi not differentiable ::: Cell C (Ex 3, diagonal ratio $- \frac{1}{2 2}$ ). Saare directional derivatives exist phir bhi not differentiable ::: Cell E (Ex 5, $D_{u} = cos^{2} θ sin θ$ ). Removable/degenerate input ko differentiable banaya ::: Cell D (Ex 4, $sin (x y) / y \to$ patch). Raw $o (∥ h ∥)$ limit se differentiability prove karna ::: Cell G (Ex 7, $E = ∥ h ∥^{2}$ ). Parameter $k$ ke liye solve karo ::: Cell H (Ex 8, $k = 0$ , phir bhi koi partials nahi).

Connections

4.4.06 Differentiability in multiple variables (Hinglish)
Partial Derivatives
Gradient and Directional Derivatives
Tangent Planes and Linear Approximation
Continuity in Multiple Variables
C1 and Smooth Functions
Chain Rule (Multivariable)
Total Derivative / Jacobian