Exercises — Differentiability in multiple variables

False. Partial derivatives sirf axis directions (East, North) ko test karte hain. Ek function dono partial derivatives rakh sakta hai aur phir bhi diagonal direction mein discontinuous ho sakta hai — aur differentiability ke liye continuity zaroori hai. Parent ka Example 2, $f=xy/(x^2+y^2)$, mein $f_x(0,0)=f_y(0,0)=0$ hai phir bhi woh continuous tak nahi hai.

$d\Rightarrow c\Rightarrow b,c\Rightarrow a.$ Strongest ==$C^1$ hai (partials exist karte hain AUR continuous hain), phir differentiable, aur differentiable do weaker consequences mein split hota hai: continuous aur partials exist==. Order: $d,c$, phir $\{a,b\}$ (dono $c$ ke consequences hain, koi ek doosre ko imply nahi karta).

$f_x=3x^2$, $f_y=2$. Dono polynomials hain, isliye everywhere continuous hain. Jis function ke partials exist karein aur continuous hon woh ==$C^1$== hai, aur $C^1\Rightarrow$ differentiable. Toh $f$ poore ${\mathbb{R}}^2$ par differentiable hai. Yeh shortcut kyun? $o(\Vert \mathbf{h}\Vert )$ limit check karna tedious hai; $C^1$ test us check ko replace kar deta hai ek aisi continuity check se jo inspection se ho sakti hai.

Step 1 (partials — why: plane ke slopes partials hote hain). $f_x=2x+y$, $f_y=x+2y$. Step 2 (evaluate karo). $f(1,2)=1+2+4=7$, $f_x(1,2)=4$, $f_y(1,2)=5$. Step 3 (plane assemble karo). $z=f(\mathbf{a})+f_x(\mathbf{a})(x-a)+f_y(\mathbf{a})(y-b)$ use karte hue: $z=7+4(x-1)+5(y-2)=4x+5y-7.$

Step 1 (base choose karo — why: hum chahte hain argument aisi value ho jise $\ln$ cleanly evaluate kare). $f(x,y)=\ln (x^2+y^2)$ lo, base $(1,1)$, toh $x^2+y^2=2$ aur $f(1,1)=\ln 2$. Step 2 (partials). $f_x=\frac{2x}{x^2+y^2}$, $f_y=\frac{2y}{x^2+y^2}$. $(1,1)$ par: $f_x=\frac{2}{2}=1$, $f_y=1$. Step 3 (step). $\mathbf{h}=(0.02,-0.03)$. Step 4 (linear approx). $f\approx \ln 2+1(0.02)+1(-0.03)=\ln 2-0.01\approx 0.6931-0.01=\mathrm{0.6831.}$ Valid kyun hai? $f$, $C^1$ hai origin se door, toh tangent plane ek legitimate first-order estimate hai.

$f_x=e^x\cos y$, $f_y=-e^x\sin y$. Dono continuous functions $e^x$ aur $\cos y$ (ya $\sin y$) ke products hain, isliye everywhere continuous hain. Toh $f\in C^1\Rightarrow$ poore ${\mathbb{R}}^2$ par differentiable. $\nabla f=(e^x\cos y,-e^x\sin y)$.

(a) Continuity — why: differentiability ko yeh chahiye, toh pehle check karo. Polar mein switch karo $x=r\cos \theta ,y=r\sin \theta$: $f=\frac{r^2{\cos }^2\theta \cdot r\sin \theta }{r^2}=r{\cos }^2\theta \sin \theta .$ Jab $r\to 0$, $|f|\le r\to 0$ regardless of $\theta$. Toh $\lim =0=f(0,0)$: continuous. (Continuity survive karti hai — parent Example 2 ke unlike.) (b) Origin par Partials (limit definition, kyunki $f$ piecewise hai). $y=0$ ke along: $f(x,0)=0$, toh $f_x(0,0)={\lim }_{h\to 0}\frac{0-0}{h}=0$. $x=0$ ke along: $f(0,y)=0$, toh $f_y(0,0)=0$. Isliye $\nabla f(0,0)=(0,0)$. (c) Differentiability — definition mein plug karo. $\nabla f(0,0)=(0,0)$ ke saath error hai $E(\mathbf{h})=f(\mathbf{h})-0-0=f(h_1,h_2)$. Test: $\frac{|E|}{\Vert \mathbf{h}\Vert }=\frac{|r{\cos }^2\theta \sin \theta |}{r}=|{\cos }^2\theta \sin \theta |.$ Yeh $0$ par nahi jaata — yeh sirf direction $\theta$ par depend karta hai, $r$ par nahi. $\theta =45^{\circ }$ ke along yeh $\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\approx 0.354\ne 0$ hai. Toh $f$ continuous hai, dono partials hain, phir bhi differentiable NAHI hai. Moral: continuity + partials kaafi nahi hai.

Direct (directional derivative ki limit definition). $D_{\mathbf{u}}f(0,0)={\lim }_{t\to 0}\frac{f(t\cos \theta ,t\sin \theta )-0}{t}={\lim }_{t\to 0}\frac{t{\cos }^2\theta \sin \theta }{t}={\cos }^2\theta \sin \theta .$ (Polar simplification use ki: numerator $t\cdot ({\cos }^2\theta \sin \theta )$ hai.) Formula prediction. $\nabla f(0,0)\cdot \mathbf{u}=(0,0)\cdot \mathbf{u}=0$. Yeh tabhi agree karte hain jab ${\cos }^2\theta \sin \theta =0$ (axes ya $\theta =90^{\circ }$). $\theta =45^{\circ }$ ke liye: direct $\frac{1}{2\sqrt{2}}\approx 0.354$ deta hai, formula $0$ deta hai. Clash kyun: identity $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}$ sirf differentiable $f$ ke liye guaranteed hai. Yahan $f$ differentiable nahi hai, toh identity legitimately break ho jaati hai — exactly parent ka doosra [!mistake].

$r=\Vert \mathbf{h}\Vert =\sqrt{h_1^2+h_2^2}$ likho. Phir $|f(\mathbf{h})|=r^2\left|\sin \frac{1}{r}\right|\le r^2$. (a) Differentiability. Pehle $0$ par partials: $y=0$ ke along, $f(x,0)=x^2\sin \frac{1}{|x|}$, toh $f_x(0,0)={\lim }_{h\to 0}\frac{h^2\sin \frac{1}{|h|}}{h}={\lim }_{h\to 0}h\sin \frac{1}{|h|}=0$ (bounded $\times$ zero). Similarly $f_y(0,0)=0$. Ab $\nabla f(0,0)=(0,0)$ ke saath definition: $\frac{|f(\mathbf{h})-0-0|}{r}=\frac{r^2|\sin \frac{1}{r}|}{r}=r\left|\sin \frac{1}{r}\right|\le r\to 0.$ Toh $f$ $(0,0)$ par differentiable hai. (b) Partials continuous nahi. $(x,y)\ne 0$ ke liye, $r=\sqrt{x^2+y^2}$ ke saath, $f_x=2x\sin \frac{1}{r}-\frac{x}{r}\cos \frac{1}{r}.$ Origin ko $x$-axis ke along approach karo ($y=0,r=|x|,x>0$): $\frac{x}{r}=1$, toh $f_x=2x\sin \frac{1}{x}-\cos \frac{1}{x}$. Pehla term $\to 0$ jaata hai, lekin $\cos \frac{1}{x}$ hamesha $-1$ aur $1$ ke beech oscillate karta rehta hai — koi limit nahi. Toh ${\lim }_{(x,y)\to 0}{f}_x$ exist nahi karta, isliye $f_x$ $0$ par continuous nahi hai. Conclusion: $f$ differentiable hai (ek genuine tangent plane exist karta hai) phir bhi $C^1$ NAHI — converse "differentiable $\Rightarrow C^1$" false hai. Isliye hierarchy arrow $C^1\Rightarrow$ differentiable ek direction mein hi point karta hai.

Differentiability se, $f(a+t{u}_1,b+t{u}_2)=f(a,b)+\nabla f(a,b)\cdot (t{u}_1,t{u}_2)+o(\Vert t\mathbf{u}\Vert )$. Kyunki $\Vert \mathbf{u}\Vert =1$, $\Vert t\mathbf{u}\Vert =|t|$, toh $g(t)-g(0)=t\nabla f\cdot \mathbf{u}+o(|t|).$ $t$ se divide karo aur $t\to 0$ let karo: $g^{\prime }(0)=\nabla f(a,b)\cdot \mathbf{u}$. Differentiability essential kyun hai: is step ne single linear approximation use ki jo har direction mein ek saath valid hai (woh $o(\Vert \mathbf{h}\Vert )$ error). Agar sirf partials exist karte, toh humein koi guarantee nahi hoti ki slanted line $\mathbf{u}$ ke along error higher-order hai — aur L3·Q2 dikhata hai ki formula genuinely fail karta hai uske bina.

Lo $f(x,y)=\{\begin{array}{ll}\frac{x^{2}y}{x^4+y^2}& (x,y)\ne (0,0)\\ 0& (0,0).\end{array}$ Har jagah directional derivatives. $\mathbf{u}=(\cos \theta ,\sin \theta )$ ke along, $x=t\cos \theta ,y=t\sin \theta$ rakho: $\frac{f(t\mathbf{u})}{t}=\frac{1}{t}\cdot \frac{t^3{\cos }^2\theta \sin \theta }{t^4{\cos }^4\theta +t^2{\sin }^2\theta }=\frac{{\cos }^2\theta \sin \theta }{t^2{\cos }^4\theta +{\sin }^2\theta }.$ Jab $t\to 0$: agar $\sin \theta \ne 0$, limit $=\frac{{\cos }^2\theta \sin \theta }{{\sin }^2\theta }=\frac{{\cos }^2\theta }{\sin \theta }$ (finite). Agar $\sin \theta =0$ ($x$-axis ke along), $f(x,0)=0$ toh derivative $=0$. Har direction finite directional derivative deta hai. ✓ Continuous bhi nahi (isliye differentiable bhi nahi). Parabola $y=x^2$ ke along approach karo (straight line nahi!): $f(x,x^2)=\frac{x^2\cdot x^2}{x^4+x^4}=\frac{x^4}{2x^4}=\frac{1}{2}\ne 0.$ Toh ${\lim }_{(x,y)\to 0}f\ne f(0,0)$: discontinuous $\Rightarrow$ differentiable nahi. Lesson: directional derivatives sirf straight lines probe karte hain; ek curved approach phir bhi function ko betray kar sakta hai. Differentiability "all directional derivatives exist" se strictly stronger hai.

Origin par Partials. $y=0$ ke along: $f(x,0)=\sqrt{|x\cdot 0|}=0$, toh $f_x(0,0)=0$. $x=0$ ke along: similarly $f_y(0,0)=0$. Toh $\nabla f(0,0)=(0,0)$. Definition test karo. $\nabla f(0,0)=0$ ke saath, error $E(\mathbf{h})=\sqrt{|h_1{h}_2|}$. $\theta =45^{\circ }$ ke along, $h_1=h_2=t>0$, $r=\Vert \mathbf{h}\Vert =t\sqrt{2}$: $\frac{|E|}{r}=\frac{\sqrt{t^2}}{t\sqrt{2}}=\frac{t}{t\sqrt{2}}=\frac{1}{\sqrt{2}}\approx 0.707\ne 0.$ Ratio ek nonzero constant hai, toh limit fail ho jaata hai: $f$ $(0,0)$ par differentiable NAHI hai. (Yeh continuous hai — $\sqrt{|xy|}\le r\to 0$ — continuous-but-not-differentiable ka ek aur specimen.)

Poori ladder ek saath ::: $C^1\Rightarrow$ differentiable $\Rightarrow$ (continuous AND all directional derivatives exist AND partials exist); koi bhi reverse arrows hold nahi karte; counterexamples: L5·Q1 (dir. derivs hain lekin discontinuous), L3·Q1 (continuous+partials hain lekin diff. nahi), L4·Q1 (diff. hai lekin $C^1$ nahi).