Exercises — Differentiability in multiple variables

False. Partial derivatives only test the two axis directions (East, North). A function can have both and still be discontinuous along a diagonal — and differentiability requires continuity. The parent's Example 2, $f=xy/(x^2+y^2)$, has $f_x(0,0)=f_y(0,0)=0$ yet is not even continuous.

$d\Rightarrow c\Rightarrow b,c\Rightarrow a.$ Strongest is ==$C^1$ (partials exist AND are continuous), then differentiable, and differentiable splits into two weaker consequences: continuous and partials exist==. Order: $d,c$, then $\{a,b\}$ (both are consequences of $c$, neither implies the other).

$f_x=3x^2$, $f_y=2$. Both are polynomials, hence continuous everywhere. A function whose partials exist and are continuous is ==$C^1$==, and $C^1\Rightarrow$ differentiable. So $f$ is differentiable on all of ${\mathbb{R}}^2$. Why this shortcut? Checking the $o(\Vert \mathbf{h}\Vert )$ limit is painful; the $C^1$ test replaces it with a continuity check you can do by inspection.

Step 1 (partials — why: the plane's slopes are the partials). $f_x=2x+y$, $f_y=x+2y$. Step 2 (evaluate). $f(1,2)=1+2+4=7$, $f_x(1,2)=4$, $f_y(1,2)=5$. Step 3 (assemble the plane). Using $z=f(\mathbf{a})+f_x(\mathbf{a})(x-a)+f_y(\mathbf{a})(y-b)$: $z=7+4(x-1)+5(y-2)=4x+5y-7.$

Step 1 (choose base — why: we want the argument to be a value $\ln$ evaluates cleanly). Let $f(x,y)=\ln (x^2+y^2)$, base $(1,1)$, so $x^2+y^2=2$ and $f(1,1)=\ln 2$. Step 2 (partials). $f_x=\frac{2x}{x^2+y^2}$, $f_y=\frac{2y}{x^2+y^2}$. At $(1,1)$: $f_x=\frac{2}{2}=1$, $f_y=1$. Step 3 (the step). $\mathbf{h}=(0.02,-0.03)$. Step 4 (linear approx). $f\approx \ln 2+1(0.02)+1(-0.03)=\ln 2-0.01\approx 0.6931-0.01=\mathrm{0.6831.}$ Why valid? $f$ is $C^1$ away from the origin, so the tangent plane is a legitimate first-order estimate.

$f_x=e^x\cos y$, $f_y=-e^x\sin y$. Both are products of the continuous functions $e^x$ and $\cos y$ (or $\sin y$), hence continuous everywhere. So $f\in C^1\Rightarrow$ differentiable on all of ${\mathbb{R}}^2$. $\nabla f=(e^x\cos y,-e^x\sin y)$.

(a) Continuity — why: differentiability needs it, so check first. Switch to polar $x=r\cos \theta ,y=r\sin \theta$: $f=\frac{r^2{\cos }^2\theta \cdot r\sin \theta }{r^2}=r{\cos }^2\theta \sin \theta .$ As $r\to 0$, $|f|\le r\to 0$ regardless of $\theta$. So $\lim =0=f(0,0)$: continuous. (Continuity survives — unlike parent Example 2.) (b) Partials at origin (limit definition, since $f$ is piecewise). Along $y=0$: $f(x,0)=0$, so $f_x(0,0)={\lim }_{h\to 0}\frac{0-0}{h}=0$. Along $x=0$: $f(0,y)=0$, so $f_y(0,0)=0$. Thus $\nabla f(0,0)=(0,0)$. (c) Differentiability — plug into the definition. With $\nabla f(0,0)=(0,0)$ the error is $E(\mathbf{h})=f(\mathbf{h})-0-0=f(h_1,h_2)$. Test: $\frac{|E|}{\Vert \mathbf{h}\Vert }=\frac{|r{\cos }^2\theta \sin \theta |}{r}=|{\cos }^2\theta \sin \theta |.$ This does not go to $0$ — it depends only on the direction $\theta$, not on $r$. Along $\theta =45^{\circ }$ it equals $\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\approx 0.354\ne 0$. So $f$ is continuous, has both partials, yet is NOT differentiable. Moral: continuity + partials is still not enough.

Direct (limit definition of directional derivative). $D_{\mathbf{u}}f(0,0)={\lim }_{t\to 0}\frac{f(t\cos \theta ,t\sin \theta )-0}{t}={\lim }_{t\to 0}\frac{t{\cos }^2\theta \sin \theta }{t}={\cos }^2\theta \sin \theta .$ (Used the polar simplification: numerator is $t\cdot ({\cos }^2\theta \sin \theta )$.) Formula prediction. $\nabla f(0,0)\cdot \mathbf{u}=(0,0)\cdot \mathbf{u}=0$. These agree only when ${\cos }^2\theta \sin \theta =0$ (axes or $\theta =90^{\circ }$). For $\theta =45^{\circ }$: direct gives $\frac{1}{2\sqrt{2}}\approx 0.354$, formula gives $0$. Why they clash: the identity $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}$ is only guaranteed for differentiable $f$. Here $f$ is not differentiable, so the identity legitimately breaks — exactly the parent's second [!mistake].

Write $r=\Vert \mathbf{h}\Vert =\sqrt{h_1^2+h_2^2}$. Then $|f(\mathbf{h})|=r^2\left|\sin \frac{1}{r}\right|\le r^2$. (a) Differentiability. First the partials at $0$: along $y=0$, $f(x,0)=x^2\sin \frac{1}{|x|}$, so $f_x(0,0)={\lim }_{h\to 0}\frac{h^2\sin \frac{1}{|h|}}{h}={\lim }_{h\to 0}h\sin \frac{1}{|h|}=0$ (bounded $\times$ zero). Likewise $f_y(0,0)=0$. Now the definition with $\nabla f(0,0)=(0,0)$: $\frac{|f(\mathbf{h})-0-0|}{r}=\frac{r^2|\sin \frac{1}{r}|}{r}=r\left|\sin \frac{1}{r}\right|\le r\to 0.$ So $f$ is differentiable at $(0,0)$. (b) Partials not continuous. For $(x,y)\ne 0$, with $r=\sqrt{x^2+y^2}$, $f_x=2x\sin \frac{1}{r}-\frac{x}{r}\cos \frac{1}{r}.$ Approach the origin along the $x$-axis ($y=0,r=|x|,x>0$): $\frac{x}{r}=1$, so $f_x=2x\sin \frac{1}{x}-\cos \frac{1}{x}$. The first term $\to 0$, but $\cos \frac{1}{x}$ oscillates between $-1$ and $1$ forever — no limit. So ${\lim }_{(x,y)\to 0}{f}_x$ does not exist, hence $f_x$ is not continuous at $0$. Conclusion: $f$ is differentiable (a genuine tangent plane exists) yet NOT $C^1$ — the converse "differentiable $\Rightarrow C^1$" is false. This is why the hierarchy arrow $C^1\Rightarrow$ differentiable points one way only.

By differentiability, $f(a+t{u}_1,b+t{u}_2)=f(a,b)+\nabla f(a,b)\cdot (t{u}_1,t{u}_2)+o(\Vert t\mathbf{u}\Vert )$. Since $\Vert \mathbf{u}\Vert =1$, $\Vert t\mathbf{u}\Vert =|t|$, so $g(t)-g(0)=t\nabla f\cdot \mathbf{u}+o(|t|).$ Divide by $t$ and let $t\to 0$: $g^{\prime }(0)=\nabla f(a,b)\cdot \mathbf{u}$. Why differentiability is essential: the step used the single linear approximation valid in every direction at once (the $o(\Vert \mathbf{h}\Vert )$ error). If only partials existed, we'd have no guarantee the error along the slanted line $\mathbf{u}$ is higher-order — and L3·Q2 shows the formula genuinely fails without it.

Take $f(x,y)=\{\begin{array}{ll}\frac{x^{2}y}{x^4+y^2}& (x,y)\ne (0,0)\\ 0& (0,0).\end{array}$ Directional derivatives everywhere. Along $\mathbf{u}=(\cos \theta ,\sin \theta )$, put $x=t\cos \theta ,y=t\sin \theta$: $\frac{f(t\mathbf{u})}{t}=\frac{1}{t}\cdot \frac{t^3{\cos }^2\theta \sin \theta }{t^4{\cos }^4\theta +t^2{\sin }^2\theta }=\frac{{\cos }^2\theta \sin \theta }{t^2{\cos }^4\theta +{\sin }^2\theta }.$ As $t\to 0$: if $\sin \theta \ne 0$, limit $=\frac{{\cos }^2\theta \sin \theta }{{\sin }^2\theta }=\frac{{\cos }^2\theta }{\sin \theta }$ (finite). If $\sin \theta =0$ (along $x$-axis), $f(x,0)=0$ so derivative $=0$. Every direction gives a finite directional derivative. ✓ Not even continuous (hence not differentiable). Approach along the parabola $y=x^2$ (not a straight line!): $f(x,x^2)=\frac{x^2\cdot x^2}{x^4+x^4}=\frac{x^4}{2x^4}=\frac{1}{2}\ne 0.$ So ${\lim }_{(x,y)\to 0}f\ne f(0,0)$: discontinuous $\Rightarrow$ not differentiable. The lesson: directional derivatives probe only straight lines; a curved approach can still betray the function. Differentiability is strictly stronger than "all directional derivatives exist."

Partials at origin. Along $y=0$: $f(x,0)=\sqrt{|x\cdot 0|}=0$, so $f_x(0,0)=0$. Along $x=0$: similarly $f_y(0,0)=0$. So $\nabla f(0,0)=(0,0)$. Test the definition. With $\nabla f(0,0)=0$, error $E(\mathbf{h})=\sqrt{|h_1{h}_2|}$. Along $\theta =45^{\circ }$, $h_1=h_2=t>0$, $r=\Vert \mathbf{h}\Vert =t\sqrt{2}$: $\frac{|E|}{r}=\frac{\sqrt{t^2}}{t\sqrt{2}}=\frac{t}{t\sqrt{2}}=\frac{1}{\sqrt{2}}\approx 0.707\ne 0.$ The ratio is a nonzero constant, so the limit fails: $f$ is not differentiable at $(0,0)$. (It is continuous — $\sqrt{|xy|}\le r\to 0$ — another continuous-but-not-differentiable specimen.)

The whole ladder in one breath ::: $C^1\Rightarrow$ differentiable $\Rightarrow$ (continuous AND all directional derivatives exist AND partials exist); none of the reverse arrows hold; counterexamples: L5·Q1 (dir. derivs but discontinuous), L3·Q1 (continuous+partials but not diff.), L4·Q1 (diff. but not $C^1$).