Prove that $f(x,y)=xy$ is differentiable at $(x_0,y_0)$ using the $\epsilon$ definition.

We can use the definition of differentiability: $$ f(x_0+\Delta x, y_0+\Delta y)-f(x_0,y_0)=f_x(x_0,y_0)\cdot\Delta x + f_y(x_0,y_0)\cdot \Delta y+\epsilon_1\cdot\Delta x+\epsilon_2\cdot\Delta y $$ According to the above formula we get: $$ x(y-y_0)-x_0(y-y_0)=(x-x_0)(y-y_0)=\epsilon_1\cdot(x-x_0)+\epsilon_2\cdot(y-y_0) $$ By comparing the coefficients we have that $\epsilon_2=x-x_0$ thus: $$ \lim_{(x,y)\to(x_0,y_0)} (x-x_0)=0 $$ But what about $\epsilon_1$? We can conclude that $\epsilon_1(x-x_0)=0$ but we can't know if $\epsilon_1$ necessarily goes to $0$.

  • $\begingroup$ why don't you change $(x,y)$ to $(r,\theta)$ where $x=r\cos \theta$ and $y=r\sin \theta$ $\endgroup$ – MAN-MADE Aug 6 '17 at 13:49

$f$ is differentiable at $(x_0,y_0)$ if the difference $\,(x-x_0)(y-y_0)\,$ between $f(x,y)-f(x_0,y_0)$ and its linear approximation $y_0(x-x_0)+x_0(y-y_0)$ is $o\bigl(\lVert(x-x_0,y-y_0)\rVert\bigr)$.

To see this, set $x-x_0=r\cos\theta$, $y-y_0=r\sin\theta$ $\;(r>0,\;0\le \theta<2\pi)$. This difference is $(x-x_0)(y-y_0)$, so we have $$\frac{\lvert(x-x_0)(y-y_0)\rvert}{\lVert(x-x_0,y-y_0)\rVert} =\frac{r^2\lvert\cos\theta\sin\theta\rvert}r=r\lvert\cos\theta\sin\theta\rvert\le r\to 0.$$

  • $\begingroup$ Does your method use $\epsilon$? If yes can you please expain what $\epsilon$ is in polar coordinates. $\endgroup$ – Yos Aug 6 '17 at 16:16
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    $\begingroup$ It uses ε implicitly: if $r=\lVert(x-x_0,y-y_0)\rVert<\varepsilon$ then $\vert xy-x_0y_0\rvert<\varepsilon\lVert(x-x_0,y-y_0)\rVert<\varepsilon^2$ $\endgroup$ – Bernard Aug 6 '17 at 16:38
  • $\begingroup$ So the formula in Cartesian coordinates is unsolvable? $\endgroup$ – Yos Aug 6 '17 at 16:40
  • $\begingroup$ I don't understand what you mean exactly. Maybe you want the full set of solutions as intervals for $x$ and $y$? $\endgroup$ – Bernard Aug 6 '17 at 17:09
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    $\begingroup$ Polar coordinates is just a useful shorter notation here. Replace $r$ with $ \sqrt{(x-x_0)^2+(y-y_0)^2}$. I'm not sure you use a correct definition of differentiability. $\endgroup$ – Bernard Aug 6 '17 at 17:18

We have that $\nabla{f}(x_0,y_0)=(y_0,x_0)$

We'll use the definition of differentiability of a function at a particular point.

So from the definition: $$\frac{|f((x_0,y_0)+(h_1,h_2))-f(x_0,y_0)-<\nabla{f}(x_0,y_0),(h_1,h_2)>|}{\sqrt{h_1^2+h_2^2}}$$ $$=\frac{|(x_0+h_1)(y_0+h_2)-x_0y_0-y_0h_1-x_0h_2|}{\sqrt{h_1^2+h_2^2}}$$ $$=\frac{|x_0y_0+x_0h_2+y_0h_1+h_1h_2-x_0y_0-y_0h_1-x_0h_2|}{\sqrt{h_1^2+h_2^2}}$$ $$=\frac{|h_1h_2|}{\sqrt{h_1^2+h_2^2}}$$

Now use the fact that $|h_1h_2| \leqslant \frac{1}{2}(h_1^2+h_2^2)$

and take $\delta=2 \epsilon$ and you are done.

Also $<.,.>$ denotes the usual inner product in $\mathbb{R}^2$.

  • $\begingroup$ Or, less fancily, use $|h_i|\le \|(h_1,h_2)\| = \sqrt{h_1^2+h_2^2}$. $\endgroup$ – Ted Shifrin Aug 7 '17 at 5:57

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