2
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$f$ 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 $$\frac{\lvert(x-x_0)(y-y_0)\rvert}{\lVert(x-x_0,y-y_0)\rVert} =\frac{r^2\lvert\cos\theta\sin\theta\rvert}r=r\lvert\cos\theta\sin\theta\rvert\le r\to 0.$$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $\mathbb{R}^2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.