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I need to find out where the function

$$f(x,y)=\left\{ \begin{array}{cc} e^{\tfrac{1}{x^2+y^2-1}}, & x^2+y^2<1 \\ 0, & x^2+y^2\geq1 \end{array} \right.$$ is differentiable. Here is my progress

For $x^2+y^2<1$, I can use the chain rule $$\frac{\partial f}{\partial x}=\frac{-2xe^{\tfrac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}, \ \ \frac{\partial f}{\partial y}=\frac{-2ye^{\tfrac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}.$$

This partial derivatives are continiuos, so $f$ is differentiable.

For $x^2+y^2>1$, $f$ is constant, then its partial derivatives are zero and $f$ is differenciable in this case too.

Finally, for $x^2+y^2=1$, the fraction $\dfrac{1}{x^2+y^2-1}$ goes to $0$ and $f(x,y)=1$ and we have
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0.$$

Therefore, $f$ is differentiable in the whole $\mathbb{R}^2$.

Is it correct?

J.G.
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Marcos Paulo
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  • If $x^2+y^2>1$ why would $f$ be constant? Also if $x^2+y^2=1$ then $\frac{1}{x^2+y^2-1}$ is not $0$ but undefined. –  Jun 25 '20 at 19:44
  • In your last case, the demoninator equals zero, so the fraction is undefined ("infinity"). Is it possible you are missing a minus sign in argument of the exponential ? – user531544 Jun 25 '20 at 19:44
  • @AndréArmatowski the 2nd sentence (I reformat the function for it), it says that $f$ is zero for $x^2+y^2\geq1$. So, in particular, for $x^2+y^2>1$, we have $f(x,y)=0$, which is constant, right?. – Marcos Paulo Jun 25 '20 at 20:03
  • @M1183 here it is the question from the book https://photos.app.goo.gl/UB2FFvrassESpjiu6 – Marcos Paulo Jun 25 '20 at 20:11
  • @AndréArmatowski why 0/0? – Marcos Paulo Jun 25 '20 at 20:33
  • Consider when $x^2+y^2$ approaches $1$ but $1-x^2-y^2 > 0$ , then $(x^2+y^2-1)^2 \to 0$ and $e^{\frac{1}{x^2+y^2-1}} \to e^{-\infty} "=" 0$ –  Jun 25 '20 at 20:39
  • @AndréArmatowski but there is no square for $x^2+y^2-1$ – Marcos Paulo Jun 25 '20 at 20:52
  • In order for the partial derivative to be continuous you need $$\lim_{x^2+y^2 \to 1} \frac{-2xe^{\frac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}$$ to be zero. –  Jun 25 '20 at 20:58
  • @AndréArmatowski understood now but is this case $x^2+y^2<1$ implies that $(x^2+y^2-1) ^2$ is never zero, so the partial derivatives are never igual to $0/0$ – Marcos Paulo Jun 25 '20 at 21:09
  • True, but the limit approaches $0/0$ which is what we are concerned with. –  Jun 25 '20 at 21:26
  • Let us continue this discussion in chat. –  Jun 25 '20 at 21:36

1 Answers1

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Let's consider $$f(t)=\left\{ \begin{array}{cc} e^{\dfrac{1}{t-1}}, & t<1 \\ 0, & t\geq1 \end{array} \right.$$ Which is differentiable and its composition with differentiable $g(x,y)=x^2+y^2$

For partial derivative let's consider any $(x,y)$ for which $x^2+y^2=1$. Then $$\lim_{\Delta x \to 0-}\frac{e^{\frac{1}{(x+ \Delta x)^2+y^2-1}}}{\Delta x}=\lim_{\Delta x \to 0-}\frac{1}{\Delta x}e^{\frac{1}{(x +\Delta x)^2- x^2}}=\lim_{\Delta x \to 0-}\frac{1}{\Delta x} e^{\frac{1}{2 x \Delta x+\Delta x^2}}=0$$

zkutch
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