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I'm being asked to show the function $f(x) = \frac{x}{y}$ is differentiable in its domain, via the definition of a differentiable function. For this I've already calculated the partial derivatives with respect to $x$ and $y$, these are: $\frac{\partial f}{\partial x} = \frac{1}{y}$ and $\frac{\partial f}{\partial y} = \frac{-x}{y^{2}}$.

It only remains to show $\lim_{(x,y) \to (x_{0}, y_{0})} \frac{|\frac{x}{y} - \frac{x_{0}}{y_{0}} - \frac{1}{y_{0}}(x-x_{0}) + \frac{x_{0}}{y_{0}^{2}}(y - y_{0})|}{||(x-x_{0},y-y_{0})||} = 0$.

However, I'm having a lot of trouble trying to factorize the numerator conveniently. Ideally, the numerator should be expressed as additions of terms of the form $|x-x_{0}|(...)$ or $|y-y_{0}|(...)$. If someone could help me just carry out this step I should be able to finish it from there. Thanks in advance.

MMMagician
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  • You don't have to check that. Partial derivatives exist and they are continuous on $D=\Bbb R^2-{0}$, so the function is differentiable on $D$. The partial derivatives are not defined at $(0,0)$, so the function is not differentiable at the origin and thus on $\Bbb R^2$. – Bob Dobbs Apr 24 '23 at 00:01
  • Try considering $X=x-x_0$ and $Y=y-y_0$ – Math Attack Apr 24 '23 at 00:02
  • @BobDobbs Thanks for the comment, however, I'm not allowed to use that theorem in this proof. – MMMagician Apr 25 '23 at 00:04
  • @FabioCaiazzo. I tried to do your suggestion, but I couldn't work out a way to use it in my advantage. – MMMagician Apr 25 '23 at 00:07

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We have $$ \lim_{(x,y) \to (x_{0}, y_{0})} \frac{|\frac{x}{y} - \frac{x_{0}}{y_{0}} - \frac{1}{y_{0}}(x-x_{0}) + \frac{x_{0}}{y_{0}^{2}}(y - y_{0})|}{||(x-x_{0},y-y_{0})||} =\lim_{(x,y) \to (x_{0}, y_{0})} \frac{|(y - y_{0})|\,|(x_0(y-y_0)-y_0(x-x_0))|}{|y|\,y_0^2\sqrt{(x-x_0)^2+(y-y_0)^2}}. $$ Now, let $x-x_0=r\cos\theta$ and $y=y_0=r\sin\theta.$ Then, we compute the limit as $$\lim_{r\rightarrow 0}\frac{r(x_0\sin\theta-y_0\cos\theta)}{y_0^2(y_0+r\sin\theta)}=\frac{0.(x_0\sin\theta-y_0\cos\theta)}{y_0^3}=0.$$ Here, $y_0\neq 0$, because the domain of the function is $D=\{(x,y)\in\Bbb R^2|y\neq 0\}$ and $(x_0,y_0)\in D$.

Note: My comment above is typo. I thought the origin is the only problematic point.

Bob Dobbs
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