Differentiability of the function $f(x,y)=\frac{x}{y}$

Question

I am having diificulty in investigating the differentiability of the following function $f (x,y)= \begin{cases} \dfrac x y & y \ne 0 \\ 0 & y = 0 \end{cases}$ .It is reqiured to check the differentiability of $f $ at $(x,y)$ where $y \ne 0$.Thanks for any help in advance.

Answer 1

Let $(x_0,y_0) \in \mathbb R^2$ such that $y_0 \ne 0$. Then show that

$f_x(x_0,y_0)=\frac{1}{y_0}$ and $f_y(x_0,y_0)=-\frac{x_0}{y_0^2}$.

Next consider the quotient

$$Q(h,k)=\frac{f(x_0+h,y_0+k)-f(x_0,y_0)-f_x(x_0,y_0)h-f_y(x_0,y_0)k}{\sqrt{h^2+k^2}}.$$

Then we have: $f$ is differentiable at $(x_0,y_0) \iff Q(h,k) \to 0$ for $(h,k) \to (0,0)$.

It is your turn to investigate if $Q(h,k) \to 0$ for $(h,k) \to (0,0)$.

Answer 2

Another possibility: let $(x_0,y_0) \in \mathbb R^2$ with $y_0 \ne 0$. Then choose a neighborhood $U$ of $(x_0,y_0)$ such that $y \ne 0$ for all $(x,y) \in U$.

It is your turn to show that the partial derivatives $f_x$ and $f_y$ are continuous on $U$.

Consequence ??