continuityderivativesmultivariable-calculus
How do I show that the following function is differentiable at $(0,0)$?
$$
\begin{cases}
\dfrac{\sin(xy)}{y}, & \text{if }y \neq 0 \\
\\
0, & \text{if }y = 0
\end{cases}
$$
I calculated the partial derivatives and
There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
You have $$x^2+y^2-2 \vert xy \vert=(\vert x \vert - \vert y \vert)^2 \ge 0$$ Hence $$\vert xy \vert \le \frac{x^2+y^2}{2}$$ and $$0 \le \frac{\vert f(x,y) \vert}{\sqrt{x^2+y^2}} = \frac{x^2y^2}{x^2+y^2} \le \frac{1}{4}(x^2+y^2)$$ As $\lim_{(x,y) \to (0,0)} x^2+y^2 = 0$, this proves that $f$ is differentiable at $(0,0)$ and that its Fréchet derivative is equal to $0$. Which means that $f_x(0,0)=f_y(0,0)=0$.
Best Answer
As you defined it your function is not even continuous at $(0,0)$. Note that $$\lim_{y\to0}{\sin(xy)\over y}=x\ .$$ I'm assuming this will be corrected. Then you can argue as follows: When $y\ne0$ you have $${\sin(xy)\over y}=\int_0^x \cos(y t)\>dt\ ,$$ whereby the RHS is obviously $C^1$ in a neighborhood of $(0,0)$.