Let $$ f(x,y) = \begin{cases} \dfrac{xy}{\sqrt{x^2+y^2}} & \text{if $(x,y)\neq(0,0)$ } \\[2ex] 0 & \text{if $(x,y)=(0,0)$ } \\ \end{cases} $$ Show that this function is continuous but not differentiable at $(0,0),$ although it has both partial derivatives existing there.
I can show this function is continous and the partial derivatives exist. But how can I show that this function is not differentiable?
Is showing that the function is differentiable similar to showing that a derivative exists?
On
I don't know if I should write my answer here but can't write all this as a comment.
This is what I tried to do:
According to the definition , $f$ is fifferentiable at $a $ if,
$\Delta$f=$f$(a$_1$+h$_1$,...,a$_n$+h$_n$)-$f$(a$_1$,a$_2$,..,a$_n$)
=l$_1$h$_1$+l$_2$h$_2$+...+l$_n$h$_n$+...+B$_1$h$_1$+...+B$_n$h$_n$
where l's are constants and $ \lim_{h1h2...hn\to 0,0,,..,0}$B$_i$=0
Here a=(0,0,).
Therefore $\Delta$f=f(0+h$_1$,0+h$_2$)-f(0,0)
=$\frac{h_1h_2}{\sqrt{h_1^2+h_2^2}}$-0
This is not in the above form .Therefore f is not differentiable at (0,0)
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 $z=f(x,y)$.