Differentiablility of a multivariable function I have the function: $f(x,y) = xy^{1/2}$
I want to check what the parital derivatives are at the origin and whether it is differentiable at the origin.
For the partial derivatives at the origin I obtained:
$f_x = y^{1/2} = 0$
$f_y = \frac{1}{2}x y^{-1/2}=0$
Please let me know if these are correct.
Where I'm struggling is showing whether the function is differentiable or not. One idea I have is to say that since $xy^{1/2}$ is a composition of two differentiable functions, it too must be differentiable. However, I'm not sure if this is correct in this case and if it is enough to prove differentiability.
Any help would be appreciated, thanks in advance.
 A: Be careful and remember the negative power reciprocal laws:
$$f_y=\frac{x}{2\sqrt{y}}=\frac{0}{0}, (x,y)=(0,0).$$
This is potentially resolvable using L'Hopital's rule, since both $x,y$ go to zero I will treat them as one variable, $u$ - this examines one of infinitely many paths of approach to the origin, but by showing that this path fails I show all paths fail (i.e if the limit does not exist for one way, then by definition it cannot exist at all):
$$\frac{d}{du}\frac{u}{2\sqrt{u}}=\frac{d}{du}\frac{\sqrt{u}}{2}=-\frac{1}{2\sqrt{u}}$$
Which is undefined at zero and further applications of the law will not help. This shows that there is one path of approaching the origin for which the partial derivative w.r.t $y$ does not exist, showing that it does not exist at the origin in general. For $f$ to be differentiable, we must have that all partial derivatives exist.
I conclude that it is not differentiable at the origin.
Look at these $3$D sketches from Wolfram Alpha of the partial derivative with respect to $y$:

See how choppy and not well defined they are at $(0,0)$?
