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Considering a function like $f(x,y) = x^{1/2} y^{1/2}$

The partial derivative with respect to x, $\frac{\partial f}{\partial x} = \frac{y^{1/2}}{2x^{1/2}} $, is undefined at (0,0), and the limit (x,y) -> (0,0) does not exist.

In this case we should then use the definition (lim h -> 0) if we want to know the value of the derivative at the origin and find that $\frac{\partial f}{\partial x} = 0 $.

With just the knowledge that the function obtained through differentiation $\frac{y^{1/2}}{2x^{1/2}}$ is undefined at (0,0) can we conclude anything about the original function? Does it tell us that f(x,y) is not differentiable at that point?

I can't off the top of my head think of other cases of continuous, non-piecwise, innocent-looking functions like the one I supplied for which the differentiation does does not equal the partial derivative. My guess is that this can happen because the function f(x,y) becomes not differential on the boundary of its domain. Is there any way to summarise this behaviour?

I couldn't find any questions that had already addressed this directly.

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First off, the partial derivative with respect to $x$ is actually $\frac{\partial f}{\partial x} = \frac{\sqrt{y}}{2\sqrt{x}}$ (The partial derivative with respect to $y$ is $\frac{\partial f}{\partial y} = \frac{\sqrt{x}}{2\sqrt{y}}$).

Secondly, the figure you're describing is an infinite elliptic cone, I don't know if that helps, but it should give you some insight on the function.

The nonexistence of a partial derivative implicates a critical point. Both $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are undefined at the point $(0,0)$. The function isn't differentiable at the point; however, that doesn't mean the function isn't continuous or vice versa.

You can view plots and get more info about the partial derivatives here and here.

For more info, check out these posts.

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  • $\begingroup$ Thanks, I forgot the factor of 2, will update the question. Not sure where you're getting the cone from, that would be $f(x,y)=\sqrt{x^2+y^2} $. $\endgroup$
    – Droobasaur
    Apr 17 '18 at 8:32
  • $\begingroup$ The partial derivatives do exist at the point and it's the contrast of that with the non-existence of $\frac{\sqrt{y}}{2\sqrt{x}}$ that I am curious about. $\endgroup$
    – Droobasaur
    Apr 17 '18 at 8:40

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