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Given the function $ f(x,y) = \sqrt[3]{y}\cdot \arctan(x)$ discuss the existence and continuity of it's partial derivatives and existence of it's total derivative.

Since the partial derivative $ \frac{\partial f}{\partial y} = \frac{\arctan(x)}{3\sqrt[3]{y^2}}$ has discontinuity at $y=0$, I tried to compute the partial derivative at $(x,0)$ using the limit, which gives: $$ \lim_{t\to0} \frac{f(x,t) - f(x,0)}{t} = \lim_{t\to 0} \frac{\sqrt[3]{t}}{t}\arctan(x) = +\infty.$$

Does this mean that the partial derivative doesn't exist at $(x,0)$? So there's no total derivative at $(x,0)$ and it can be said that the function is not differentiable at $\mathbb{R}^2$ ?

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up vote 1 down vote accepted

If x=o the partial derivates at (0.0) are both 0. If x is not zero then at (x,0) the partial derivate with respect to the variable y does not exist and then the function is not differentable at those points .

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