For $$
f(x,y)=\begin{cases}
y^2 sin\left(\frac{x}{y}\right) & \text{if } y\neq0 \\
0 & \text{if } y=0
\end{cases}$$
i've shown that it is continuous and that the partial derivatives exist in $\mathbb{R}^2$. However, it appears that the partials are not continuous at $(0,0)$; is this sufficient to show that $f$ is not differentiable at $(0,0)$?
Friday 27 June 2014
real analysis - differentiability at a point (0,0) based on partial derivatives
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