Consider $$ f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} & , \text{if } \ (x,y)\neq(0,0) \\
\hspace{0.5cm}0 &, \text{if } \ (x,y)=(0,0) \end{cases} $$
Computing the limit at zero
$$
\left\{
\begin{array}{c}
x=rcos\theta \\
y=rsin\theta \\
\end{array}
\right.
$$
$$ (x,y) \to (0,0) \Longleftrightarrow r \to 0 $$
$$ \lim_{ r \to \ 0 } \frac{ r^2cos\theta sin\theta }{ r^2( cos^2 \theta + sin^2 \theta) } = cos \theta sin \theta $$
Then the limit does not exist. Hence this function is not continuos at zero.
$$ \nabla f(x,y) = \begin{cases} \left( \frac{y(y^2-x^2)}{(x^2+y^2)^2} , \frac{x(x^2-y^2)}{(x^2+y^2)^2} \right) & , \text{if } \ (x,y)\neq(0,0) \\
\hspace{0.5cm} (0,0) &, \text{if } \ (x,y)=(0,0) \end{cases} $$
So it doesn't matter that the function is not continuous at zero? or this function does not differentiable at zero?
Best Answer
If a function $f$ is differentiable at a point $p$, then $\nabla f(p)$ exists. However, the existence of $\nabla f(p)$ only requires the existence of partial derivatives of $f$ at $p$, which is not enough to assure that $f$ is differentiable at $p$.