Is the following function continuous, differentiable, continuously differentiable? \[ f: \RR^2 &\to \RR \\ f(x, y) &= \begin{cases} {xy \over \sqrt{x^2 + y^2}} & (x, y) \neq (0, 0) \\ 0 & \text{else}. \end{cases} \]
Solution
Away from \((0,0)\) the function is smooth. At the origin, \[ |f(x,y)|=\frac{|xy|}{\sqrt{x^2+y^2}} \le \frac{x^2+y^2}{2\sqrt{x^2+y^2}} =\frac12\sqrt{x^2+y^2}, \] so \(f(x,y)\to0=f(0,0)\). Hence \(f\) is continuous everywhere.
The partial derivatives at the origin are both \(0\), so if \(f\) were differentiable there, its derivative would be the zero map. Differentiability would then require \[ \frac{|f(x,y)|}{\sqrt{x^2+y^2}}\to0. \] But along \(y=x\ne0\), \[ \frac{|f(x,x)|}{\sqrt{2x^2}} =\frac{x^2/(\sqrt2|x|)}{\sqrt2|x|} =\frac12. \] Thus \(f\) is not differentiable at \((0,0)\).
Therefore \(f\) is continuous everywhere, differentiable on \(\mathbb R^2\setminus\{0\}\) but not at the origin, and consequently is not continuously differentiable on \(\mathbb R^2\).