How to show $f(x,y)=|xy|^\alpha\log(x^2+y^2)$ is differentiable at $(0,0)$?

Question

Show that if $\alpha > 1/2$, then

$$f(x, y)=\begin{cases} |xy|^\alpha\log(x^2+y^2), ~(x, y) \ne (0,0)\\\\ ~~~0, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{otherwise}\;. \end{cases}$$

is differentiable at $(0,0)$.

Answer 1

Note that $|x| = \sqrt{x^2} \leq \sqrt{x^2+y^2}$ and similarly $|y| = \sqrt{y^2} \leq \sqrt{x^2+y^2}$. Thus $|xy|^\alpha \leq (x^2+y^2)^\alpha$.

Then $$ \left|\frac{f(x,y)}{\sqrt{x^2+y^2}}\right| \leq (x^2+y^2)^{\alpha-\frac12} \log(x^2+y^2) = (r^2)^{\alpha-\frac12} \log(r^2) $$ Of course $(x,y) \to 0$ if and only if $r^2 = x^2+y^2 \to 0$. Since $\alpha>1/2$ the limit is zero by an elementary argument using L'Hospital's rule. It follows that $f$ is differentiable at $(0,0)$.