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Let $f: \mathbb R^2 \to \mathbb R$ be the function

$$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$

where $(x,y) \neq (0,0)$ and $f(0,0)=0$.

Is $f$ differentiable at $(0,0)$ and if so, how can I prove it?

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up vote 2 down vote accepted

We will prove that $f$ is differentiable at $(0,0)$ with derivative $df_{(0,0)}=0$.

Using $|\sin z|\leq |z|$, we have $$ \frac{|f(x,y)|}{\sqrt{x^2+y^2}}\leq \frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}+\frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}. $$

Now using Cauchy-Schwarz, we find $$ \frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}\leq \frac{|x|^3\sqrt{2}\sqrt{x^2+y^2}}{(x^2+y^2)^{3/2}}=\frac{|x|^3\sqrt{2}}{x^2+y^2}\leq \frac{\sqrt{2}|x| (x^2+y^2)}{x^2+y^2}=\sqrt{2}|x|\leq \sqrt{2}\sqrt{x^2+y^2} $$ So this term converges to $0$ as $\sqrt{x^2+y^2}$ tends to $0$.

Next observe that $$ \frac{|y|}{(x^2+y^2)^{1/2}}\leq \frac{\sqrt{x^2+y^2}}{(x^2+y^2)^{1/2}}=1 $$ and $$ \frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}=\frac{y^2}{x^2+y^2}\cdot|y\ln(x^2+y^2)|\leq \sqrt{x^2+y^2}\ln(x^2+y^2). $$

Finally, $$ \frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}\leq \sqrt{x^2+y^2}\ln(x^2+y^2) $$ so this term also converges to $0$ as $\sqrt{x^2+y^2}$ tends to $0$, using $\lim_{u\rightarrow 0^+}\sqrt{u}\ln u=0$.

So $$ \lim_{(x,y)\rightarrow(0,0)} \frac{|f(x,y)-f(0,0)-0|}{\sqrt{x^2+y^2}}=0 $$ which proves exactly our claim.

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Fantastic. Thank you very much. –  user59917 Feb 3 '13 at 22:37
    
You're welcome. –  1015 Feb 3 '13 at 23:01
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