I've of $f(x,y)=\sqrt{x^2+y^2} \sin (2 \arctan {y\over x})$ for $x \ne 0$ and $0$ for $x=0$
The function is continuous in all $R^2$.
In the points $(0,y_0)$ with $y_0 \ne 0$ the $\partial x f(0,y_0)$ is $2$ if $y_0>0$ and
$-2$ if $y_0<0$?
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1you might like to correct the typo in your question. it sounds rude without the "n" in "function" ;-) – David Holden Nov 10 '14 at 15:14
2 Answers
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it may help to observe the following:
for $x \ne 0$ let $t=\frac{y}{x}$ and $\theta=\arctan(t)$
then $$\sin 2\theta = 2 \sin \theta \cos \theta = \frac{4t(t^2-1)}{(1+t^2)^2} $$ so $$ \sqrt{x^2+y^2} \sin (2 \arctan t) = 4xy(y^2-x^2)(y^2+x^2)^{-\frac32} $$
David Holden
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Note that for all $\tau\in{\mathbb R}$ one has $$\sin\bigl(2\arctan \tau\bigr)={2\tau\over 1+t^2}\ .$$ Therefore your function $f$ can be written as $$f(x,y)=\sqrt{x^2+y^2}{2{y\over x}\over 1+\left({y\over x}\right)^2}={2xy\over r}\ ,\tag{1}$$ where $r:=\sqrt{x^2+y^2}$. It is easy to check that $(1)$ produces the desired value also when $x=0$. It follows that for all $(x,y)\in\dot{\mathbb R}^2$ one has $$f_x(x,y)={2y\over r}-{2xy\over r^2} r_x={2y\over r}-{2xy\over r^2} {x\over r}=2{y^3\over (x^2+y^2)^{3/2}}\ ,$$ and similarly $$f_y(x,y)={2x\over r}-{2xy\over r^2} r_y={2x\over r}-{2xy\over r^2} {y\over r}=2{x^3\over (x^2+y^2)^{3/2}}\ .$$ Since $f_x$ and $f_y$ both are continuous on all of $\dot{\mathbb R}^2$ the functioon $f$ is differentiable in this domain.
In particular for $y\ne0$ one has $$f_x(0,y)=2{y^3\over|y|^3}=2\>{\rm sgn}\ y\ .$$
Christian Blatter
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