1. ## [SOLVED] Partial derivative

$\displaystyle f(x,y)=\left\{\begin{array}{cc}y^2arctan\frac{x}{y },&\mbox{ if } y\neq 0\\0, & \mbox{ if } y=0\end{array}\right.$

Show that $\displaystyle f \in C^1.$

$\displaystyle y = 0: \frac{\partial{f}}{\partial{x}} = \lim_{h\to 0} \frac{f(a+h, 0) - f(a,0)}{h} = \frac{0-0}{h} = 0\ \forall x$

Correct?

$\displaystyle y \neq 0: \frac{\partial{f}}{\partial{x}} = \frac{y^3}{y^2 + x^2}, \frac{\partial{f}}{\partial{x}} \in C^n$

2. Originally Posted by Spec
$\displaystyle f(x,y)=\left\{\begin{array}{cc}y^2arctan\frac{x}{y },&\mbox{ if } y\neq 0\\0, & \mbox{ if } y=0\end{array}\right.$

Show that $\displaystyle f \in C^1.$
There is still one more thing you need to show. Remember the theorem says that if $\displaystyle \partial_1 f,\partial_2 f$ exist and are continous and $\displaystyle f$ is continous then $\displaystyle f$ is differenciable. You want to show that $\displaystyle f$ is continous at any point $\displaystyle (a,b)$. Say that $\displaystyle b=0$ then $\displaystyle f(a,b) = 0$. And so $\displaystyle |f(x,y) - f(a,b)| = |f(x,y)|$. We want to show we can make this sufficiently small in the punctured disk $\displaystyle 0<\sqrt{(x-a)^2+(y-b)^2} < \delta$. We can safely ignore the case when $\displaystyle y=b$ because then $\displaystyle |f(x,y)| = 0 < \epsilon$ for any $\displaystyle \epsilon > 0$. Note that $\displaystyle |y|=\sqrt{ 0 + (y-0)^2} \leq \sqrt{(x-a)^2+(y-b)^2} < \delta$, so, $\displaystyle |f(x,y)| = \left| y^2 \tan^{-1} \frac{y}{x} \right| \leq \frac{\pi}{2}y^2 < \frac{\pi}{2}\delta^2$, and this quantity can be made sufficiently small. A similar argument applies when $\displaystyle b\not = 0$. And so $\displaystyle f$ is continous at any point.