# Check for differentiability, continuitity and existence of partial derivatives/directional derivative

The exercise is :

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be as follows : $$f(x,y) = u(x,y)\cdot(x^2+y^2)$$ where $$u(x,y) = \begin{cases} 1 &\text{for } x=0 \vee y=0 \\0 &\text{for other cases }\end{cases}$$ Check for continuity of $f$ in $0$, existence of partial derivatives of $f$ in $0$, existence of directional derivatives of $f$ in $0$ and for differentiability of $f$ in $0$.

How should I approach each step of this? I know that to check continuity I could check the iterated limits (which I did, both 0) but does does this prove that its continuous? For the partial derivatives should I do it from the definition or just count each one and check if its $\in\mathbb{R}$? Same goes for the directional derivative and differentiability. Or should I perhaps check differentiability first since when its differentiable its also continuous.

Thanks for your tips!

• I have a small problem with your notation; $u$ is a function of $x$, but in its definition you involve also the $y$-coordinate. I do not fully understand it... – Avitus Jun 18 '13 at 9:17
• Yes, sorry, $u$ should be a function of $x$ and $y$ – darenn Jun 18 '13 at 9:36

## 1 Answer

Some tipps. By definition, $f=x^2$ if $(x,y)=(x,0)$, $f=y^2$ if $(x,y)=(0,y)$ and $f=0$ otherwise. Then, you should check that along the $x$-axis and $y$-axis the fucntion approaches $0$, when $x\rightarrow 0$ and $y\rightarrow 0$, independently on the direction along the axis. It should be clear that the function is now continuous at $(0,0)$. If not, then use polar coordinates and check that the limit is independent on the way you choose to approach $(0,0)$. If you think about it for a while, in polar coordinates the function $f$ depends only on the radius $r=\sqrt{x^2+y^2}$.

On directional derivatives; for any $v=(v_1,v_2)$ s.t. $v_1\neq 0 \cap v_2\neq 0$, the directional derivative at $(0,0)$ is s.t. $D_v(f)(0,0)=0$. (Why? Because for any $t\neq 0$ we have $f(tv_1,tv_2)=0$, as well as $f(0,0)$). We are left with the directional derivatives in $(0,0)$ along the $x$-axis and $y$-axis. They are known as partial derivatives of $f$ at $(0,0)$.

For the $x$-axis choose a vector $v=(v_1,0)$ with $v_1\neq 0$ and compute $D_v(f)(0,0)$; similarly for the other case. All you need is to write explicitly

$$f(tv_1,0)$$

and

$$f(0,tv_2).$$

For differentiability you can use definitions and compute the gradient of $f$ at $(0,0)$.