# Thread: For a rational/irrational function, determine where f is differentiable

1. ## For a rational/irrational function, determine where f is differentiable

This relates to a previous question I asked, so we can somewhat pick up from there.

For the function $f(x,y)=\left\{\begin{array}{cc}x^2+y^2,&x,y\mbox{ both rational }\\0, & \mbox{ otherwise, }\end{array}\right$
determine at which points $f$ is differentiable.

I already know that $f$ is continuous ONLY at the point $(0,0)$, so that's the only potential location for where $f$ could be differentiable. Showing whether or not it is, however, is the trick.

Would using this trick work?
$u(h,k)=\frac{f(x+h,y+k)-f(x,y)-f_x(x,y)h-f_y(x,y)k}{\sqrt{h^2+k^2}}$ when $(h,k)\ne (0,0)$
I only need to determine whether or not $u(h,k)\rightarrow 0$ as $(h,k)\rightarrow (0,0)$ to prove whether or not the function is differentiable at $(0,0)$.

2. Originally Posted by Runty
$u(h,k)=\frac{f(x+h,y+k)-f(x,y)-f_x(x,y)h-f_y(x,y)k}{\sqrt{h^2+k^2}}$ when $(h,k)\ne (0,0)$
If you define u(h,k) in this way, you need to say what $f_x$ and $f_y$ are. And if you say they are partial derivatives, you need to explain at least why they exist.

In any case, if the derivative (a linear functional $\mathbb{R}^2\to\mathbb{R}$) exists, it is zero, so we have to say whether $\displaystyle\lim_{(h,k)\to(0,0)}\frac{f(h,k)}{\sq rt{h^2+k^2}}=0$. Well, $\displaystyle\frac{f(h,k)}{\sqrt{h^2+k^2}}\le\sqrt {h^2+k^2}$, so yes.