I know that there is a proof here showing that such a function $f(x,y) = \sqrt{|x|+|y|}$ is not differentiable at $x \neq 0$ or $y \neq 0$, but how do I show that $f(x,y)$ is differentiable at points such that both $x \neq 0$ and $y \neq 0$?
I can use the chain rule to take the partials, but how do I know that this derivative exists at a point that does not have zero in its coordinates? Specifically, the chain rule yields the first partial as $\frac{1}{2\sqrt{|x|+|y|}} $. Then how do I show that
$$\lim_{t\to0} \frac{1}{t}\left[f(x + te_j,y) - f(x,y) - \frac{1}{2\sqrt{|x|+|y|}} \right] = 0$$ when $x \neq 0$ and $y \neq 0$?
The set in the x,y plane of points with x$\neq$0 and y$\neq$0 is the union of 4 quadrants ;namely x>0 and y>0 ;x>0 and y<0 ;x<0 and y>0 ; x<0 and y<0 . In each quadrant the function z=g(x,y)= |x|+|y| is positive valued and continuously differentiable . The function h(z)=$\sqrt{z}$ is continuously differentiable on the positive reals . Thus the composition f(x,y)= h(g(x,y)) is continuosly differentiable .By continuously differentiable I mean that its partial derivatives are continuous . It follows by a theorem of advanced calculus that f is differentiable .In fact an induction argument can be used to show f is infinitely differentiable .