# Show that $\sqrt{|x| + |y|}$ is differentiable at nonzero points

by user1691278   Last Updated January 21, 2018 04:20 AM

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

$\lim_{t\to0} \frac{1}{t}\big[f(x + te_j,y) - f(x,y) - \frac{1}{2\sqrt{|x|+|y|}} \big] = 0$ when $x \neq 0$ and $y \neq 0$?

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