# Directional Derivatives of $\sqrt{|xy|}$

Let $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y):= \sqrt{|xy|}$. In order to calculate the directional derivative in the direction $u = (a,b)\in\mathbb{R}^2$ I've made the following inequality (considering $t\neq 0$, $x\neq 0$ and $y\neq 0$)

$$\frac{\sqrt{|(x+ta)(y+tb)|}-\sqrt{|xy|}}{t} = \frac{|(x+ta)(y+tb)| - |xy|}{t[\sqrt{|(x+ta)(y+tb)|}+\sqrt{|xy|}]} =$$

$$= \frac{|xy + tbx + tay + t^2 ab| - |xy|}{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]} \leq \frac{\big ||xy + tbx + tay + t^2 ab| - |xy|\big|}{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]}\leq\frac{| tbx + tay + t^2 ab| }{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]} = \frac{|t|| bx + ay + tab| }{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]} \leq \frac{| bx + ay + tab| }{\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}} \to \frac{|ax + by|}{2\sqrt{|xy|}}\text{ as } t\to 0$$

It's right that $$\operatorname{D}_{u}f(x,y) = \frac{|ax + by|}{2\sqrt{|xy|}}?\;\;\;\;\;\; (*)$$ And if it is, what is the other inequality I can use in to prove (*) by squeezing it?

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What you can do is looking at $$\left| \frac{\sqrt{|(x+ta)(y+tb)|}-\sqrt{|xy|}}{t} - \frac{|ax+by|}{2\sqrt{|xy|}}\right|$$ instead of trying to find a lower limit. You can then use one inequality to prove that it converges to zero. – vanna Oct 6 '12 at 19:21
$D_{u} f(x,y)= \nabla f . u$ – Mykolas Oct 6 '12 at 19:34
@Mykolas, that is not always true – Paulo Henrique Oct 6 '12 at 19:51
@Fëanor when it wouldn't be? – Mykolas Oct 6 '12 at 19:53
@Mykolas, for instance let $u = (1/ \sqrt{2}, 1/ \sqrt{2}) and$f(x,y) = frac{x^2 y}{x^2 + y^2}$if$(x,y)\neq (0,0)$and$f(0,0) = 0\$ at the origin. – Paulo Henrique Oct 6 '12 at 20:00