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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

1 Answer 1

up vote 2 down vote accepted

The function f(x,y) = sqrt(abs(x*y)) is not differentiable at points of either axis. But at other points you can use the usual definition del(f) dot u, mentioned by Mykolas. Things go smoother if you use that, for any real t, abs(t) = ( t^2 )^(1/2). For your function this gives f(x,y) = (x^2*y^2)^(1/4), and the usual differentiation rules can be applied to this.

When I did that I got del(f) = (1/4)(x^2*y^2)^(-3/4)*(2x*y)*(y,x). Note how x and y are reversed here, so that dotting with u=(a,b) gives the extra factor (ay+bx) rather than (ax+by) the OP had mentioned.

We can replace the term (x^2*y^2)^(-3/4) by |xy|^(-3/2), but we can't absorb the x*y part of the other factor 2*x*y into the absolute value, since it's sign depends on the quadrant.

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