Register to reply 
Sqrt(xy) derivatives at (0,0) 
Share this thread: 
#1
Apr2312, 10:09 PM

P: 757

[tex] f(x,y)=\sqrt{xy} [/tex]
Do the partial derivatives of f exist at x=0, y=0? 


#2
Apr2312, 10:23 PM

P: 606

Let's see: [itex]\displaystyle{f_x(0,0):=\lim_{x\to 0}\frac{f(x,0)f(0,0)}{x}=\lim_{x\to 0}\frac{0}{x}}=0[/itex] , and of course the same is true for [itex]f_y(0,0)[/itex] DonAntonio 


#3
Apr2312, 10:27 PM

P: 757

Mathematica could not do it.
And if I write the function as ((xy)^2)^(1/4) then the partial derivative is [tex]\frac{x y^2}{2 \left(x^2 y^2\right)^{3/4}}[/tex] which is undefined at (0,0). It means it is not differentiable at 0,0 correct? 


#4
Apr2312, 10:36 PM

Mentor
P: 18,231

Sqrt(xy) derivatives at (0,0)



#5
Apr2312, 10:49 PM

P: 606

What's not clear in what I wrote to you? I think it is pretty straightforward, though I can be wrong, of course. Check this. DonAntonio 


#6
Apr2412, 12:52 AM

P: 757

Even though partials exist at (0,0) as DonAntonio showed, the function is still not differentiable. So I'm trying to find out to what extent this is true: "If the "usual way" of taking partials does not yield the correct result at a point, the function is not differentiable at that point". 


#7
Apr2412, 04:51 AM

P: 606

There is only one way to calculate the (partial or not) derivatives wrt the definition. Now, if these derivatives exist AND they're continuous at some point then the function is differentiable at that point. This is precisely what does NOT happen in this case as the partial derivatives aren't cont. at (0,0), and to call something "the usual way" invites confusion and problems when dealing with mathematics, as what is usual for you might not be so for me and the other way around. DonAntonio 


#8
Apr2412, 02:19 PM

P: 898

Just to clarify for the OP: This does not mean that if the partial derivatives are not continuous at a point, then the function is not differentiable there, which is the converse statement. There are many functions whose derivatives exist everywhere, but the derivative function is not continuous. Likewise, there are nondifferentiable functions whose partial derivatives exist everywhere. They are rarely more than pathological curiosities, however, as the derivative in these cases fail to meet the ideal of being a linear approximation to the function.
In summary, the existence or nonexistence of the partial derivatives at a point will not tell you whether the function is differentiable there. If they exist and are continuous, however, then you have a derivative. 


#9
Apr2412, 10:16 PM

P: 757

The simple question I was asking is this: if the partial derivatives at a point DISAGREES with what is obtained by using the partial derivative formulas (the way we learned in highschool with all the differentiation rules etc etc) then does that mean that the function is not differentiable at that point?
In this example the partial derivatives are both 0 at (0,0), but using the usual "formula" I get [tex]\frac{x y^2}{2 \left(x^2 y^2\right)^{3/4}}[/tex] which is undefined at (0,0). So "undefined" and "0" are not the same thing, so there is disagreement, and "coincidentally" the function happens to be nondifferentiable at (0,0). Is this a general result? Will the partial derivative "formulas" always fail at nondifferentiable points? Its really a simple question but I dont know if I'm explaining it very well. 


#10
Apr2512, 12:11 AM

P: 898

When you take the partial derivative of f(x, y) = ((xy)^2)^(1/4) at the point x = 0, y = 0, with respect to x, you applied the chain rule to the singlevariable function f(x) = h(g(x)) where h(g) = g^(1/4) and g(x) = (xy)^2, treating y as a constant. The chain rule states that if f(x) = h(g(x)) and h is differentiable at the point g(0) and g is differentiable at the point x = 0, then f'(0) = h'(g(0))*g'(0). However, h(g) is not differentiable at g(0), so the chain rule is not applicable. You have to use another method to find the partial derivative of f there, such as the definition of the partial derivative, as applied by DonAntonio. 


#11
Apr2512, 08:21 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,491




#12
Apr2512, 08:28 AM

Mentor
P: 18,231

This might be relevant: https://docs.google.com/viewer?a=v&q...M8EzDZzAHQSo5g



Register to reply 
Related Discussions  
Prove that the limit of sin(sqrt(x+1))sin(sqrt(x1)) at infinity doesn't exist  Calculus & Beyond Homework  10  
Prove series is divergent (sqrt(n+1)  sqrt(n))/sqrt(n)  Calculus & Beyond Homework  5  
Iterative square root? sqrt(2+sqrt(2+sqrt(...  General Math  6  
Is there a way to show that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?  General Math  4  
Proof that sqrt(6)sqrt(2)sqrt(3) is irrational  General Math  10 