Question

let f(x,y,z) = \[\frac{xyz }{ x^{4} + y ^{4} + z ^{4}}\] and f(0,0,0) = 0. Show that \[D_{x}f\], \[D_{y}f\] and \[D_{z}f\] all exist everywhere. Find the points for which Df exists.

Answer 1

@perl

Answer 2

I would help but I have no idea

Answer 3

Well x, y, and z are all the symmetrical, so all you have to do is show it for one variable, and that should imply that the rest of the problem is true automatically.

So where are you stuck? Show me how far you can get on taking the partial derivative of this with respect to x.

Answer 4

respect to x : \[\frac{ yz(-3x ^{4} + y ^{4} +z ^{4} )}{ (x ^{4} + y ^{4} + z ^{4} )^{2}}\]

Answer 5

That's almost right but you left a couple terms out of the top. Remember the quotient rule is: \[\large \left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\] For comparison what you did was left out two terms on the top like this: \[\large \left( \frac{f(x)}{g(x)} \right)' \ne \frac{f'(x)g'(x)}{[g(x)]^2}\]

Answer 6

hmmm i think you misslooked. if you combine the terms on top you will get what i got.

Answer 7

Can you show me your work?

Answer 8

take the derivative is no problem to me . question i have is that how to show that all three exists everywhere?

Answer 9

Yeah, once you have the derivative, you have a function that exists for all values of x,y, and z as long as the domain is not restricted, like for instance because there's a zero in the denominator.

Answer 10

so x^4 + y^4 + z^4 = 0 iff x=y=z=0, and f(0,0,0) = 0, meaning it has no restriction. thus it exists everywhere?

Answer 11

Well you're given that f(0,0,0)=0 we don't really have enough to say that df/dx(0,0,0)=0 I think yet.

Answer 12

$$
\Large D_x= \frac{-y~z(3~x^4-y^4-z^4)}{(x^4+y^4+z^4)^2}

$$

Answer 13

i got the derivative part covered

Answer 14

ok :)

Answer 15

like i said above. how can i say that it exists everywhere?

Answer 16

now you see that Dx exists everywhere, since when f(0,0,0) = 0
you have no points undefined

Answer 17

i.e. there is no 'division by zero'

Answer 18

there is no reason to restrict x,y,z from being all reals

Answer 19

so i can say the same to y, z as well

Answer 20

yes , Dx, Dy, Dz are defined for all x,y,z . The only problem is the division by zero, but that is taken care by defining f(0,0,0) as zero

Answer 21

cool. got confused a little bit but now everything is clear. thx