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f(x,y)= (xy)/[{(x^2 + y^2)}^(1/2)] if x^2 + y^2=0
= 0 if x=y=0.

2007-02-19 18:47:13 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

Sorry, one big mistake:
f(x,y)={(xy)/[(x^2+y^2)]^(1/2) if (x^2 + y^2) not=0

2007-02-19 19:20:58 · update #1

5 answers

f(x,y) = (xy) / [{(x^2 + y^2)}^(1/2)] if x^2 + y^2 not =0
= 0 if x=y=0.

For above function, fx and fy exist everywhere except the origin (0,0). Better picture is shown by the partial derivatives. Like
∂f/∂x = y^3/(x^2+y^2)^3/2 and ∂f/∂y = x^3/(x^2+y^2)^3/2

These partial derivatives will exist for all values of (x,y) EXCEPTING that for which (x^2+y^2) =0. These are not continuous everywhere. Therefore first part is proved.

If differentiated at (x.y) then
(dy/dx)x,y = ∂f/∂x * ∂y/∂f = y^3 / x^3

For origin (dy/dx)0,0 = 0 / 0 (indeterminate)

so the function possesses partial derivatives if (x^2+y^2) NOT =0. But it is not differentiable at origin.

2007-02-20 03:08:06 · answer #1 · answered by anil bakshi 7 · 0 0

If I'm reading the second line right, then they explicitly define f(0,0)=0. So we can't say that the function is undefined there. But it might not be continuous at that point.

To see if it's continuous at the origin, see if the limit exists there. Keeping y as a constant and looking at x:
Lim(x goes to 0) of f(x,y) = 0/0, so use l'Hopital's rule where df/dx is taken:
Lim ( y / (1/2)(x²+y²)^(-1/2)(2x) ) =
Lim ( y / (-x)(x²+y²)^(-1/2) ) =
Lim ( - y√(x²+y²) / x ) = -y²/0 = undefined

Taking the limit as y goes to 0 when x is a constant also shows that the limit is undefined. Since it's not continuous either way, it can't be differentiable at the origin.

2007-02-19 19:33:59 · answer #2 · answered by Anonymous · 0 0

Just show that
lim (f((0+σ),0))/σ
lim (f(0-σ),0))/σ
lim (f(0,0+σ))/σ and
lim (f(0,0-σ))/σ
are not all equal as σ -> 0+
(It's called the Cauchy-Riemann condition if you're in the complex plane)

HTH ☺


Doug

Edit: RTFQ charro. It *is* defined (to be 0) if x² + y² = 0
so f *is* continuous in the sense that it contains all of it's own limit points. But the partials are different depending on the direction from which you are *approaching* 0.

2007-02-19 19:06:26 · answer #3 · answered by doug_donaghue 7 · 0 0

the element (a million,2) lies on f(x) = 2x the element (a million,2) does no longer lie on f(x) = 2 The left by-product is 0, the main appropriate by-product is two.for this reason the function has a leap discontinuity on the element (2,a million). This piecewise function is differentiable in any respect values of x different than a million.

2016-11-24 19:39:13 · answer #4 · answered by ? 4 · 0 0

cause is undefined at 0, 1/0!!done

2007-02-19 18:54:36 · answer #5 · answered by Anonymous · 0 0

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