Question

Verify Clairaut's theorem for:
\[xe^{-x^2y^2}\]

Answer 1

and hows this a theorem?

Answer 2

I think you need to understand partial derivatives before you can understand and verify clairaut's theorem, @shamil98
and I know you're not in Multivariable Calculus.

Answer 3

I understand partial derivatives, i've been doing calc 3 -.-

Answer 4

And the theorem is:
\[f_{xy} (a,b) = f_{yx} (a,b)\]

Answer 5

Then this should be very easy. You find \(u_x,~u_y\), and then the second fixed partials, as you stated above.

Answer 6

Yes, it should hold true that \(\sf \color{red}{f_{xy}=f_{yx}}\)

Answer 7

The first derivatives are:
\[f_x (x,y) =e^{-x^2y^2} - 2x^2y^2e^{-x^2y^2}\]
\[f_y (x,y) = -2yx^3e^{-x^2y^2}\]

Answer 8

Yes, now find \(f_{xy}\) and \(f_{yx}\).
You'll see that they are in fact both equal.

Answer 9

uh
\[f_{xy} (x,y) = -2yx^2e^{-x^2y^2} - 4x^2ye^{-x^2y^2} + 4x^4y^3e^{-x^2y^2}\]
\[f_{yx} (x,y) = -6x^2ye^{-x^2y^2}+4y^3x^4e^{-x^2y^2}\]

Answer 10

the first one simplifies to
\[f_{xy} (x,y) = -6x^2ye^{-x^2y^2} + 4x^2y^3e^{-x^2y^2}\]

Answer 11

+4x^4****

Answer 12

Yessss........

Answer 13

kk