can someone show me how to take the partial derivative of tan^-1[(2xy)/(x^2-y^2)] with respect to X and Y. I need to prove that it satisfies the Laplace equation. I understand the partial of X and the partial of y must equal zero. I have been working on the partials for hours and it is not working out. can someone please help!

Question

can someone show me how to take the partial derivative of tan^-1[(2xy)/(x^2-y^2)] with respect to X and Y. I need to prove that it satisfies the Laplace equation.  I understand the partial of X and the partial of y must equal zero. I have been working on the partials for hours and it is not working out.  can someone please help!

anonymous · Answer

good grief what algebra.  let me get pencil and paper and see if i can get the partials

anonymous · Answer

thank you... 4 pages of work, what a mess

anonymous · Answer

did you get 
$$d/dx = -\frac{2y}{x^2+y^2}$$

anonymous · Answer

and other partial is similar
$$\frac{2x}{x^2+y^2}$$

anonymous · Answer

the algebra is a big fat mess

anonymous · Answer

d/dx=[(-2y(x^2+2y)]/(x^2+y^2)^2 ... not sure if it correct

zarkon · Answer

Satellite has the correct answers

anonymous · Answer

hard to write all algebra.  you have to use 
$$\frac{1}{(\frac{2xy}{x^2-y^2})^2+1}\times \frac{\delta}{\delta x}\frac{2xy}{x^2-y^2}$$

anonymous · Answer

& then do the same for y. Answers should add up to 0.

anonymous · Answer

second term there is
$$-\frac{2y(x^2+y^2)}{(x^2-y^2)^2}$$

anonymous · Answer

i am sorry i cannot be more help. it is almost impossible to write this out. fairly certain of the answer though, so maybe you can work towards it

anonymous · Answer

and other partial is similar.  first term is the same, and second term is 
$$\frac{2x(x^2+y^2)}{(x^2-y^2)^2}$$

anonymous · Answer

brb