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!

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

the algebra is a big fat mess

Answer 6

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

Answer 7

Satellite has the correct answers

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

brb