Work Check On a Complicated Partial Derivative

Liquid7800
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Homework Statement



Hello, I was given this complicated partial derivative to work out:

z = √( 1-( (x+y)/(xy) )2 ) + arcsin (x+y)/(x-y)

find:

fx and fy

and this is my final answer taking the partial derivative with respect to 'x' only right now...which is rather nasty looking (sorry LaTex would not render out for me) :

z= -[ (x+y)/xy * (-x-y/x2y + 1/xy) ] / [ √( 1 - (x+y / xy)2) ] + [ -x-y/(x-y)2 + 1/x+y ] / [ √( 1 - (x+y)2 / (x-y)2 ) ]

The complicated part (or peculiar) is that I really wanted to make sure my algebra etc. was correct. I tried using a CAS to check my work...but I noticed that depending how you entered the problem I got different answers ...
(somehow depends on if you enter the full problem in...or by how the denominator 'xy' is entered),
...so I didnt feel confident just using a CAS to check my work.

I would be very grateful if someone were to assist me.

Let me know if something is unclear... thanks again
 
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I think you just have one sign error - your 1/(x+y) should be 1/(x-y). The whole thing should be
\frac{\frac{(x+y)^2}{x^3 y^2}-\frac{(x+y)}{x^2 y^2}}{\sqrt{1-\frac{(x+y)^2}{x^2<br /> y^2}}}+\frac{\frac{1}{x-y}-\frac{x+y}{(x-y)^2}}{\sqrt{1-\frac<br /> {(x+y)^2}{(x-y)^2}}}
 
Thanks very much for the reply, I hadnt simplified the whole thing yet...as I wanted to see if I was on the right track...and thanks for pointing that error out...that was a typo on my part.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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