Partial Derivatives

[KLS2111]

Hello,
The problem is F(x,y)= y/(x^2+y^2) i - x/( x^2+y^2) j

C1 is and c2 are the semicircular paths from (1,0) to (-1,0) given by
C1 x=cos t y= sin t
C2 x=cost t y=-sint

It says show that the partial with respect to y (y/(x^2+y^2)) = the partial with respect to x (-x/(x^2+y^2))

I think I am doing something wrong because for the partial with repect to y I ended up with:
y^2 (x^2+y^2)^-3/2 + (x^2+y^2)^-1/2
and with respect to x
-x^2 (x^2+y^2)^-3/2 + (x^2+y^2)^-1/2

These are not equal, so clearly I am doing something wrong.

 

[Deleted member 4993]

Hello,
The problem is F(x,y)= y/(x^2+y^2) i - x/( x^2+y^2) j

C1 is and c2 are the semicircular paths from (1,0) to (-1,0) given by
C1 x=cos t y= sin t
C2 x=cost t y=-sint

It says show that the partial with respect to y (y/(x^2+y^2)) = the partial with respect to x (-x/(x^2+y^2))

I think I am doing something wrong because for the partial with repect to y I ended up with:
y^2 (x^2+y^2)^-3/2 + (x^2+y^2)^-1/2
and with respect to x
-x^2 (x^2+y^2)^-3/2 + (x^2+y^2)^-1/2

These are not equal, so clearly I am doing something wrong.

\(\displaystyle \frac{d}{dy}\left [\frac{y}{x^2 + y^2}\right ] \ = \ \frac{1 \cdot (x^2+y^2) - 2y\cdot y}{(x^2+y^2)^2} \ = \ \frac{x^2 \ - \ y^2}{(x^2+y^2)^2}\)

\(\displaystyle \frac{d}{dx}\left [\frac{x}{x^2 + y^2}\right ] \ = \ \frac{1 \cdot (x^2+y^2) - 2x\cdot x}{(x^2+y^2)^2} \ = \ \frac{-x^2 \ + \ y^2}{(x^2+y^2)^2}\)