If cos−1(x2−y2x2+y2)=k (a constant), then dydx=
cos−1x2−y2x2+y2=k
x2−y2x2+y2=cos(k)
On differentiating both sides, we get
(2x−2ydydx)(x2+y2)−(2x+2ydydx)(x2−y2)(x2+y2)2=0
(2x3+2xy2−2x2ydydx−2y3dydx)−(2x3−2xy2+2x2ydydx−2y3dydx)=0
4xy2−4yx2dydx=0
dydx=yx