If cos−1(x2−y2x2+y2)=a then dydx=
Consider the given equation,
cos−1(x2−y2x2−y2)=a
x2−y2x2−y2=cosa
Differentiate with respect to x
ddx(x2−y2x2+y2)=ddxcosa=0
(x2+y2)ddx(x2−y2)−(x2−y2)ddx(x2+y2)(x2+y2)2=0
(x2+y2)(2x−2ydydx)−(x2−y2)(2x+2ydydx)=0
(x2+y2)(2x−2ydydx)=(x2−y2)(2x+2ydydx)
2x3−2x2ydydx+2xy2−2y3dydx=2x3+2x2ydydx−2xy2−2y3dydx
−2x2ydydx+2xy2=2x2ydydx−2xy2
4x2ydydx=4xy2
dydx=yx
Hence, this is the answer.