Answer:
Option D
Explanation:
We have,
y=√x+√y+√x+√y+..........∞
y=√x+√y+y⇒y2=x+√2y
⇒ y2−x=√2y⇒(y2−x)2=2y
⇒ y4−2xy2+x2=2y
⇒ y4−2xy2−2y+x2=0
On differentiating w.r.t to x , we get
4y3dydx−2y2−4yxdydx−2dydx+2x=0
dydx(4y3−4xy−2)=−2x+2y2
⇒ dydx=y2−x2y3−2xy−1