Answer is (b) yy'' + (y')2 + 1 = 0
Explanation:
The given expression is x2 + y2 = 1
Differentiating w.r.t. x, we get
2x + 2y(dy/dx) = 0
x + y(dy/dx) = 0
Again differentiating, w.r.t. x, we get
1 + (dy/dx)2 + y(d2y/dx2) = 0
1 + y'2 + y.y'' = 0
y.y'' + (y')2 + 1 = 0