Q. If $x^{2} + y^{2} = 1$, then $\left( y^{'} = \frac{dy}{dx},\ y^{''} = \frac{d^{2}y}{dx^{2}} \right)$
A) $yy^{''} - \left( 2y^{'} \right)^{2} + 1 = 0$
B) $yy^{''} + \left( y^{'} \right)^{2} + 1 = 0$
C) $y^{''} - \left( y^{'} \right)^{2} - 1 = 0$
D) $y^{''} + 2\left( y^{'} \right)^{2} + 1 = 0$
Explanation
(b)
Given, $x^{2} + y^{2} = 1$ ….(i)
On differentiating $w.r.t.x,$ we get
$2x + 2yy^{'} = 0\ \Rightarrow x + yy^{'} = 0$
Again, on differentiating $w.r.t.x,$ we get
$1 + \left( y^{'} \right)^{2} + yy^{''} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.