If x^2 + y^2 = 1 then ( y' = dy dx ,y'' = d^2 y d x^2 ) - Vidyadip

MCQ

If ${x^2} + {y^2} = 1$ then $\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)$

A
$yy'' - 2{(y')^2} + 1 = 0$
✓
$yy'' + {(y')^2} + 1 = 0$
C
$yy'' - {(y')^2} - 1 = 0$
D
$yy'' + 2{(y')^2} + 1 = 0$

✓

Answer

Correct option: B.

$yy'' + {(y')^2} + 1 = 0$

b
(b) Differentiating $w.r.t$. $x,$ $2x + 2yy' = 0$ or $x + yy' = 0$

Differentiating again $w.r.t$. $x,$ $1 + {y'^2} + yy'' = 0$.

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