If y ^ 2 + _ e ( ^ 2 x) = y, x ∈ (- (π)/(2), (π)/(2)) then :

Putting, x = 0 in ${y}^{2} + \log_{e}\left(\cos^{2}x\right) = y$ we get $y = 0, 1$

$2y \cdot y ' + \dfrac{1}{\cos^{2}x} \cdot 2\cos x\left(- \sin x\right) = y '$

$⇒ 2y \cdot y ' - 2\tan x = y ' \ldots \ldots \left(1\right)$

$y ' \left(0\right) = 0$ for $y = 0 \text{ and } y = 1$.

Differentiating \left(1\right)$,$2y \cdot y ' ' + 2{\left(y '\right)}^{2} - 2se{c}^{2}x = y ' '$,</p><p>${y}^{' '}\left(0\right) = - 2$for$y = 0$</p><p>${y}^{' '}\left(0\right) = 2$, for$y = 1$</p><p>$∴ \left|y ' ' \left(0\right)\right| = 2