If y 2 + log e cos 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 $y=0,1$

$2y·y\text{'}+\frac{1}{{\cos }^{2}x}·2\cos x\left(-\sin x\right)=y\text{'}$

$\Rightarrow 2y·y\text{'}-2\tan x=y\text{'}\dots \dots \left(1\right)$

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

Differentiating $\left(1\right)$, $2y·y\text{'}\text{'}+2{\left(y\text{'}\right)}^2-2se{c}^{2}x=y\text{'}\text{'}$,

$y^{\text{'}\text{'}}\left(0\right)=-2$ for $y=0$

$y^{\text{'}\text{'}}\left(0\right)=2$, for $y=1$

$\therefore \left|y\text{'}\text{'}\left(0\right)\right|=2$