If y² + log_e (cos²x) = y,
$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$, then :

Given y² + log_e (cos²x) = y .....(1) <br><br>Put x = 0, we get <br><br>y² + log_e (1) = y <br><br>$\Rightarrow$ y² = y <br><br>$\Rightarrow$ y = 0, 1 <br><br>Differentiating (1) we get <br><br>2yy' + ${1 \over {\cos x}}\left( { - \sin x} \right)$ = y' <br><br>$\Rightarrow$ 2yy' - 2tanx = y' ....(2) <br><br>From (2) when x = 0, y = 0 then y'(0) = 0 <br><br>From (2) when x = 0, y = 1 then <br><br>2y' = y' <br><br>$\Rightarrow$ y'(0) = 0 <br><br>Again differentiating (2) we get <br><br>2(y')² + 2yy'' – 2sec²x = y'' <br><br>from (2) when x = 0, y = 0, y’(0) = 0 then <br><br>y”(0) = -2 <br><br>Also from (2) when x = 0, y = 1, y’(0) = 0 then <br><br>y”(0) = 2 <br><br>$\therefore$ |y''(0)| = 2