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Q.
Let $y =\ln (1+\cos x )^2$ then the value of $\frac{ d ^2 y }{ dx ^2}+\frac{2}{ e ^{ y / 2}}$ equals
Continuity and Differentiability
Solution:
$ y=2 \ln (1+\cos x)$
$y _1=\frac{-2 \sin x }{1+\cos x } $
$y _2=-2\left[\frac{(1+\cos x ) \cos x -\sin x (-\sin x )}{(1+\cos x )^2}\right]=-2\left[\frac{\cos x +1}{(1+\cos x )^2}\right]=\frac{-2}{(1+\cos x )}$
$\therefore 2 e ^{- y / 2}=2 \cdot e ^{-\frac{\ln (1+\cos x )^2}{2}}=\frac{2}{(1+\cos x )} $
$\therefore y _2+\frac{2}{ e ^{ y / 2}}=0$