Question

Let $y=\ln (1+\cos x{)}^2$, then the value of $\frac{d^{2}y}{d{x}^2}+\frac{2}{e^{y/2}}$ equals

• A. $0$
• B. $\frac{2}{1+\cos x}$
• C. $\frac{4}{1+\cos x}$
• D. $\frac{-4}{(1+\cos x{)}^2}$

Answer 1

The correct option is A $0$

Given:
$y=\ln (1+\cos x{)}^2\Rightarrow y=2\ln (1+\cos x)\Rightarrow \frac{dy}{dx}=\frac{-2\sin x}{1+\cos x}\Rightarrow \frac{d^{2}y}{d{x}^2}=-2\left[\frac{(1+\cos x)\cos x-\sin x(-\sin x)}{(1+\cos x{)}^2}\right]\Rightarrow \frac{d^{2}y}{d{x}^2}=-2\left[\frac{1+\cos x}{(1+\cos x{)}^2}\right]\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2}{1+\cos x}$
Now,
$e^{y/2}=e^{\ln (1+\cos x)}=1+\cos x$
So,
$\frac{d^{2}y}{d{x}^2}+\frac{2}{e^{y/2}}=\frac{-2}{1+\cos x}+\frac{2}{1+\cos x}=0$