Question

Let $y=ln{\left(1+cosx\right)}^2$ , then the value of $\frac{d^{2}y}{d{x}^2}+\frac{2}{e^{y/2}}$ is equal to:

Answer 1

Answer: 0

Given,
$y=\ln (1+\cos (x){)}^2$
Differentiating the above equation with respect to $x$ we get,
$\frac{dy}{dx}=\frac{d}{dx}\left\{\ln (1+\cos (x){)}^2\right\}$
By using differentiation of composite functions we get,
$\frac{dy}{dx}=\frac{1}{(1+\cos (x{)}^2}\frac{d}{dx}(1+\cos (x){)}^2$
$\because \frac{d}{dt}\{\ln t\}=\frac{1}{t}$
$\frac{dy}{dx}=\frac{2(1+\cos x)d}{(1+\cos x{)}^{2}dx}(\cos x)$
$\frac{dy}{dx}=\frac{-2\sin x}{(1+\cos x)}$
Differentiating the above equation with respect to $x$,
$\frac{d}{dx}\left\{\frac{dy}{dx}\right\}=\frac{d}{dx}\left\{\frac{-2\sin x}{(1+\cos x)}\right\}$
By using the quotient rule of differentiation we get,
$\frac{d^{2}y}{d{x}^2}=-2\left[\frac{(1+\cos x)\frac{d}{dx}(\sin (x))-\sin (x)\frac{d}{dx}(1+\cos (x))}{(1+\cos (x{)}^2}\right]$
$\frac{d^{2}y}{d{x}^2}=-2\left[\frac{(1+\cos (x))\cos (x)-\sin (x)(-\sin (x))}{(1+\cos (x){)}^2}\right]$
$\frac{d^{2}y}{d{x}^2}=-2\left[\frac{\cos (x)+{\cos }^2(x)+{\sin }^2(x)}{(1+\cos (x){)}^2}\right]$
As we know, ${\cos }^2(\theta )+{\sin }^2(\theta )=1$.
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{-2(1+\cos (x))}{(1+\cos (x){)}^2}\text{ or }\frac{d^{2}y}{d{x}^2}=\frac{-2}{(1+\cos (x))}$
Now we have to find the value of $\frac{d^{2}y}{d{x}^2}+\frac{2}{e^{\frac{y}{2}}}$.
Let us find the value of $e^{\frac{\nu }{2}}$.
$e^{\frac{y}{2}}=e^{\left\{\frac{\ln (1+\cos (x){)}^2}{2}\right\}}$
$e^{\frac{y}{2}}=e^{\left\{\frac{2\ln (1+\cos (x))}{2}\right\}}\left\{\because \ln (a{)}^b=b\ln (a)\right\}$
$e^{\frac{y}{2}}=e^{\ln (1+\cos (x))}$
$e^{\frac{y}{2}}=1+\cos (x)\left\{\because a^{{\log }_a(b)}=b\right\}$
$\therefore \frac{d^{2}y}{d{x}^2}+\frac{2}{e^{\frac{1}{2}}}=\frac{-2}{1+\cos (x)}+\frac{2}{1+\cos (x)}$
$\frac{d^{2}y}{d{x}^2}+\frac{2}{e^{\frac{y}{2}}}=0$