Question

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

Answer 1

Given,
$y=\ln (1+\cos (x))^{2}$
Differentiating the above equation with respect to $x$ we get,
$\frac{d y}{d x}=\frac{d}{d x}\left\{\ln (1+\cos (x))^{2}\right\}$
By using differentiation of composite functions we get,
$\frac{d y}{d x}=\frac{1}{\left(1+\cos (x)^{2}\right.} \frac{d}{d x}(1+\cos (x))^{2} $
$\because \frac{d}{d t}\{\ln t\}=\frac{1}{t} $
$\frac{d y}{d x}=\frac{2(1+\cos x) d}{(1+\cos x)^{2} d x}(\cos x) $
$\frac{d y}{d x}=\frac{-2 \sin x}{(1+\cos x)}$
Differentiating the above equation with respect to $x$,
$\frac{d}{d x}\left\{\frac{d y}{d x}\right\}=\frac{d}{d x}\left\{\frac{-2 \sin x}{(1+\cos x)}\right\}$
By using the quotient rule of differentiation we get,
$\frac{d^{2} y}{d x^{2}}=-2\left[\frac{(1+\cos x) \frac{d}{d x}(\sin (x))-\sin (x) \frac{d}{d x}(1+\cos (x))}{\left(1+\cos (x)^{2}\right.}\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\} $
$\therefore \frac{d^{2} y}{d x^{2}}+\frac{2}{e^{\frac{1}{2}}}=\frac{-2}{1+\cos (x)}+\frac{2}{1+\cos (x)}$
$\frac{d^{2} y}{d x^{2}}+\frac{2}{e^{\frac{y}{2}}}=0$