Let y=1n(1+cos⁡x)2.Then the value of d2ydx2+2ey/2 equals

y=2ln⁡(1+cos⁡x)dydx=−2sin⁡x1+cos⁡x⇒d2ydx2=−2(1+cos⁡x) Now, 2e−y/2=2.e−ln⁡(1+cos⁡x)22=2(1+cos⁡x)∴d2ydx2+2ey/2=0

Let y=1n(1+cos⁡x)2.Then the value of d2ydx2+2ey/2 equals

y=2ln⁡(1+cos⁡x)dydx=−2sin⁡x1+cos⁡x⇒d2ydx2=−2(1+cos⁡x) Now, 2e−y/2=2.e−ln⁡(1+cos⁡x)22=2(1+cos⁡x)∴d2ydx2+2ey/2=0

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Let $y = 1 n (1 + \cos x)^{2} .$ Then the value of $\frac{d^{2} y}{d x^{2}} + \frac{2}{e^{y / 2}}$ equals

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$\frac{2}{1 + \cos x}$

$\frac{4}{1 + \cos x}$

$\frac{- 4}{(1 + \cos x)^{2}}$

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$y = 2 \ln (1 + \cos x)$

$\frac{d y}{d x} = \frac{- 2 \sin x}{1 + \cos x} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- 2}{(1 + \cos x)} Now,$

$2 e^{- y / 2} = 2 . e^{- \frac{\ln (1 + \cos x)^{2}}{2}} = \frac{2}{(1 + \cos x)} ∴ \frac{d^{2} y}{d x^{2}} + \frac{2}{e^{y / 2}} = 0$

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Let y=1n(1+cos⁡x)2.Then the value of d2ydx2+2ey/2 equals

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Let $y = 1 n (1 + \cos x)^{2} .$ Then the value of $\frac{d^{2} y}{d x^{2}} + \frac{2}{e^{y / 2}}$ equals