Let `y=1n(1+cosx)^2`. Then the value of `(d^2y)/(dx^2)+2/(e^(y/2))` equal

Options

A. `0`
B. `2/(1+cosx)`
C. `4/(1+cosx)`
D. `(-4)/((1+cos"x")^2)`

Answer

To solve the problem, we need to find the value of \(\frac{d^2y}{dx^2} + \frac{2}{e^{y/2}}\) where \(y = \ln((1 + \cos x)^2)\).

### Step 1: Differentiate \(y\) with respect to \(x\)

Given:
\[ y = \ln((1 + \cos x)^2) \]
Using the property of logarithms, we can simplify:
\[ y = 2 \ln(1 + \cos x) \]

Now, differentiate \(y\):
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + \cos x} \cdot (-\sin x) = \frac{-2 \sin x}{1 + \cos x} \]

### Step 2: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)

Now, we need to differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-2 \sin x}{1 + \cos x}\right) \]

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(1 + \cos x)(-2 \cos x) - (-2 \sin x)(-\sin x)}{(1 + \cos x)^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{-2 \cos x (1 + \cos x) + 2 \sin^2 x}{(1 + \cos x)^2} \]
Using the identity \(\sin^2 x + \cos^2 x = 1\):
\[ \sin^2 x = 1 - \cos^2 x \]
Thus:
\[ \frac{d^2y}{dx^2} = \frac{-2 \cos x - 2 \cos^2 x + 2 - 2 \cos^2 x}{(1 + \cos x)^2} = \frac{2 - 2 \cos x - 4 \cos^2 x}{(1 + \cos x)^2} \]

### Step 3: Calculate \(e^{y/2}\)

Now we calculate \(e^{y/2}\):
\[ y = 2 \ln(1 + \cos x) \implies \frac{y}{2} = \ln(1 + \cos x) \]
Thus:
\[ e^{y/2} = 1 + \cos x \]

### Step 4: Substitute into the expression

Now we substitute back into the expression:
\[ \frac{d^2y}{dx^2} + \frac{2}{e^{y/2}} = \frac{2 - 2 \cos x - 4 \cos^2 x}{(1 + \cos x)^2} + \frac{2}{1 + \cos x} \]

To combine these, we need a common denominator:
\[ \frac{2 - 2 \cos x - 4 \cos^2 x + 2(1 + \cos x)}{(1 + \cos x)^2} = \frac{2 - 2 \cos x - 4 \cos^2 x + 2 + 2 \cos x}{(1 + \cos x)^2} \]
This simplifies to:
\[ \frac{4 - 4 \cos^2 x}{(1 + \cos x)^2} = \frac{4(1 - \cos^2 x)}{(1 + \cos x)^2} = \frac{4 \sin^2 x}{(1 + \cos x)^2} \]

### Step 5: Final Result

Since \(4 \sin^2 x\) is always non-negative, we conclude that:
\[ \frac{d^2y}{dx^2} + \frac{2}{e^{y/2}} = 0 \]

### Final Answer:
The value of \(\frac{d^2y}{dx^2} + \frac{2}{e^{y/2}}\) is \(0\).

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