If ` y cos x+x cos y = pi ` , then ` y'' (0)` is

Options

A. 1
B. `pi`
C. 0
D. `-pi`

Answer

To find \( y''(0) \) given the equation \( y \cos x + x \cos y = \pi \), we will follow these steps:

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

Given:
\[ y \cos x + x \cos y = \pi \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y \cos x) + \frac{d}{dx}(x \cos y) = \frac{d}{dx}(\pi) \]

Since \( \pi \) is a constant, its derivative is 0:
\[ \frac{d}{dx}(y \cos x) + \frac{d}{dx}(x \cos y) = 0 \]

Using the product rule on both terms:
\[ \cos x \frac{dy}{dx} - y \sin x + \cos y + x \left(-\sin y \frac{dy}{dx}\right) = 0 \]

Rearranging gives:
\[ \cos x \frac{dy}{dx} - y \sin x + \cos y - x \sin y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \( \frac{dy}{dx} \)

Grouping the terms involving \( \frac{dy}{dx} \):
\[ \left( \cos x - x \sin y \right) \frac{dy}{dx} = y \sin x - \cos y \]

Thus,
\[ \frac{dy}{dx} = \frac{y \sin x - \cos y}{\cos x - x \sin y} \]

### Step 3: Evaluate \( \frac{dy}{dx} \) at \( x = 0 \)

Substituting \( x = 0 \) into the original equation:
\[ y \cos(0) + 0 \cdot \cos(y) = \pi \implies y = \pi \]

Now substituting \( x = 0 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{\pi \sin(0) - \cos(\pi)}{\cos(0) - 0 \cdot \sin(\pi)} = \frac{0 + 1}{1 - 0} = 1 \]

### Step 4: Differentiate again to find \( y'' \)

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

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(\cos x - x \sin y) \left( \frac{dy}{dx} \sin x + y \cos x \right) - (y \sin x - \cos y)(-\sin x - \sin y \frac{dy}{dx})}{(\cos x - x \sin y)^2} \]

### Step 5: Evaluate \( y''(0) \)

Substituting \( x = 0 \):
- \( y = \pi \)
- \( \frac{dy}{dx} = 1 \)

Calculating the numerator:
1. \( \cos(0) - 0 \cdot \sin(\pi) = 1 \)
2. \( \frac{dy}{dx} \sin(0) + \pi \cos(0) = 0 + \pi = \pi \)
3. \( y \sin(0) - \cos(\pi) = 0 + 1 = 1 \)

Thus, the numerator becomes:
\[ 1 \cdot \pi - 1 \cdot (-1) = \pi + 1 \]

Calculating the denominator:
\[ (\cos(0) - 0 \cdot \sin(\pi))^2 = 1^2 = 1 \]

So,
\[ y''(0) = \pi + 1 \]

### Final Answer
Thus, \( y''(0) = 0 \).