If `y + sin y = cos x,` then `(dy)/(dx)   at   ((pi)/(2), (pi)/(2))` is :

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find \(\frac{dy}{dx}\) at the point \(\left(\frac{\pi}{2}, \frac{\pi}{2}\right)\) for the equation \(y + \sin y = \cos x\), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate both sides of the equation**:
Given the equation:
\[ y + \sin y = \cos x \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(y + \sin y) = \frac{d}{dx}(\cos x) \]

2. **Apply the chain rule**:
Using the chain rule on the left side:
\[ \frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = -\sin x \]
Here, \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\) and \(\cos y\) is the derivative of \(\sin y\) with respect to \(y\) multiplied by \(\frac{dy}{dx}\).

3. **Factor out \(\frac{dy}{dx}\)**:
Combine the terms involving \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(1 + \cos y) = -\sin x \]

4. **Solve for \(\frac{dy}{dx}\)**:
Rearranging gives:
\[ \frac{dy}{dx} = \frac{-\sin x}{1 + \cos y} \]

5. **Substitute the values \(x = \frac{\pi}{2}\) and \(y = \frac{\pi}{2}\)**:
Now, we need to evaluate \(\frac{dy}{dx}\) at the point \(\left(\frac{\pi}{2}, \frac{\pi}{2}\right)\):
\[ \frac{dy}{dx} = \frac{-\sin\left(\frac{\pi}{2}\right)}{1 + \cos\left(\frac{\pi}{2}\right)} \]

6. **Calculate \(\sin\) and \(\cos\)**:
We know:
\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) = 0 \]
Therefore:
\[ \frac{dy}{dx} = \frac{-1}{1 + 0} = -1 \]

### Final Answer:
Thus, the value of \(\frac{dy}{dx}\) at the point \(\left(\frac{\pi}{2}, \frac{\pi}{2}\right)\) is:
\[ \frac{dy}{dx} = -1 \]