Find `(dy)/(dx)`, when:
`sin^(2)x+2cosy+xy=0`

Answer

To find \(\frac{dy}{dx}\) for the equation \( \sin^2 x + 2 \cos y + xy = 0 \), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate the equation**:
\[ \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(2 \cos y) + \frac{d}{dx}(xy) = 0 \]

2. **Differentiate \(\sin^2 x\)**:
Using the chain rule:
\[ \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin(2x) \]

3. **Differentiate \(2 \cos y\)**:
Using the chain rule:
\[ \frac{d}{dx}(2 \cos y) = -2 \sin y \frac{dy}{dx} \]

4. **Differentiate \(xy\)**:
Using the product rule:
\[ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y \]

5. **Combine all differentiated parts**:
Putting it all together:
\[ \sin(2x) - 2 \sin y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0 \]

6. **Rearranging the equation**:
Group the terms involving \(\frac{dy}{dx}\):
\[ (-2 \sin y + x) \frac{dy}{dx} = -\sin(2x) - y \]

7. **Solve for \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{-\sin(2x) - y}{-2 \sin y + x} \]
Simplifying, we get:
\[ \frac{dy}{dx} = \frac{\sin(2x) + y}{2 \sin y - x} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\sin(2x) + y}{2 \sin y - x} \]