If sin ` (x + y) = log (x+y) `, then ` dy //dx` =

Options

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

Answer

To find \( \frac{dy}{dx} \) given the equation \( \sin(x + y) = \log(x + y) \), we will differentiate both sides with respect to \( x \).

### Step-by-Step Solution:

1. **Differentiate both sides**:
\[ \frac{d}{dx}[\sin(x + y)] = \frac{d}{dx}[\log(x + y)] \]

2. **Apply the chain rule on the left side**:
- The derivative of \( \sin(u) \) is \( \cos(u) \cdot \frac{du}{dx} \), where \( u = x + y \).
- Thus, we have:
\[ \cos(x + y) \cdot \frac{d}{dx}(x + y) = \cos(x + y) \cdot (1 + \frac{dy}{dx}) \]

3. **Differentiate the right side**:
- The derivative of \( \log(v) \) is \( \frac{1}{v} \cdot \frac{dv}{dx} \), where \( v = x + y \).
- Thus, we have:
\[ \frac{1}{x + y} \cdot \frac{d}{dx}(x + y) = \frac{1}{x + y} \cdot (1 + \frac{dy}{dx}) \]

4. **Set the derivatives equal**:
\[ \cos(x + y) \cdot (1 + \frac{dy}{dx}) = \frac{1}{x + y} \cdot (1 + \frac{dy}{dx}) \]

5. **Rearrange the equation**:
- Move all terms involving \( \frac{dy}{dx} \) to one side:
\[ \cos(x + y) \cdot (1 + \frac{dy}{dx}) - \frac{1}{x + y} \cdot (1 + \frac{dy}{dx}) = 0 \]

6. **Factor out \( (1 + \frac{dy}{dx}) \)**:
\[ (1 + \frac{dy}{dx}) \left( \cos(x + y) - \frac{1}{x + y} \right) = 0 \]

7. **Set each factor to zero**:
- From \( 1 + \frac{dy}{dx} = 0 \):
\[ \frac{dy}{dx} = -1 \]
- The other factor \( \cos(x + y) - \frac{1}{x + y} = 0 \) does not give us a value for \( \frac{dy}{dx} \).

### Final Answer:
\[ \frac{dy}{dx} = -1 \]