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

Options

A. `tan(x+y)`
B. `log(x+y)`
C. `-1`
D. 1

Answer

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

### Step-by-Step Solution:

1. **Differentiate both sides**:
We start with the equation:
\[ \sin(x+y) = \log(x+y) \]
Now, we differentiate both sides with respect to \( x \).

2. **Differentiate the left side**:
Using the chain rule, the derivative of \( \sin(x+y) \) is:
\[ \frac{d}{dx}[\sin(x+y)] = \cos(x+y) \cdot \left(1 + \frac{dy}{dx}\right) \]

3. **Differentiate the right side**:
The derivative of \( \log(x+y) \) is:
\[ \frac{d}{dx}[\log(x+y)] = \frac{1}{x+y} \cdot \left(1 + \frac{dy}{dx}\right) \]

4. **Set the derivatives equal**:
Now we set the derivatives from both sides equal to each other:
\[ \cos(x+y) \cdot \left(1 + \frac{dy}{dx}\right) = \frac{1}{x+y} \cdot \left(1 + \frac{dy}{dx}\right) \]

5. **Factor out \( (1 + \frac{dy}{dx}) \)**:
We can factor \( (1 + \frac{dy}{dx}) \) from both sides:
\[ (1 + \frac{dy}{dx}) \left(\cos(x+y) - \frac{1}{x+y}\right) = 0 \]

6. **Solve for \( \frac{dy}{dx} \)**:
This gives us two cases:
- Case 1: \( 1 + \frac{dy}{dx} = 0 \) which leads to \( \frac{dy}{dx} = -1 \)
- Case 2: \( \cos(x+y) - \frac{1}{x+y} = 0 \) which leads to \( \cos(x+y) = \frac{1}{x+y} \)

However, we are primarily interested in the first case for \( \frac{dy}{dx} \).

### Final Result:
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -1 \]