If `log(x+y)=2xy,` then `y'(0)` is

Options

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

Answer

To solve the problem, we need to find \( y'(0) \) given the equation \( \log(x+y) = 2xy \). We will differentiate both sides of the equation with respect to \( x \) and then evaluate at \( x = 0 \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \log(x+y) = 2xy \]

2. **Differentiate both sides with respect to \( x \):**
Using implicit differentiation, we differentiate the left side and the right side:
\[ \frac{d}{dx}[\log(x+y)] = \frac{d}{dx}[2xy] \]

3. **Differentiate the left side:**
Using the chain rule:
\[ \frac{1}{x+y} \cdot \left(1 + \frac{dy}{dx}\right) \]

4. **Differentiate the right side:**
Using the product rule:
\[ 2 \left( y + x \frac{dy}{dx} \right) \]

5. **Set the derivatives equal to each other:**
\[ \frac{1 + \frac{dy}{dx}}{x+y} = 2 \left( y + x \frac{dy}{dx} \right) \]

6. **Substitute \( x = 0 \) to find \( y \):**
When \( x = 0 \):
\[ \log(0 + y) = 2(0)(y) \implies \log(y) = 0 \implies y = 1 \]

7. **Substitute \( x = 0 \) and \( y = 1 \) into the differentiated equation:**
\[ \frac{1 + \frac{dy}{dx}}{0 + 1} = 2(1 + 0 \cdot \frac{dy}{dx}) \implies 1 + \frac{dy}{dx} = 2 \]

8. **Solve for \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = 2 - 1 = 1 \]

9. **Thus, we find:**
\[ y'(0) = 1 \]

### Final Answer:
\[ y'(0) = 1 \]