If `y` is a function of `x and log (x+y)=2xy`, then the value of `y'(0)` is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Differentiate both sides**:
We start with the equation:
\[ \log(x + y) = 2xy \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}[\log(x + y)] = \frac{d}{dx}[2xy] \]

2. **Apply the chain rule on the left side**:
Using the chain rule, we have:
\[ \frac{1}{x + y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x\frac{dy}{dx} \]

3. **Rearranging the equation**:
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{1 + \frac{dy}{dx}}{x + y} = 2y + 2x\frac{dy}{dx} \]
Multiply both sides by \( x + y \):
\[ 1 + \frac{dy}{dx} = (2y + 2x\frac{dy}{dx})(x + y) \]

4. **Substituting \( x = 0 \)**:
To find \( y'(0) \), we first need to find the value of \( y \) when \( x = 0 \). Substitute \( x = 0 \) into the original equation:
\[ \log(0 + y) = 2(0)(y) \implies \log(y) = 0 \implies y = 1 \]

5. **Substituting \( y = 1 \) and \( x = 0 \)**:
Now, substitute \( x = 0 \) and \( y = 1 \) into the differentiated equation:
\[ 1 + \frac{dy}{dx}\bigg|_{x=0} = (2(1) + 2(0)\frac{dy}{dx})(0 + 1) \]
Simplifying this gives:
\[ 1 + \frac{dy}{dx}\bigg|_{x=0} = 2 \]

6. **Solving for \( \frac{dy}{dx}\bigg|_{x=0} \)**:
Rearranging the equation:
\[ \frac{dy}{dx}\bigg|_{x=0} = 2 - 1 = 1 \]

### Final Answer:
Thus, the value of \( y'(0) \) is:
\[ \boxed{1} \]