`log (x+y)- 2xy =0` then ` y'(0) =`

Options

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

Answer

To solve the equation \( \log(x+y) - 2xy = 0 \) and find \( y'(0) \), we will follow these steps:

### Step 1: Differentiate the equation implicitly
Starting with the equation:
\[ \log(x+y) - 2xy = 0 \]
We differentiate both sides with respect to \( x \).

Using the chain rule for the logarithm:
\[ \frac{d}{dx}[\log(x+y)] = \frac{1}{x+y} \cdot \left(1 + \frac{dy}{dx}\right) \]
For the second term, we apply the product rule:
\[ \frac{d}{dx}[-2xy] = -2\left(x \frac{dy}{dx} + y\right) \]
Thus, differentiating the entire equation gives:
\[ \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) - 2\left(x \frac{dy}{dx} + y\right) = 0 \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{1 + \frac{dy}{dx}}{x+y} = 2\left(x \frac{dy}{dx} + y\right) \]
Multiplying through by \( x+y \) to eliminate the fraction:
\[ 1 + \frac{dy}{dx} = 2(x \frac{dy}{dx} + y)(x+y) \]
Expanding the right-hand side:
\[ 1 + \frac{dy}{dx} = 2xy \frac{dy}{dx} + 2y(x+y) \]

### Step 3: Collecting terms involving \( \frac{dy}{dx} \)
Rearranging gives:
\[ \frac{dy}{dx} - 2xy \frac{dy}{dx} = 2y(x+y) - 1 \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(1 - 2xy) = 2y(x+y) - 1 \]
Thus,
\[ \frac{dy}{dx} = \frac{2y(x+y) - 1}{1 - 2xy} \]

### Step 4: Evaluate at \( x = 0 \)
Next, we need to find \( y \) when \( x = 0 \). Substituting \( x = 0 \) into the original equation:
\[ \log(0+y) - 2(0)(y) = 0 \implies \log(y) = 0 \implies y = 1 \]
So at \( x = 0 \), \( y = 1 \).

Now substituting \( x = 0 \) and \( y = 1 \) into the derivative:
\[ \frac{dy}{dx} = \frac{2(1)(0+1) - 1}{1 - 2(0)(1)} = \frac{2 - 1}{1} = 1 \]

### Final Answer
Thus, \( y'(0) = 1 \).

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