If `e^(x + y) = y^2` then `(d^2y)/(dx^2)` at (`-1, 1`) is equal to :

Options

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

Explanation

To find the second derivative \(\frac{d^2y}{dx^2}\) at the point \((-1, 1)\) given the equation \(e^{x+y} = y^2\), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
Start with the equation:
\[ e^{x+y} = y^2 \]
Taking the natural logarithm on both sides gives:
\[ \ln(e^{x+y}) = \ln(y^2) \]
Using the property \(\ln(e^a) = a\) and \(\ln(a^b) = b \ln(a)\), we can simplify this to:
\[ x + y = 2 \ln(y) \]

### Step 2: Differentiate both sides with respect to \(x\)
Now, differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x + y) = \frac{d}{dx}(2 \ln(y)) \]
This gives:
\[ 1 + \frac{dy}{dx} = 2 \cdot \frac{1}{y} \cdot \frac{dy}{dx} \]
Rearranging this, we have:
\[ 1 + \frac{dy}{dx} = \frac{2}{y} \frac{dy}{dx} \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation:
\[ 1 = \frac{2}{y} \frac{dy}{dx} - \frac{dy}{dx} \]
Factoring out \(\frac{dy}{dx}\):
\[ 1 = \left(\frac{2}{y} - 1\right) \frac{dy}{dx} \]
Thus,
\[ \frac{dy}{dx} = \frac{1}{\frac{2}{y} - 1} = \frac{y}{2 - y} \]

### Step 4: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)
Now we need to differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{y}{2 - y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(2 - y) \frac{dy}{dx} - y \left(-\frac{dy}{dx}\right)}{(2 - y)^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{(2 - y) \frac{dy}{dx} + y \frac{dy}{dx}}{(2 - y)^2} = \frac{2 \frac{dy}{dx}}{(2 - y)^2} \]

### Step 5: Substitute \((-1, 1)\) into the derivatives
At the point \((-1, 1)\):
1. Calculate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{2 - 1} = 1 \]
2. Substitute this into the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{2 \cdot 1}{(2 - 1)^2} = \frac{2}{1^2} = 2 \]

### Conclusion
Thus, the value of \(\frac{d^2y}{dx^2}\) at the point \((-1, 1)\) is:
\[ \boxed{2} \]