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

### Step 1: Differentiate the given equation implicitly.
We start with the equation:
\[ e^{x+y} = y^2 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(e^{x+y}) = \frac{d}{dx}(y^2) \]
Using the chain rule on the left side:
\[ e^{x+y} \left(1 + \frac{dy}{dx}\right) = 2y \frac{dy}{dx} \]

### Step 2: Solve for \(\frac{dy}{dx}\).
Rearranging the equation:
\[ e^{x+y} + e^{x+y} \frac{dy}{dx} = 2y \frac{dy}{dx} \]
Factoring out \(\frac{dy}{dx}\):
\[ e^{x+y} = \frac{dy}{dx} (2y - e^{x+y}) \]
Thus, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{e^{x+y}}{2y - e^{x+y}} \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at the point \((-1, 1)\).
First, we need to find \(e^{x+y}\) at \((-1, 1)\):
\[ e^{-1 + 1} = e^0 = 1 \]
Now substituting \(y = 1\):
\[ \frac{dy}{dx} = \frac{1}{2(1) - 1} = \frac{1}{1} = 1 \]

### Step 4: Differentiate again to find \(\frac{d^2y}{dx^2}\).
Now we differentiate \(\frac{dy}{dx} = \frac{e^{x+y}}{2y - e^{x+y}}\) again using the quotient rule:
Let \(u = e^{x+y}\) and \(v = 2y - e^{x+y}\):
\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \(\frac{du}{dx}\):
\[ \frac{du}{dx} = e^{x+y} \left(1 + \frac{dy}{dx}\right) = 1(1 + 1) = 2 \]
Calculating \(\frac{dv}{dx}\):
\[ \frac{dv}{dx} = 2 \frac{dy}{dx} - \frac{du}{dx} = 2(1) - 2 = 0 \]

### Step 5: Substitute into the second derivative formula.
Now substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{(2y - e^{x+y})(2) - (e^{x+y})(0)}{(2y - e^{x+y})^2} \]
At the point \((-1, 1)\):
\[ \frac{d^2y}{dx^2} = \frac{(2(1) - 1)(2)}{(2(1) - 1)^2} = \frac{(2 - 1)(2)}{(2 - 1)^2} = \frac{1 \cdot 2}{1^2} = 2 \]

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