If `e^(y) +xy = e`, the ordered pair ` ((dy)/(dx),(d^(2)y)/(dx^(2)))` at x = 0 is equal to

Options

A. `(1/e,1/e^(2))`
B. `(-1/e,1/e^(2))`
C. `(1/e,1/e^(2))`
D. `(-1/e, -1/e^(2))`

Explanation

To solve the problem, we need to find the ordered pair \(\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)\) at \(x = 0\) for the equation \(e^y + xy = e\).

### Step 1: Differentiate the equation with respect to \(x\)

Starting from the equation:
\[ e^y + xy = e \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(e^y) + \frac{d}{dx}(xy) = \frac{d}{dx}(e) \]
Using the chain rule and product rule, we get:
\[ e^y \frac{dy}{dx} + \left(x \frac{dy}{dx} + y\right) = 0 \]
This simplifies to:
\[ e^y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)

Rearranging the equation gives:
\[ \left(e^y + x\right) \frac{dy}{dx} + y = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{y}{e^y + x} \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \(x = 0\)

Substituting \(x = 0\) into the equation:
\[ \frac{dy}{dx} = -\frac{y}{e^y} \]

### Step 4: Differentiate again to find \(\frac{d^2y}{dx^2}\)

Now we differentiate \(\frac{dy}{dx} = -\frac{y}{e^y + x}\) again with respect to \(x\):
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(e^y + x)(-\frac{dy}{dx}) - (-y)(1)}{(e^y + x)^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{(e^y + x)\left(-\frac{dy}{dx}\right) + y}{(e^y + x)^2} \]

### Step 5: Substitute \(\frac{dy}{dx}\) into \(\frac{d^2y}{dx^2}\)

Substituting \(\frac{dy}{dx} = -\frac{y}{e^y + x}\):
\[ \frac{d^2y}{dx^2} = \frac{(e^y + x)\left(\frac{y}{e^y + x}\right) + y}{(e^y + x)^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{y + y}{(e^y + x)^2} = \frac{2y}{(e^y + x)^2} \]

### Step 6: Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 0\)

Substituting \(x = 0\):
\[ \frac{d^2y}{dx^2} = \frac{2y}{(e^y)^2} = \frac{2y}{e^{2y}} \]

### Step 7: Final ordered pair

Thus, the ordered pair \(\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)\) at \(x = 0\) is:
\[ \left(-\frac{y}{e^y}, \frac{2y}{e^{2y}}\right) \]