If `x^2e^y+2xye^x+13=0` then `(dy)/(dx)=`

Options

A. `(-2xe^(y-x)-2y(x-1))/(x(xe^(y-x)+2))`
B. `(2xe^(x-y)-2y(x-1))/(x(xe^(y-x)+2))`
C. `(2xe^(x-y)+2y(x-1))/(x(xe^(y-x)+2))`
D. None of these

Answer

To find \(\frac{dy}{dx}\) for the equation \(x^2 e^y + 2xy e^x + 13 = 0\), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate both sides of the equation with respect to \(x\).
Given:
\[ x^2 e^y + 2xy e^x + 13 = 0 \]
Differentiating both sides:
\[ \frac{d}{dx}(x^2 e^y) + \frac{d}{dx}(2xy e^x) + \frac{d}{dx}(13) = 0 \]

### Step 2: Apply the product rule and chain rule.
1. For \(x^2 e^y\):
- Using the product rule:
\[ \frac{d}{dx}(x^2 e^y) = e^y \frac{d}{dx}(x^2) + x^2 \frac{d}{dx}(e^y) = e^y (2x) + x^2 e^y \frac{dy}{dx} \]
Therefore,
\[ \frac{d}{dx}(x^2 e^y) = 2x e^y + x^2 e^y \frac{dy}{dx} \]

2. For \(2xy e^x\):
- Again using the product rule:
\[ \frac{d}{dx}(2xy e^x) = 2y e^x + 2x e^x \frac{dy}{dx} + 2xy e^x \]
Therefore,
\[ \frac{d}{dx}(2xy e^x) = 2y e^x + 2xy e^x + 2x e^x \frac{dy}{dx} \]

3. The derivative of \(13\) is \(0\).

### Step 3: Combine the derivatives.
Now, substituting back into the differentiated equation:
\[ (2x e^y + x^2 e^y \frac{dy}{dx}) + (2y e^x + 2xy e^x + 2x e^x \frac{dy}{dx}) + 0 = 0 \]

### Step 4: Rearrange the equation.
Combining like terms:
\[ x^2 e^y \frac{dy}{dx} + 2x e^x \frac{dy}{dx} + 2x e^y + 2y e^x + 2xy e^x = 0 \]
Factoring out \(\frac{dy}{dx}\):
\[ \left(x^2 e^y + 2x e^x\right) \frac{dy}{dx} + (2x e^y + 2y e^x + 2xy e^x) = 0 \]

### Step 5: Solve for \(\frac{dy}{dx}\).
Isolate \(\frac{dy}{dx}\):
\[ \left(x^2 e^y + 2x e^x\right) \frac{dy}{dx} = - (2x e^y + 2y e^x + 2xy e^x) \]
Thus,
\[ \frac{dy}{dx} = -\frac{2x e^y + 2y e^x + 2xy e^x}{x^2 e^y + 2x e^x} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{2x e^y + 2y e^x + 2xy e^x}{x^2 e^y + 2x e^x} \]