if ` x^2 e^y + 2 xye^x + 13 =0` then ` dy //dx` =

Options

A. `-(2 xe^(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^(x-y)+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 differentiate both sides of the equation with respect to \( x \) using implicit differentiation.

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

### Step 2: Apply the product rule and chain rule
1. For the first term \( x^2 e^y \):
- Using the product rule: \( u = x^2 \) and \( v = e^y \)
- \( \frac{d}{dx}(x^2 e^y) = x^2 \frac{d}{dx}(e^y) + e^y \frac{d}{dx}(x^2) \)
- \( = x^2 e^y \frac{dy}{dx} + e^y (2x) \)
- So, \( \frac{d}{dx}(x^2 e^y) = x^2 e^y \frac{dy}{dx} + 2x e^y \)

2. For the second term \( 2xy e^x \):
- Using the product rule: \( u = 2xy \) and \( v = e^x \)
- \( \frac{d}{dx}(2xy e^x) = 2xy \frac{d}{dx}(e^x) + e^x \frac{d}{dx}(2xy) \)
- \( = 2xy e^x + e^x (2y + 2x \frac{dy}{dx}) \)
- So, \( \frac{d}{dx}(2xy e^x) = 2xy e^x + 2y e^x + 2x e^x \frac{dy}{dx} \)

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

### Step 3: Combine the derivatives
Putting it all together:
\[ x^2 e^y \frac{dy}{dx} + 2x e^y + 2xy e^x + 2y e^x + 2x e^x \frac{dy}{dx} = 0 \]

### Step 4: Collect terms involving \( \frac{dy}{dx} \)
Group the terms with \( \frac{dy}{dx} \):
\[ \left( x^2 e^y + 2x e^x \right) \frac{dy}{dx} + 2x e^y + 2xy e^x + 2y e^x = 0 \]

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

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = - \frac{2x e^y + 2xy e^x + 2y e^x}{x^2 e^y + 2x e^x} \]