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^(y-x)+2y(x-1))/(x(xe^(y-x)+2))]`
C. `(2xe^(y-x)+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 are the steps:

### Step 1: Differentiate both sides with respect to \(x\)

Differentiate the equation \(x^2 e^y + 2xy e^x + 13 = 0\):

\[ \frac{d}{dx}(x^2 e^y) + \frac{d}{dx}(2xy e^x) + \frac{d}{dx}(13) = 0 \]

### Step 2: Apply the product rule and chain rule

For the first term \(x^2 e^y\), we apply the product rule:

\[ \frac{d}{dx}(x^2 e^y) = e^y \frac{d}{dx}(x^2) + x^2 \frac{d}{dx}(e^y) = e^y (2x) + x^2 e^y \frac{dy}{dx} \]

For the second term \(2xy e^x\), we again apply the product rule:

\[ \frac{d}{dx}(2xy e^x) = 2y e^x \frac{d}{dx}(x) + 2x e^x \frac{d}{dx}(y) + 2xy \frac{d}{dx}(e^x) = 2y e^x + 2x e^x \frac{dy}{dx} + 2xy e^x \]

The derivative of a constant (13) is 0.

### Step 3: Combine the derivatives

Now we combine the derivatives we computed:

\[ e^y (2x) + x^2 e^y \frac{dy}{dx} + 2y e^x + 2xy e^x + 2x e^x \frac{dy}{dx} = 0 \]

### Step 4: Collect terms involving \(\frac{dy}{dx}\)

Rearranging gives:

\[ x^2 e^y \frac{dy}{dx} + 2x e^x \frac{dy}{dx} = - (2x e^y + 2y e^x + 2xy e^x) \]

Factoring out \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} (x^2 e^y + 2x e^x) = - (2x e^y + 2y e^x + 2xy e^x) \]

### Step 5: Solve for \(\frac{dy}{dx}\)

Now, we can isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{- (2x e^y + 2y e^x + 2xy e^x)}{x^2 e^y + 2x e^x} \]

### Step 6: Simplify the expression if possible

We can factor out common terms in the numerator:

\[ \frac{dy}{dx} = \frac{-2 (x e^y + y e^x + xy e^x)}{x^2 e^y + 2x e^x} \]

### Final Answer

Thus, the final expression for \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{-2 (x e^y + y e^x + xy e^x)}{x^2 e^y + 2x e^x} \]

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