Differentiate xy+xe^(-y)+ye^(x)=x^2?

To differentiate the given equation, we will use the product rule and the chain rule.

Given equation: xy + xe^(-y) + ye^(x) = x^2

1. Differentiate the first term xy:
d/dx(xy) = y + x(dy/dx)

2. Differentiate the second term xe^(-y):
d/dx(xe^(-y)) = e^(-y) + x(-e^(-y))(dy/dx)

3. Differentiate the third term ye^(x):
d/dx(ye^(x)) = e^(x) + y(e^(x))(dy/dx)

4. Differentiate the right side of the equation x^2:
d/dx(x^2) = 2x

Now, we can rewrite the equation with the derivatives:

y + x(dy/dx) + e^(-y) - xe^(-y)(dy/dx) + e^(x) + ye^(x)(dy/dx) = 2x

Now, we can solve for dy/dx by isolating it on one side of the equation:

dy/dx(x - xe^(-y) + ye^(x)) = 2x - y - e^(-y) - e^(x)

dy/dx = (2x - y - e^(-y) - e^(x)) / (x - xe^(-y) + ye^(x))

This is the derivative of the given equation xy + xe^(-y) + ye^(x) = x^2.

Differentiate with respect to x:

```
$\frac{d}{dx}(xyxe^{-y}ye^{x})=x^2$
```

Product rule:

```
$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$
```

Applying the product rule:

```
$\frac{d}{dx}(xyxe^{-y}ye^{x}) = x\frac{d}{dx}(yxe^{-y}ye^{x})+yxe^{-y}ye^{x}\frac{d}{dx}(x)$
```

Calculating each derivative:

```
$\frac{d}{dx}(yxe^{-y}ye^{x}) = y(xe^{-y}ye^{x}+e^{-y}ye^{x})+yxe^{-y}ye^{x}(1)$
```

Simplifying:

```
$\frac{d}{dx}(yxe^{-y}ye^{x}) = yx^2e^{-y}+y^2xe^{-y}+yxe^{-y}$
```

```
$\frac{d}{dx}(x)=1$
```

Substituting the derivatives back into the original equation:

```
$x\frac{d}{dx}(yxe^{-y}ye^{x})+yxe^{-y}ye^{x}\frac{d}{dx}(x) = x^2$
```

Simplifying:

```
$xy^2xe^{-y}+y^2xe^{-y}+xyxe^{-y}=x^2$
```

Combining like terms:

```
$2xy^2xe^{-y}+y^2xe^{-y}=x^2$
```

Dividing both sides by $xye^{-y}$:

```
$2y+y^2=x$
```

Therefore, the derivative of the given function is $2y+y^2=x$.