Differentiate x^e + x^x + e^x + e^e with respect to x.

We use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$, the derivative of $e^x$ is $e^x$, and the derivative of a constant is 0. For $x^x$, we use logarithmic differentiation.
Let $y_1 = x^x$. Then $\ln y_1 = x \ln x$.
Differentiating with respect to $x$:

$$\frac{1}{y_1} \frac{dy_1}{dx} = \frac{d}{dx}(x) \cdot \ln x + x \cdot \frac{d}{dx}(\ln x)$$ $$\frac{1}{y_1} \frac{dy_1}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x}$$ $$\frac{1}{y_1} \frac{dy_1}{dx} = \ln x + 1$$ $$\frac{dy_1}{dx} = y_1 (1 + \ln x) = x^x (1 + \ln x)$$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$$\frac{d}{dx}(x^e) = e x^{e-1}$$

Since $e^e$ is a constant:

$$\frac{d}{dx}(e^e) = 0$$

$$\frac{d}{dx}(x^e + x^x + e^x + e^e) = \frac{d}{dx}(x^e) + \frac{d}{dx}(x^x) + \frac{d}{dx}(e^x) + \frac{d}{dx}(e^e)$$ $$\frac{d}{dx}(x^e + x^x + e^x + e^e) = e x^{e-1} + x^x (1 + \ln x) + e^x + 0$$

The derivative is:

$$\boxed{e x^{e-1} + x^x (1 + \ln x) + e^x}$$