Find the derivative of the function f(x) = (x2 + 2x + 4) / (x2 + 2x).

$$f(x) = \frac{x^2 + 2x + 4}{x^2 + 2x}$$

We can split the fraction:

$$f(x) = \frac{x^2 + 2x}{x^2 + 2x} + \frac{4}{x^2 + 2x}$$ $$f(x) = 1 + \frac{4}{x^2 + 2x}$$

Now, express the second term with a negative exponent:

$$f(x) = 1 + 4(x^2 + 2x)^{-1}$$

The derivative of a constant is 0. For the second term, we use the chain rule $\frac{d}{dx}[c \cdot g(x)^n] = c \cdot n \cdot g(x)^{n-1} \cdot g'(x)$.
Here, $c=4$, $n=-1$, and $g(x) = x^2 + 2x$.
First, find $g'(x)$:

$$g'(x) = \frac{d}{dx}(x^2 + 2x) = 2x + 2$$

$$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(4(x^2 + 2x)^{-1})$$ $$f'(x) = 0 + 4 \cdot (-1)(x^2 + 2x)^{-1-1} \cdot (2x + 2)$$ $$f'(x) = -4(x^2 + 2x)^{-2} (2x + 2)$$

$$f'(x) = \frac{-4(2x + 2)}{(x^2 + 2x)^2}$$ $$f'(x) = \frac{-8x - 8}{(x^2 + 2x)^2}$$

We can factor out $-8$ from the numerator:

$$f'(x) = \frac{-8(x + 1)}{(x^2 + 2x)^2}$$

The derivative $f'(x)$ is $\boxed{\frac{-8(x + 1)}{(x^2 + 2x)^2}}$.