Find the gradient of the given curves at the indicated points.

The numerator $x^2 + 4x + 4$ is a perfect square trinomial, which can be factored as $(x+2)^2$.

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

Assuming $x \neq -2$, we can cancel out one $(x+2)$ term from the numerator and denominator.

$$y = x + 2$$

Now, we find the derivative of $y = x+2$.

$$\frac{dy}{dx} = \frac{d}{dx}(x + 2)$$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the constant rule $\frac{d}{dx}(c) = 0$:

$$\frac{dy}{dx} = 1 \cdot x^{1-1} + 0$$ $$\frac{dy}{dx} = 1 \cdot x^0 + 0$$ $$\frac{dy}{dx} = 1 \cdot 1 + 0$$ $$\frac{dy}{dx} = 1$$