The derivative of $\dfrac{2}{x + 1}$ is $\boxed{-\dfrac{2}{x^2+2x+1}}$.

To find the derivative of $\dfrac{2}{x+1}$, we will apply the chain rule. The chain rule is:

\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)) \end{align*}\]

Notice that we may factor out the constant term $2$. This leaves us with the two functions $f(x) = \dfrac1x$ and $g(x) = x + 1$. Finally, in our computations, we will require the power rule. The power rule is:

\[\begin{align*} \dfrac{d}{dx} x^n &= nx^{n – 1} \end{align*}\]

Using both the chain rule and power rule, we get:

\[\begin{align*} \dfrac{d}{dx} \dfrac{2}{x+1} &= 2 \cdot \dfrac{d}{dx} \dfrac{1}{x + 1}\\ f(x) &= \dfrac1x\\ g(x) &= x+1\\ f'(x) &= \dfrac{d}{dx}(x^{-1}) = -\dfrac{1}{x^2}\\ g'(x) &= 1\\ f'(g(x)) & = -\dfrac{1}{(x + 1)^2} \cdot 1\\ f'(g(x)) &= -\dfrac{1}{x^2+2x+1}\\ \end{align*}\]

Multiplying our answer by the $2$ we factored out before our calculations, we get the end result that $\boxed{\dfrac{d}{dx}\dfrac{2}{x+1} = -\dfrac{2}{x^2+2x+1}}$.