### Using the Quotient Rule

One way to find the derivative of $\dfrac{1}{x + 2}$ is by using the quotient rule. Indeed, when you see a quotient, it is only natural to use the quotient rule. The quotient rule states that for functions $f(x)$ and $g(x)$,

$$\dfrac d{dx}\left(\frac {f(x)}{g(x)}\right) = \frac{f'(x)g(x) – f(x)g'(x)}{g^2(x)}$$

If we let $f(x) = 1$ and $g(x) = x+2$, then $f'(x) = 0$ and $g'(x) = 1$. The quotient rule yields

$$\frac d{dx} \left(\frac1{x+2}\right) =\frac{0(x+2) – 1\cdot(1)}{(x+2)^2} = \boxed{-\frac2{(x+2)^2}}$$

### Using the Chain Rule

Alternatively, we may use the chain rule. Recall that the chain rule states that for two functions $f(x)$ and $g(x)$

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

We will let $f(x) = \dfrac 1x$ and $g(x) = x+2$. Then $f'(x) = -\dfrac1{x^2}$ and $g'(x) = 1$. Then

$$\frac d{dx} \left(\frac1{x+2}\right) =\frac{0(x+2) – 1\cdot(1)}{(x+2)^2} = \boxed{-\frac2{(x+2)^2}}$$

Note that we can find the derivative of $\dfrac1x$ similar way, which is described here.

### Generalizations

Both of these methods will work for calculating derivatives of the expressions of the form $\dfrac1{x+k}$ for any constant $k$. In fact, we can use both of these methods for calculating the derivatives of the expressions like $\dfrac 3{x+5}$ or $\dfrac{11}{2x+3}$. For example,

$$\dfrac d{dx} \frac{3}{x+5} = -\frac3{(x+5)^2}$$

However, when the denominator is more complicated, we must apply the chain rule to the denominator. For example,

$$\dfrac d{dx}\frac3{2x+5} = -\dfrac3{(2x+5)^2}\cdot 2$$

The “2” appears from using the chain rule because it is the derivative of $2x + 5$. The denominators can be even more complicated and the derivatives will follow the same pattern:

$$\dfrac d{dx}\frac3{2x^2+5x} = -\dfrac3{(2x^2+5x)^2}\cdot (4x+5)$$