To find the derivative of 1/x, we rewrite it as $x^{-1}$. We may apply the power rule $\dfrac{d}{dx} x^n = nx^{n – 1}$:

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

We get the end result that $\boxed{\dfrac{d}{dx} \dfrac1x = \frac{d}{dx} x^{-1} = -\dfrac1{x^2}}$.

In fact, this technique generalizes to any reciprocal, like $\dfrac1{x+1}$, or $\dfrac1{x+4}$, or $\dfrac1{3x-4}$, since we can rewrite each of the above functions as follows:

$$\dfrac1{x+1} = (x+1)^{-1}$$

$$\dfrac1{x+4} = (x+4)^{-1}$$

$$\dfrac1{3x-4} = (3x-4)^{-1}$$

Now we can apply the power rule in all three cases to get the derivatives

$$\dfrac{d}{dx}\dfrac1{x+1} = -\dfrac1{(x+1)^2}$$

$$\dfrac{d}{dx}\dfrac1{x+4} = -\dfrac1{(x+4)^2}$$

Can you guess what the derivative of $(3x-4)^{-1}$ will be? If you guessed

$$\dfrac{d}{dx}\dfrac1{x+4} = -\dfrac1{(3x-4)^2}$$

then you fell into a trap! We cannot always apply the same pattern and hope for the same result. So how is $(3x-4)^{-1}$ different from the other fractions? The different part is the coefficient of $x$. Because of this coefficient we must use the chain rule, which often scares students, but in this case it is quite simple. You just need to put the coefficient of $x$ in the numerator. Namely,

$$\dfrac{d}{dx}\dfrac3{3x-4} = -\dfrac3{(3x-4)^2}$$

So what do you think the derivative of $\dfrac1{5x+7}$ should be? If you guessed that

$$\dfrac{d}{dx}\dfrac1{5x+7} = -\dfrac5{(5x+7)^2}$$

then you got it! Good job. These types of fractions are very common and if you remember this trick, you will be able to find derivatives with ease in a lot of cases.