At first glance, it looks like we should use the quotient rule to differentiate $\dfrac2x$. Indeed, we do see a fraction. Recall the quotient rule. 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 $f(x) = 2$ and $g(x) = x$, then $f'(x) = 0$ and $g'(x) = 1$. The quotient rule yields

$$\frac d{dx} \left(\frac2x\right) =\frac{0(x) – 2\cdot1{x^2} = \boxed{-\frac2{x^2}}$$

However, is there a shorter way to get to the same answer? The numerator of $\dfrac 2x$ is a constant, so maybe there is a way to get to the answer faster. Indeed, if we use the laws of exponents, we may write

$$\frac2x = 2x^{-1}$$

Now it is clear that we can use the power rule to find the derivative:

$$\frac d{dx} \left(\frac2x\right) = \frac d{dx} \left(2x^{-1}\right) = -2x^{-2} = \boxed{-\frac2{x^2}}$$

which is the same answer. However, we arrived at the same answer much faster.

### Other examples

The takeaway is that you should always try to use the laws of exponents in conjunction with the power rule when differentiating (or integrating!) expressions that do not contain any special functions. For example, an expression like

$$\frac{\sqrt x}{x^2}$$

might look like a prime candidate for the quotient rule. Yet, if you use the laws of exponents, you can rewrite it as

$$\frac{\sqrt x}{x^2} = \frac {x^{\frac12}}{x^2} = x^{\frac12} x^{-2} = x^{-\frac32}$$

At this point, it is quite clear that you can use the power rule to differentiate this expression. Similarly, you can use the reverse power rule to integrate this expression as well!