$$\boxed{\dfrac{d}{dx}\dfrac{x}{2}=\dfrac{1}{2}}$$

To find the derivative of x/2, or $\dfrac{d}{dx}\dfrac{x}{2}$, we’ll use two important derivative rules, the constant multiple rule and the power rule.

The first rule we’ll use is the constant multiple rule. It states that $\dfrac{d}{dx}cx=c\cdot\dfrac{d}{dx}$, for any constant $c$. We can, therefore, write

$$ \dfrac{d}{dx}\dfrac{x}{2} = \dfrac{d}{dx}(\dfrac{1}{2}\cdot x) = \dfrac{d}{dx}(\dfrac{1}{2}\cdot x). $$
In the first step, we just rewrote the expression

We now have the expression in a form where we can easily use our second rule, the power rule. The power rule states $\dfrac{d}{dx}x^n=nx^{n-1}$. We can use this because we can rewrite $x$ as $x^1$.
$$ \dfrac{1}{2}\cdot \dfrac{d}{dx}x^1=\dfrac{1}{2}\cdot(1\cdot x^0)=\dfrac{1}{2}\cdot(1\cdot 1)=\boxed{\dfrac{1}{2}}. $$
So, the derivative of x/2 is $\dfrac{1}{2}$. As you can see, by factoring and expressing $x$ in exponential form, we can make an manipulate an expression into a form where it’s much easier to apply derivative rules. And therefore easier to compute.

You might be tempted to use the quotient rule whenever you see a fraction. It will work, but in this case it is completely unnecessary. For example, in this case, we can use the quotient rule. Let’s see what happens.

Recall that the quotient rule states that we can find the derivative of a quotient as follows:
$$ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x) g(x) – g'(x) f(x)}{g^2(x)}. $$
If we apply this rule to $\dfrac x2$, then $f(x) = x$, $f'(x) = 1$, $g(x) = 2$, $g'(x) = 0$. Thus, the derivative will be
$$ \frac{f'(x) g(x) – g'(x) f(x)}{g^2(x)} = \frac{1\cdot2 – 0\cdot x}{2^2} = \frac{2}{4} = \boxed{\frac12}. $$

As you can see, the result is the same, but using the constant multiple rule results in the same answer much quicker!