Result:
$$\boxed{\dfrac{d}{dx}\dfrac{x}{2}=\dfrac{1}{2}}$$
At first, when you see this problem, you might be tempted to use the quotient rule, since the expression contains a quotient. Indeed, using the quotient rule would work, but here it is not necessary as 2 is a constant.
To find the derivative of x/2, or $\dfrac{d}{dx}\dfrac{x}{2}$, we will use the constant multiple rule and the power rule.
Constant multiple rule states $\dfrac{d}{dx}cx=c\cdot\dfrac{d}{dx}$, for any constant $c$.
We can, therefore, write
$$ \dfrac{d}{dx}\dfrac{x}{2} = \dfrac{d}{dx}(\dfrac{1}{2}\cdot x) = \dfrac{1}{2}\cdot \dfrac{d}{dx}(x). $$
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 $\dfrac{1}{2}$.
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!
The takeaway is that deriving a fraction does not always require the quotient rule. If the denominator is a constant, then just factor it out.
For example, the derivative of $\dfrac x5$ would be $\dfrac15$ and the derivative of $\dfrac{3x}4$ would be $\dfrac 34$.
It is natural to ask whether this property generalizes further and, in fact, it does. In most cases, it is convenient to factor out the constant. For example, if you would like to differentiate the polynomial
\begin{align*} \frac{x^2}4 – \frac{3x}5 + 9\end{align*}
power rule will yield the answer
\begin{align*} 2\cdot\frac x4 – \frac35 = \frac x2 – \frac35.\end{align*}
Similarly, to find the derivative of $\dfrac{\ln x}2$, factor out the constant (the derivative of $\ln x$ is shown with proof here).
\begin{align*} \frac12 \cdot \dfrac{d}{dx} \ln x\end{align*}
and at this point, the derivative is clearly
\begin{align*} \frac1{2x}.\end{align*}
