To find the derivative of $\dfrac{x-2}{x-1}$, we use the quotient rule, which states that for two functions $f(x)$ and $g(x)$, provided that $g(x)$ is not equal to $0$ and that both derivatives of $f(x)$ and $g(x)$ exist,

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

We will use the fact that

$$\dfrac{d}{dx} (x+1) = \dfrac{d}{dx} x+\dfrac{d}{dx}1 = 1 + 0 = 1$$

and that, similarly,

$$\dfrac{d}{dx} (x+2) = \dfrac{d}{dx} x+\dfrac{d}{dx}2 = 1 + 0 = 1$$

We are now ready to set $f(x) = (x-2)$ and $g(x) = x-1$ and use the quotient rule:

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

We get that the derivative of $\dfrac{x-2}{x-1}$ is $\boxed{\dfrac1{(x – 1)^2}}$.

In fact, note that we could have used the exact same technique if the fraction was, instead, say, $\dfrac{x-6}{x+4}$. The result would have been similar. For example, if you want to differentiate $\dfrac{x-6}{x+4}$, you would get $\dfrac{10}{(x +4)^2}$. Can you guess what would happen if you differentiated $\dfrac{x-6}{x+12}$? If you guessed $\dfrac{18}{(x +12)^2}$, then you have the correct intuition!

If we wish to go a bit deeper into the reason why this happens, we will need to break down the original fraction into two fractions as follows:

$$\dfrac{x-2}{x-1} = \dfrac{x-1}{x-1}-\dfrac{1}{x-1} = 1 -\dfrac{1}{x-1}$$

When differentiating the expression on the right, we immediately get the answer $\dfrac1{(x – 1)^2}$ by either using the quotient rule on the second fraction or by rewriting it as $-(x-1)^{-1}$ and differentiating to get $(x-1)^{-2}$. This technique easily generalizes to, say, the fraction

$$\dfrac{x-6}{x+4} = \dfrac{x+4}{x+4} – \dfrac{10}{x+4} = 1- \dfrac{10}{x+4}$$

Deriving this expression yields the expected $\dfrac{10}{(x +4)^2}$.