To find the derivative of $\dfrac{1}{x-1}$, we have to use the power rule and the chain rule.

Step 1. Rewrite

First, it is easier to rewrite $\dfrac{1}{x-1}$ as $(x-1)^{-1}$. This makes differentiation simpler because we can now apply the power rule directly.

Step 2. Power Rule

According to the power rule, $\dfrac{d}{dx} x^n = n x^{n-1}$. This rule allows us to find the derivative of any function where $x$ is raised to a power $n$.

Step 3. Chain Rule

According to the chain rule, $\dfrac{d}{dx} f(g(x)) = f'(g(x)) g'(x)$. The chain rule is used when dealing with composite functions, where one function is inside another.

For our example, the outer function, $f(x)$, is $x^{-1}$. To find $f'(x)$, we will take the derivative of $x^{-1}$ using the power rule:

$f'(x) = \dfrac{d}{dx} x^{-1} = -1 \cdot x^{-1 – 1} = -1 \cdot x^{-2} = -x^{-2}$.

Next, we substitute the inner function, $g(x)$, which is $x – 1$, into $f'(x)$. So the first part of the chain rule, $f'(g(x))$, becomes:

$f'(g(x)) = – (x – 1)^{-2}$.

The next step is to take the derivative of the inner function, $g(x)$, which is $x – 1$. Since the derivative of $x$ is $1$ and the derivative of a constant is $0$, we have:

$g'(x) = \dfrac{d}{dx} (x – 1) = 1 – 0 = 1$.

Conclusion

Combining these results using the chain rule to get $f'(g(x)) g'(x)$, we find the derivative of $\dfrac{1}{x-1}$:

$\dfrac{d}{dx} \dfrac{1}{x-1} = f'(g(x)) g'(x) = – (x – 1)^{-2} \cdot 1 = – (x – 1)^{-2}$.

Thus, the derivative of $\dfrac{1}{x-1}$ is $\boxed{ – (x – 1)^{-2} }$.

Similar Example:

Let’s find the derivative of $\dfrac{1}{x+1}$ using the same method.

First, rewrite $\dfrac{1}{x+1}$ as $(x+1)^{-1}$.

Again, we use the power rule and the chain rule.

The outer function is $f(x) = x^{-1}$, so $f'(x) = -x^{-2}$.

Substitute the inner function $g(x) = x + 1$ into $f'(x)$:

$f'(g(x)) = – (x + 1)^{-2}$.

The derivative of the inner function is:

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

Now, apply the chain rule:

$\dfrac{d}{dx} \dfrac{1}{x+1} = f'(g(x)) g'(x) = – (x + 1)^{-2} \cdot 1 = – (x + 1)^{-2}$.

Therefore, the derivative of $\dfrac{1}{x+1}$ is $\boxed{ – (x + 1)^{-2} }$.