To find the derivative of $\dfrac{1}{x – 1}$, we will apply the chain rule

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

on the functions $f(x) = \dfrac1x$ and $g(x) = x – 1$. First, we need to find the derivatives of $f(x)$ and $g(x)$:

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

and

$$g'(x) = \dfrac{d}{dx}(x-1) = 1$$

We are now ready to apply the chain rule. Note that when we evaluate $f'(g(x))$, we are substituting the function $g(x)$ into the derivative of $f(x)$:

$$f'(g(x))g'(x) = -\frac{1}{(x-1)^2} \cdot 1 = \boxed{-\frac{1}{(x-1)^2}}$$

We could arrive at the same answer by using the quotient rule

$$\dfrac{d}{dx}\left(\frac{p(x)}{q(x)}\right) = \frac{p'(x)q(x) -q'(x)p(x)}{q^2(x)}$$

by letting $p(x) = 1$ and $q(x) = x-1$. Then

$$p'(x) = 0$$

and

$$q'(x) = 1$$

Substituting into the quotient rule will result in the same answer:

$$ \frac{p'(x)q(x) -q'(x)p(x)}{q^2(x)} = \frac{0\cdot(x-1) – 1\cdot1}{(x-1)^2}$$

You just saw two different ways to arrive at the same answer. Usually, only one of the two methods will work. Sometimes both methods need to be used together. See if you know which method you can use to derive

$$\frac{5x^3}{7x+1}$$

Here the quotient rule will do the trick. If you chose it, then you have good intuition! The function has a numerator and a denominator, which suggest the quotient rule. In this case, the chain rule cannot be used because the numerator and denominator aren’t easily separated. You wouldn’t be able to use the chain rule, since there is no clear inner and outer functions. Now try

$$\frac{5}{\ln x}$$

In this problem, both the chain rule and the quotient rule will work, although the quotient rule will probably get you the answer quicker. We hope that you enjoyed reading this fun article!