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

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

on the functions $f(x) = \dfrac1x$ and $g(x) = 1-x$. 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

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

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 $f(x)$:

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

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 $q(x) = 1-x$. Then

$$p'(x) = 0$$

and

$$q'(x) = -1$$

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

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

Naturally, you may wonder which way is better. As you can see, both methods are quite fast and take roughly the same amount of time. However, let’s see a few examples where one of the two methods is faster. Try to guess which method is better to derive

$$\dfrac{x-3}{x+4}$$

If you chose quotient rule, good job. Chain rule would not help much here, since there is no inner and outer function to apply the chain rule. Now try the derivative of

$$\dfrac{\sin^2x}x$$

This was a trick question. In order to find the derivative of this function, you will need both the chain rule and the quotient rule. Now that you learned about the chain rule and quotient rule, we hope that your future calculus adventures will be more enjoyable!