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

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

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 $f(x)$:

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

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:

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

You might now 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. Which would you choose to find the derivative of

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

If you chose quotient rule, good job. Chain rule would not help much here, since it is not clear what is the inner and what is the outer function. Now how about the derivative of

$$\dfrac{e^{x^2}}7$$

If you said chain rule, then you got it right. Even though the function is a fraction, the denominator is a constant. However, the exponent of $e^{x^2}$ is itself a function $x^2$, so you would have to use the chain rule. Finally, what would you use to derive

$$\dfrac{\ln x + 1}{\sin(x^2) + 2}$$

This is a trick question, since you will have to use both the chain rule and the quotient rule. We hope that you enjoyed this fun article!