A Commonly Missed Shortcut: The Reciprocal Rule for Derivatives

In differential calculus, there are a number of times when you will know the derivative of a function, but need to find the derivative of the reciprocal of a function.

For example, you may be able to rattle off what the derivative of f(x) = -6x³ + 2x² - 5x + 3 without much thought, but have a relative amount of difficulty in figuring out the derivative of g(x) = 1/f(x) = 1 / (-6x³ + 2x² - 5x + 3). There is a short-cut to finding the derivative of the reciprocals of functions, but because of time restrictions it's rarely taught in calculus courses.
Here we're going to take a look at this rule, why it works, and how it can save you a ton of time. First let's take a moment to derive our shortcut using the quotient rule. Let's say we're taking the derivative of 1/g(x) where we know the derivative of g(x) already. If we let 1 = f(x), then we're taking the derivative of f(x)/g(x), which we know from the quotient rule is (f'(x)g(x) - f(x)g'(x))/(g(x))².
Since f(x) = 1, we know that f'(x) = 0, which greatly simplifies our equation! We have the following after the substitution: (f'(x)g(x) - f(x)g'(x)) / (g(x))² (0 * g(x) - 1 * g'(x)) / (g(x))² - g'(x) / (g(x))² And this gives us the reciprocal rule for derivatives! The derivative of 1/g(x) is simply -g'(x)/(g(x))².
If you'd like to say it out loud, you could say "the derivative of a reciprocal is the derivative of the bottom divided by the square of the bottom".
This makes it easier to remember for a lot of people.

So how can we put this to use? Let's start with the complicated polynomial derivative we looked at earlier, f(x) = -6x³ + 2x² - 5x + 3.
Through our basic understanding of how the derivatives of polynomials work, we know that our derivative of f(x) is f'(x) = -18x² + 4x - 5.

Using only that information and our newly-found reciprocal rule, we can find the derivative of g(x) = 1/f(x) = 1 / (-6x³ + 2x² - 5x + 3) as follows: d/dx (1/f(x)) = - f'(x) / (f(x))² d/dx (1/f(x)) = -(-18x² + 4x - 5) / (-6x³ + 2x² - 5x + 3)² And except for basic simplifying, we're completely finished with finding the derivative in one simple easy-to-follow step! This process works with any differentiable functions, not just polynomials.

Take for example the difficult-to-remember derivatives of trigonometric functions like secant (x) or cosecant (x).
If you can remember the reciprocal rule for derivatives and the basic derivatives of sine (x) and cosine (x), then it's easy to find all of these derivatives when you need them instead of memorizing a bunch of formulas that you will rarely use. So let's find the derivative of f(x) = sec(x). If you remember, sec(x) = 1/cos(x), so we're really finding the derivative of f(x) = 1/cos(x), which fits the pattern f(x) = 1/g(x), making g(x) = cos(x). We know that the derivative of cos(x) is -sin(x), so g'(x) = -sin(x). Since f'(x) = -g'(x)/(g(x))², f'(x) = sin(x)/cos²(x), which simplifies to f'(x) = sec(x)tan(x). By learning this quick and easy-to-remember short-cut, not only can you speed up the process of taking a lot of derivatives that match the pattern, but you can also avoid getting bogged down with remembering a bunch of rarely-used formulas.