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## A Commonly Missed Shortcut: The Reciprocal Rule for Derivatives

In calculus, there are several times that you will know the derivative of a function, but you will need to find the derivative of the reciprocal of a function. For example, you might be able to figure out the derivative of f(x) = -6x³ + 2x² – 5x + 3 without much thought, but have a relative amount of difficulty figuring out the derivative of g(x) = 1 /f (x) = 1 / (-6x³ + 2x² – 5x + 3). There is a shortcut to finding the derivative of reciprocals of functions, but due to time constraints it is rarely taught in calculus courses. Here we take a look at this rule, why it works and how it can save you a lot of time.

First, let’s take a moment to derive our shortcut using the quotient rule. Suppose we take the derivative of 1/g(x) where we already know the derivative of g(x). If we let 1 = f(x), then we are 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 replacement:

(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 of derivatives! The derivative of 1/g(x) is simply -g'(x)/(g(x))². If you want to say it out loud, you can say “the derivative of a reciprocal is the derivative of the bottom divided by the square of the bottom”. This makes it easier for many people to remember.

So how can we use this? Let’s start with the complicated polynomial derivative we looked at earlier, f(x) = -6x³ + 2x² – 5x + 3. Using our basic understanding of how derivatives of polynomials work, we know that our derivative of f(x) is f ‘(x) = -18x² + 4x – 5. Using only this 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 aside from the basic simplification, we’re completely done with finding the derivative in one simple, easy-to-follow step!

This process works with any differentiable function, not just polynomials. Take for example the hard-to-remember derivatives of trigonometric functions such as secant (x) or cosecant (x). If you remember the reciprocal derivative rule and the basic derivatives of sine(x) and cosine(x), it’s easy to find all those derivatives when you need them instead of memorizing a bunch of formulas you’ll rarely use.

So we look for 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 shortcut, you’ll not only be able to speed up the process of adopting many pattern-matching derivatives, but you’ll also be able to avoid getting stuck remembering a bunch of rarely-used formulas. .

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