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The Quotient Rule

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

A rule for derivative division

We learned a rule for multiplying two derivatives, now we learn about how to take the derivative of a function divided by another function, $\frac{f (x)}{g (x)}$ .However, before we delve deeply into this rule, let’s start with something easier, called the reciprocal rule. This rule allows us to find the derivatives of functions that are written in the form $F (x) = \frac{1}{g (x)}$
The reciprocal rule tells us that the derivative of the function above is equal to,

$F^{'} (x) = - \frac{g^{'} (x)}{g (x)^{2}}$
Using this rule, let’s revisit some old derivatives.

Example Problem 4.2.1

Prove that the derivative of sec(x) is sec(x)tan(x) using the reciprocal rule.

Here, we know two things that can help us:
$1) \sec x = \frac{1}{\cos x}$

$2) \frac{d}{d x} \cos x = - \sin x$
Plugging these into the reciprocal rule formula gives us:

$\frac{d}{d x} \frac{1}{\cos x} = - \frac{- \sin x}{\cos^{2} x}$

$= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$

$= \sec x \cdot \tan x$

That’s it! Easy, huh.

Example Problem 4.2.2

Using the reciprocal rule, prove that the derivative of csc(x) is -csc(x)•cot(x)

We know that $\csc x = \frac{1}{\sin x}$ . Let’s plug this into the reciprocal rule to find its derivative.
$\frac{d}{d x} \csc x = - \frac{\frac{d}{d x} \sin x}{\sin^{2} x}$

$\frac{d}{d x} \csc x = - \frac{\cos x}{\sin^{2} x}$

$\frac{d}{d x} \csc x = - \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$
$\frac{d}{d x} \csc x = - \cot x \cdot \csc x$
We proved it! Now if you forget any of these trigonometric derivatives during an exam, you can go back and derive it with ease using the reciprocal rule.

The quotient rule

Now that we have gotten comfortable using the reciprocal rule, we can do a very simple derivation of the quotient rule using the product rule because we know that any function, F(x), in the form of:
$F (x) = \frac{f (x)}{g (x)} = f (x) \cdot \frac{1}{g (x)}$
Therefore, we can derive the quotient rule using a combination of the product rule and the reciprocal rule!

Note: I am doing this derivation because the quotient rule is a bit hard to memorize, whereas memorizing the product rule and reciprocal rule individually is a much less daunting task. Perhaps this derivation will help you remember how to use the quotient rule more effectively.

$F^{'} (x) = \frac{d}{d x} f (x) \cdot \frac{1}{g (x)} + f (x) \cdot \frac{d}{d x} (\frac{1}{g (x)})$ (Product Rule)

$F^{'} (x) = \frac{f^{'} (x)}{g (x)} + f (x) \cdot (- \frac{g^{'} (x)}{g (x)^{2}})$ (Reciprocal Rule)
$F^{'} (x) = \frac{g (x)}{g (x)} \cdot \frac{f^{'} (x)}{g (x)} - \frac{f (x) \cdot g^{'} (x)}{g (x)^{2}}$
After combining like fractions, we are left with the quotient rule!

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

Example Problem 4.2.3

Find the derivative of the function $F (x) = \frac{3 x^{4} + 2 x^{2} + \frac{1}{x}}{e^{x}}$

We must find the derivative using the quotient rule because one function is divided by another. First let’s write out the quotient rule.

$\frac{d}{d x} \frac{f (x)}{g (x)} = \frac{f^{'} (x) \cdot g (x) - f (x) \cdot g^{'} (x)}{g (x)^{2}}$
In this case, $f (x) = 3 x^{4} + 2 x^{2} + \frac{1}{x}$ and $g (x) = e^{x}$ . $= \frac{\frac{d}{d x} (3 x^{4} + 2 x^{2} + \frac{1}{x}) \cdot e^{x} - (3 x^{4} + 2 x^{2} + \frac{1}{x}) \cdot \frac{d}{d x} e^{x}}{(e^{x})^{2}}$

$= \frac{(12 x^{3} + 4 x - \frac{1}{x^{2}}) \cdot e^{x} - (3 x^{4} + 2 x^{2} + \frac{1}{x}) \cdot e^{x}}{(e^{x})^{2}}$
Here, we can factor out an e^x from the numerator since it is present in both terms, and divide it by one of the e^x in the denominator.

$= \frac{e^{x} (3 x^{4} + 12 x^{3} + 2 x^{2} + 4 x + \frac{1}{x} - \frac{1}{x^{2}}}{(e^{x})^{2}}$

$= \frac{3 x^{4} + 12 x^{3} + 2 x^{2} + 4 x + \frac{1}{x} - \frac{1}{x^{2}}}{e^{x}}$

Example Problem 4.2.4

Prove using the quotient rule that the derivative of cot(x) is -csc²(x).

We know that $\cot (x) = \frac{\cos x}{\sin x}$
Here, let’s try to skip a step and do the derivatives in our heads.
$\frac{d}{d x} \cot (x) = \frac{\sin x \cdot (- \sin x) - \cos x \cdot \cos x}{\sin^{2} x}$
$= \frac{- \sin^{2} x - \cos^{2} x}{\sin^{2} x} = - \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x}$
$= - \frac{1}{\sin^{2} x}$
$= - \csc^{2} x$ . That was easy!

With this, we now have the tools to derive the derivative of any trigonometric function in case we forget.

Glossary and deeper reading · 2 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.
Read more: 2.2 · Derivative →
Function
A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.

Pop Quiz 4.2.1 The Reciprocal and Quotient Rules

Derivative

Function

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