# Quotient Rule of Derivatives – Formula and Examples

The Quotient Rule is one of the major principles used in Differential Calculus (or Calculus I). It is commonly applied in deriving a function that involves the division arithmetic operation. The quotient rule was proven and developed using the backbone of Calculus, which is the limits.

In this article, we will discuss everything about the quotient rule. We will cover its definition, formula, and application usage. We will also look at some examples and practice problems to apply the principles of the quotient rule.

##### CALCULUS

Relevant for

Learning about the quotient rule with examples.

See formula

##### CALCULUS

Relevant for

Learning about the quotient rule with examples.

See formula

## The quotient rule and its formula

### What is the quotient rule?

The quotient rule is a rule that states that a quotient of functions can be derived by taking the denominator g(x) multiplied by the derivative of the numerator f(x) subtracted from the numerator f(x) multiplied by the derivative of the denominator g(x), all divided by the square of the denominator g(x).

The first function f(x) is the dividend or numerator of the problem to be derived, while the second function g(x) is the divisor or denominator.

### The formula for the quotient rule of derivatives

The formula for the quotient rule is:

$$\left(\frac{f}{g}\right)'(x) = \frac{g(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} f'(x) \hspace{2.3 pt} – \hspace{2.3 pt} f(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} g'(x)}{( \hspace{1.15 pt} g(x) \hspace{1.15 pt} )^2}$$

where

• $latex u =$ first function $latex f(x)$ or the numerator/dividend
• $latex v =$ second function $latex g(x)$ or the denominator/divisor

Or in other forms, it can be written as:

$$\frac{d}{dx}(F(x)) = \frac{g(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} f'(x) \hspace{2.3 pt} – \hspace{2.3 pt} f(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} g'(x)}{( \hspace{1.15 pt} g(x) \hspace{1.15 pt} )^2}$$

or

$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

which is the most commonly used form of the quotient rule formula where

$latex u = f(x)$
$latex v = g(x)$

and $latex \frac{d}{dx}(\frac{u}{v})$ can also be $latex y’$, $latex F'(x)$, $latex \Upsilon’$, or other letters used to denote functions with the apostrophe symbol.

## Quotient rule with three or more terms

Since we mentioned that the quotient rule can be used to derive a quotient of functions, what if there are three or more functions to divide? In this case, instead of adjusting the quotient rule formula, it will be more efficient to simply apply the algebraic rules for fractions and then continue using the product and quotient rules.

For example, we have

$$F(x) = \frac{\left( \frac{f(x)}{g(x)} \right)}{\left( \frac{h(x)}{j(x)} \right)}$$

or to be simpler,

$$F(x) = \frac{\left( \frac{s}{u} \right)}{\left( \frac{v}{w} \right)}$$

To differentiate F(x), we first apply the rules of fractions

$$\frac{\left( \frac{A}{B} \right)}{\left( \frac{C}{D} \right)} = \frac{AD}{BC}$$

Therefore, we have

$$F(x) = \frac{sw}{uv}$$

And from this equation, we can derive F(x) using the quotient rule and then apply the product rule to derive the numerator and denominator individually. In doing so, we obtain

$$F(x) = \frac{\left( \frac{s}{u} \right)}{\left( \frac{v}{w} \right)} = \frac{sw}{uv}$$

$$\frac{d}{dx} \left[\frac{\left( \frac{s}{u} \right)}{\left( \frac{v}{w} \right)} \right] = \frac{d}{dx}\left( \frac{sw}{uv} \right)$$

$$\frac{d}{dx} \left[\frac{\left( \frac{s}{u} \right)}{\left( \frac{v}{w} \right)} \right] = \frac{uv \cdot \frac{d}{dx}(sw) – sw \cdot \frac{d}{dx}(uv)}{(uv)^2}$$

Therefore, the quotient rule formula for functions with multiple divisions $latex F(x) = \frac{\left( \frac{s}{u} \right)}{\left( \frac{v}{w} \right )}$ is:

$$F'(x) = \frac{uv \cdot (sw’+ws’) \hspace{2.3 pt} – \hspace{2.3 pt} sw \cdot (uv’+vu’)}{(uv)^2}$$

This is a more efficient and practical formula for deriving complex multiple division functions.

## How to use the quotient rule, a step-by-step tutorial

Suppose we have to derive

$$f(x) = \frac{\sin{(x)}}{x}$$

As you can see, this given function is a fraction, but it cannot be divided algebraically or simplified. But to derive this problem, we can use the quotient rule as shown in the following steps:

#### 1. Identify numerator/dividend and denominator/divisor.

We denote the numerator/dividend as $latex u$ and denote the denominator/divisor as $latex v$:

$latex u = \sin{(x)}$
$latex v = x$

#### 2. Derive $latex u$ and $latex v$ individually.

In this case, we have $latex u’ = \cos{(x)}$ and $latex v’ = 1$.

#### 3. Apply the quotient rule formula.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{(x) \cdot (\cos{(x)}) \hspace{2.3 pt} – \hspace{2.3 pt} (\sin{(x)}) \cdot (1)}{(x)^2}$$

#### 4. Simplify the derivative.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{x \cos{(x)} \hspace{2.3 pt} – \hspace{2.3 pt} \sin{(x)}}{x^2}$$

$$f'(x) = \frac{x \cos{(x)} \hspace{2.3 pt} – \hspace{2.3 pt} \sin{(x)}}{x^2}$$

For formality purposes, it is recommended to use $latex f'(x), y’,$ or $latex \frac{d}{dx}(f(x))$ as the derivative symbol on the left-hand side of the final answer derived instead of $latex (\frac{u}{v})’$ or $latex \frac{d}{dx}(\frac{u}{v})$.

## Quotient rule – Examples with answers

### EXAMPLE 1

Derive the following function:

$$f(x) = \frac{\sqrt[3]{x^2}}{x^3}$$

Step 1: Identify which is the numerator/dividend and which is the denominator/divisor. The numerator/dividend will be denoted as $latex u$ while the denominator/divisor will be $latex v$.

Therefore, we have

$latex u = \sqrt[3]{x^2}$
$latex v = x^3$

Step 2: Derive $latex u$ and $latex v$ individually.

Then, we have

$latex u = \sqrt[3]{x^2}$
$latex u = x^{\frac{2}{3}}$
$latex u’ = \frac{2}{3} x^{-\frac{1}{3}}$

$latex v = x^3$
$latex v’ = 3x^2$

Step 3: Substitute $latex u$, $latex u’$, $latex v$ and $latex v’$ into the quotient rule formula.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{(x^3) \cdot (\frac{2}{3} x^{-\frac{1}{3}}) \hspace{2.3 pt} – \hspace{2.3 pt} (x^{\frac{2}{3}}) \cdot (3x^2)}{(x^3)^2}$$

Step 4: Simplify algebraically.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{(x^3) \cdot (\frac{2}{3} x^{-\frac{1}{3}}) \hspace{2.3 pt} – \hspace{2.3 pt} (x^{\frac{2}{3}}) \cdot (3x^2)}{(x^3)^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\frac{2}{3} x^{\frac{8}{3}} \hspace{2.3 pt} – \hspace{2.3 pt} 3x^{\frac{8}{3}})}{x^6}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-\frac{7}{3} x^{\frac{8}{3}}}{x^6}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-\frac{7}{3} x^{\frac{8}{3}}}{x^6}$$

$$\frac{d}{dx}(\frac{u}{v}) = -\frac{7}{3x^{\frac{10}{3}}}$$

$$f'(x) = -\frac{7}{3 \sqrt[3]{x^{10}}}$$

### EXAMPLE 2

Find the derivative of the given function:

$$f(x) = \frac{\cos{(x^3)}}{\sin{(x^3)}}$$

Step 1: The numerator/dividend will be denoted as $latex u$ while the denominator/divisor will be $latex v$.

Therefore, we have

$latex u = \cos{(x^3)}$
$latex v = \sin{(x^3)}$

Step 2: Differentiate $latex u$ and $latex v$.

Then, we have

$latex u = \cos{(x^3)}$
$latex u’ = -3x^2 \sin{(x^3)}$

$latex v = \sin{(x^3)}$
$latex v’ = 3x^2 \cos{(x^3)}$

Step 3: Substitute $latex u$, $latex u’$, $latex v$ and $latex v’$ in the formula for the quotient rule.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{(\sin{(x^3)}) \cdot (-3x^2 \sin{(x^3)}) \hspace{2.3 pt} – \hspace{2.3 pt} (\cos{(x^3)}) \cdot (3x^2 \cos{(x^3)})}{(\sin{(x^3)})^2}$$

Step 4: Simplify algebraically and apply trigonometric identities if applicable.

$$\frac{d}{dx}(\frac{u}{v}) =\frac{(\sin{(x^3)}) \cdot (-3x^2 \sin{(x^3)}) \hspace{2.3 pt} – \hspace{2.3 pt} (\cos{(x^3)}) \cdot (3x^2 \cos{(x^3)})}{(\sin{(x^3)})^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-3x^2 \sin^{2}{(x^3)} \hspace{2.3 pt} – \hspace{2.3 pt} 3x^2 \cos^{2}{(x^3)}}{\sin^{2}{(x^3)}}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-3x^2 \sin^{2}{(x^3)} \hspace{2.3 pt} – \hspace{2.3 pt} [3x^2 (1-\sin^{2}{(x^3)})]}{\sin^{2}{(x^3)}}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-3x^2 \sin^{2}{(x^3)} \hspace{2.3 pt} – \hspace{2.3 pt} (3x^2 – 3x^2\sin^{2}{(x^3)})}{\sin^{2}{(x^3)}}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-3x^2 \sin^{2}{(x^3)} – 3x^2 + 3x^2\sin^{2}{(x^3)}}{\sin^{2}{(x^3)}}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{-3x^2}{\sin^{2}{(x^3)}}$$

$$\frac{d}{dx}(\frac{u}{v}) = -3x^2 \cdot \csc^{2}{(x^3)}$$

$latex f'(x) = -3x^2 \csc^{2}{(x^3)}$

### EXAMPLE 3

Derive the following function:

$$f(x) = \frac{\ln{(x)}}{5^x}$$

Step 1: We can write as follows:

$latex u = \ln{(x)}$
$latex v = 5^x$

Step 2: Find the derivatives of $latex u$ and $latex v$:

$latex u = \ln{(x)}$
$latex u’ = \frac{1}{x}$

$latex v = 5^x$
$latex v’ = 5^x \ln{(5)}$

Step 3: Use the quotient rule with the expressions for $latex u$, $latex u’$, $latex v$ and $latex v’$:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{(5^x) \cdot (\frac{1}{x}) \hspace{2.3 pt} – \hspace{2.3 pt} (\ln{(x)}) \cdot (5^x \ln{(5)}}{(5^x)^2}$$

Step 4: Simplifying, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\frac{5^x}{x} \hspace{2.3 pt} – \hspace{2.3 pt} 5^x \ln{(x)} \ln{(5)}}{25^x}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\frac{5^x}{x} \hspace{2.3 pt} – \hspace{2.3 pt} \frac{5^x x \ln{(x)} \ln{(5)}}{x}}{25^x}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{5^x \hspace{2.3 pt} – \hspace{2.3 pt} 5^x x \ln{(x)} \ln{(5)}}{25^x x}$$

$$f'(x) = -\frac{5^x (1 – x \ln{(x)} \ln{(5)})}{25^x x}$$

### EXAMPLE 4

What is the derivative of the given function?

$$f(x) = \frac{\tan^{-1}{(x)}}{\cot^{-1}{(x)}}$$

Step 1: The numerator/dividend will be denoted as $latex u$ while the denominator/divisor will be $latex v$:

$latex u = \tan^{-1}{(x)}$
$latex v = \cot^{-1}{(x)}$

Step 2: Differentiate $latex u$ and $latex v$, we have:

$latex u = \tan^{-1}{(x)}$
$latex u’ = \frac{1}{x^2+1}$

$latex v = \cot^{-1}{(x)}$
$latex v’ = -\frac{1}{x^2+1}$

Step 3: Substitute $latex u$, $latex u’$, $latex v$ and $latex v’$ in the formula for the quotient rule.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) =\frac{(\cot^{-1}{(x)}) \cdot \left(\frac{1}{x^2+1} \right) \hspace{2.3 pt} – \hspace{2.3 pt} (\tan^{-1}{(x)}) \cdot \left(-\frac{1}{x^2+1} \right)}{(\cot^{-1}{(x)})^2}$$

Step 4: Simplify algebraically and apply inverse trigonometric identities if applicable.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\frac{\cot^{-1}{(x)}}{x^2+1} \hspace{2.3 pt} – \hspace{2.3 pt} \left(-\frac{\tan^{-1}{(x)}}{x^2+1} \right)}{(\cot^{-1}{(x)})^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\frac{\cot^{-1}{(x)}}{x^2+1} + \frac{\tan^{-1}{(x)}}{x^2+1}}{(\cot^{-1}{(x)})^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\frac{\cot^{-1}{(x)} + \tan^{-1}{(x)}}{x^2+1}}{(\cot^{-1}{(x)})^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{\cot^{-1}{(x)} + \tan^{-1}{(x)}}{(x^2+1) \cdot (\cot^{-1}{(x)})^2}$$

$$f'(x) = \frac{\cot^{-1}{(x)} + \tan^{-1}{(x)}}{x^2 \left(\cot^{-1}{(x)})^2 + (\cot^{-1}{(x)})^2 \right)}$$

### EXAMPLE 5

Derive the following function:

$$f(x) = \frac{\frac{x}{\ln{(x)}}}{\frac{x}{e^x}}$$

In this example, both our dividend and our divisor are fractions. Therefore, we must first apply the fraction rules before deriving them using the quotient formula:

$$f(x) = \frac{\frac{x}{\ln{(x)}}}{\frac{x}{e^x}} = \frac{xe^x}{x \ln{(x)}}$$

Then, with this comes our quotient formula for functions with multiple divisions:

$$F'(x) = \frac{uv \cdot (sw’+ws’) \hspace{2.3 pt} – \hspace{2.3 pt} sw \cdot (uv’+vu’)}{(uv)^2}$$

Step 1: Since we have multiple divisions in our given function, we will represent the numerator of the dividend as $latex s$, the denominator of the dividend as $latex u$, the numerator of the divisor as $latex v$ and the denominator of the divisor as $latex w$.

Therefore, we have

$$f(x) = \frac{\frac{s}{u}}{\frac{v}{w}} = \frac{\frac{x}{\ln{(x)}}}{\frac{x}{e^x}}$$

Applying the rules of fractions:

$$f(x) = \frac{sw}{uv} = \frac{xe^x}{x \ln{(x)}}$$

Then, we have

$latex s = x$
$latex u = \ln{(x)}$
$latex v = x$
$latex w = e^x$

Step 2: Differentiate $latex s$, $latex u$, $latex v$ and $latex w$.

Thus, we have

$latex s = x$
$latex s’ = 1$

$latex u = \ln{(x)}$
$latex u’ = \frac{1}{x}$

$latex v = x$
$latex v’ = 1$

$latex w = e^x$
$latex w’ = e^x$

Step 3: Substitute $latex s$, $latex u$, $latex v$ and $latex w$ in the formula for the quotient rule for functions with multiple divisions:

$$F'(x) = \frac{uv \cdot (sw’+ws’) \hspace{2.3 pt} – \hspace{2.3 pt} sw \cdot (uv’+vu’)}{(uv)^2}$$

$$\frac{d}{dx}f(x) = \frac{\left[(x \ln{(x)}) \cdot ((x) \cdot (e^x) + (e^x) \cdot (1)) \right] \hspace{2.3 pt} – \hspace{2.3 pt} \left[(xe^x) \cdot \left((\ln{(x)}) \cdot (1) + (x) \cdot \left(\frac{1}{x} \right) \right) \right]}{(x \ln{(x)})^2}$$

Step 4: Simplify algebraically:

$$\frac{d}{dx}f(x) = \frac{[(x \ln{(x)}) \cdot (xe^x + e^x)] \hspace{2.3 pt} – \hspace{2.3 pt} [(xe^x) \cdot (\ln{(x)} + 1)]}{(x \ln{(x)})^2}$$

$$\frac{d}{dx}f(x) = \frac{(x^2 e^x \ln{(x)} + xe^x \ln{(x)}) \hspace{2.3 pt} – \hspace{2.3 pt} (xe^x \ln{(x)} + xe^x)}{(x \ln{(x)})^2}$$

$$\frac{d}{dx}f(x) = \frac{x^2 e^x \ln{(x)}}{(x \ln{(x)})^2} + \frac{xe^x \ln{(x)}}{(x \ln{(x)})^2} – \left( \frac{xe^x \ln{(x)}}{(x \ln{(x)})^2} + \frac{xe^x}{(x \ln{(x)})^2} \right)$$

$$\frac{d}{dx}f(x) = \frac{e^x}{\ln{(x)}} + \frac{e^x}{x \ln{(x)}} – \frac{e^x}{x \ln{(x)}} – \frac{e^x}{x(\ln{(x)})^2}$$

$$\frac{d}{dx}f(x) = \frac{e^x}{\ln{(x)}} – \frac{e^x}{x(\ln{(x)})^2}$$

$$\frac{d}{dx}f(x) = \frac{xe^x \ln{(x)}}{x(\ln{(x)})^2} – \frac{e^x}{x(\ln{(x)})^2}$$

$$f'(x) = \frac{xe^x \ln{(x)} – e^x}{x(\ln{(x)})^2}$$

## Quotient rule – Practice problems

Quotient rule quiz
You have completed the quiz!

#### Find the derivative of the following function and determine the value of $latex f^{\prime}(1)$. $$f(x) = \left( \frac{x-3}{\sqrt{x}} \right)^2$$

Write the answer in the input box.

$latex f^{\prime}(1)=$