# Quotient Rule of Derivatives – Examples with Answers

Derivation exercises that involve the quotient of functions can be solved using the quotient rule formula. This formula allows us to derive a quotient of functions such as but not limited to $latex \frac{f}{g} (x) = \frac{f(x)}{g(x)}$.

Here, we will look at the summary of the quotient rule. Additionally, we will explore several examples with answers to understand the application of the quotient rule formula.

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Exploring the quotient rule of derivatives with examples.

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##### CALCULUS

Relevant for

Exploring the quotient rule of derivatives with examples.

See examples

## Summary of the quotient rule

The quotient rule is a very useful formula for deriving quotients of functions. It is a rule that states that the derivative of a quotient of two functions is equal to the function in 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).

This gives us the formula for the quotient rule as:

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

or in a shorter form, it can be illustrated as:

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

where $latex u = f(x)$ is the numerator/dividend of the given problem and $latex v = g(x)$ is the denominator/divisor of the given problem.

You can use either of these two forms of the product rule formula according to your preference.

We use this formula to derive functions that have the following form:

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

or

$$F(x) = \frac{u}{v}$$

where $latex f(x)$ or $latex u$ is the numerator/divider while $latex g(x)$ and $latex v$ is the denominator/divisor of the given problem.

## Quotient rule – Examples with answers

### EXAMPLE 1

Derive the following:

$$f(x) = \frac{x^3}{x-5}$$

We have $latex x^3$ as numerator/dividend and $latex x-5$ as denominator/divisor.

Based on the quotient rule formula, $latex u$ is the numerator and $latex v$ is the denominator. Therefore, we have

$latex u = x^3$
$latex v = x-5$
$latex f(x) = \frac{u}{v}$

Then, we derive $latex u$ and $latex v$ individually and then substitute by the quotient rule formula below:

$latex u = x^3$
$latex u’ = 3x^2$

$latex v = x-5$
$latex v’ = 1$

Substituting $latex u$, $latex v$, $latex u’$ and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(x-5) \cdot (3x^2) – (x^3) \cdot (1)}{(x-5)^2}$$

Simplifying algebraically, we get

$$f'(x) = \frac{(3x^3 – 15x^2) – (x^3)}{(x^2-10x+25)}$$

$$f'(x) = \frac{2x^3 – 15x^2}{x^2-10x+25}$$

### EXAMPLE 2

What is the derivative of the following?

$$f(x) = \frac{6x^3}{\ln{(x)}}$$

Based on the given, we have $latex 6x^3$ as the numerator/dividend and $latex \ln{(x)}$ as the denominator/divisor. Therefore, we have

$latex u = 6x^3$
$latex v = \ln{(x)}$
$latex f(x) = \frac{u}{v}$

Next, we derive $latex u$ and $latex v$ individually:

$latex u = 6x^3$
$latex u’ = 18x^2$

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

Substituting $latex u$, $latex v$, $latex u’$ and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\ln{(x)}) \cdot (18x^2) – (6x^3) \cdot (\frac{1}{x})}{(\ln{(x)})^2}$$

Simplifying, we obtain

$$f'(x) = \frac{18x^2 \ln{(x)} – \frac{6x^3}{x}}{(\ln{(x)})^2}$$

$$f'(x) = \frac{18x^2 \ln{(x)} – 6x^2}{(\ln{(x)})^2}$$

### EXAMPLE 3

Derive the following function:

$$f(x) = \frac{5x^5-x^4}{30x-12x^2}$$

$latex u$ is the numerator and $latex v$ is the denominator. Therefore, we have

$latex u = 5x^5-x^4$
$latex v = 30x-12x^2$
$latex f(x) = \frac{u}{v}$

Then, we derive $latex u$ and $latex v$ individually and then substitute by the quotient rule formula below:

$latex u = 5x^5-x^4$
$latex u’ = 25x^4-4x^3$

$latex v = 30x-12x^2$
$latex v’ = 30-24x$

Using the quotient rule, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(30x-12x^2) \cdot (25x^4-4x^3) – (5x^5-x^4) \cdot (30-24x)}{(30x-12x^2)^2}$$

Simplifying, we obtain

$$f'(x) = \frac{(-300x^6+798x^5-120x^4) – (-120x^6+174x^5-30x^4)}{9x^4-72x^3+144x^2}$$

$$f'(x) = \frac{-180x^6+624x^5-90x^4}{9x^4-72x^3+144x^2}$$

$$f'(x) = \frac{-(180x^6-624x^5+90x^4)}{9x^4-72x^3+144x^2}$$

$$f'(x) = -\frac{180x^6-624x^5+90x^4}{9x^4-72x^3+144x^2}$$

### EXAMPLE 4

Find the derivative of the following function:

$$f(x) = \frac{x^2}{e^{2x}}$$

Based on the given, we have $latex u=x^2$ as numerator/dividend and $latex v=e^{(2x)}$ as denominator/divisor. Therefore, we have

$latex u = x^2$
$latex v = e^{2x}$
$latex f(x) = \frac{u}{v}$

The derivatives of $latex u$ and $latex v$ are:

$latex u = x^2$
$latex u’ = 2x$

$latex v = e^{2x}$
$latex v’ = 2e^{2x}$

Substituting $latex u$, $latex v$, $latex u’$ and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

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

Simplifying, we have:

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

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

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

$$f'(x) = \frac{2x – 2x^2}{e^{2x}}$$

### EXAMPLE 5

What is the derivative of f(x)?

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

$latex u$ is the numerator and $latex v$ is the denominator. Therefore, we have

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

Next, we derive $latex u$ and $latex v$ individually:

$latex u = \sin{(x)}$
$latex u’ = \cos{(x)}$

$latex v=\tan{(x)}$
$latex v’ = \sec^{2}{(x)}$

Now, we can use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\sin{(x)}) \cdot (\sec^{2}{(x)}) – (\tan{(x)}) \cdot (\cos{(x)})}{(\tan{(x)})^2}$$

Simplifying algebraically and applying trigonometric identities, we get

$$\frac{d}{dx}f(x) = \frac{(\sin{(x)}) \cdot (\frac{1}{cos^{2}{(x)}}) – (\frac{\sin{(x)}}{\cos{(x)}}) \cdot (\cos{(x)})}{\tan^{2}{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{(\frac{\sin{(x)}}{\cos{(x)}}) \cdot (\frac{1}{cos{(x)}}) – (\sin{(x)}}{\tan^{2}{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{\tan{(x)} \cdot \sec{(x)} – \sin{(x)}}{\tan^{2}{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{\tan{(x)} \sec{(x)}}{\tan^{2}{(x)}} – \frac{\sin{(x)}}{\tan^{2}{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{\tan{(x)} \sec{(x)}}{\tan^{2}{(x)}} – \frac{\sin{(x)}}{(\frac{\sin{(x)}}{\cos{(x)}})^2}$$

$$\frac{d}{dx}f(x) = \frac{sec{(x)}}{\tan{(x)}} – \frac{\sin{(x)} \cos^{2}{(x)}}{\sin^{2}{(x)}}$$

$$\frac{d}{dx}f(x) = \left[\frac{\frac{1}{\cos{(x)}}}{\frac{\cos{(x)}}{\sin{(x)}}}\right] – \left[\sin{(x)} \cdot \frac{cos^{2}{(x)}}{sin^{2}{(x)}}\right]$$

$$\frac{d}{dx}f(x) = \frac{\cos{(x)}}{\cos{(x)} \cdot \sin{(x)}} – \frac{cos^{2}{(x)}}{sin{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{1}{\sin{(x)}} – \frac{\sin^{2}{(x)}-1}{sin{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{1 – \sin^{2}{(x)} – 1}{sin{(x)}}$$

$$\frac{d}{dx}f(x) = \frac{-\sin^{2}{(x)}}{sin{(x)}}$$

$latex f'(x) = -sin{(x)}$

### EXAMPLE 6

Derive the function f(x):

$$f(x) = \frac{\sqrt{x^3}}{x^5+3x^2-4x}$$

Based on the given, we have $latex u=\sqrt{x^3}$ as the numerator/dividend and $latex v=x^5+3x^2-4x$ as the denominator/divisor.

$latex u = \sqrt{x^3}=x^{\frac{3}{5}}$

$latex v = x^5+3x^2-4x$

$latex f(x) = \frac{u}{v}$

Next, let us derive $latex u$ and $latex v$ individually and then substitute by the quotient rule formula below:

$$u = x^{\frac{3}{5}}$$
$$u’ = \frac{3}{5} x^{-\frac{2}{5}}$$

$latex v = x^5+3x^2-4x$
$latex v’ = 5x^4+6x-4$

Now, we can use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(x^5+3x^2-4x) \cdot \left( \frac{3}{5} x^{-\frac{2}{5}} \right) – \left(x^{\frac{3}{5}} \right) \cdot (5x^4+6x-4)}{(x^5+3x^2-4x)^2}$$

Simplifying, we have:

$$\frac{d}{dx}f(x) = \frac{\left( \frac{3}{5} x^{\frac{23}{5}}+\frac{9}{5} x^{\frac{8}{5}}-\frac{12}{5} x^{\frac{3}{5}} \right) – \left( 5x^{\frac{23}{5}}+6x^{\frac{8}{5}}-4x^{\frac{3}{5}} \right)}{(x^5+3x^2-4x)^2}$$

$$\frac{d}{dx}f(x) = \frac{\frac{22}{5} x^{\frac{23}{5}}-\frac{21}{5} x^{\frac{8}{5}}+\frac{8}{5} x^{\frac{3}{5}}}{(x^5+3x^2-4x)^2}$$

$$\frac{d}{dx}f(x) = \frac{\frac{-22 x^{\frac{23}{5}} – 21 x^{\frac{8}{5}} + 8 x^{\frac{3}{5}}}{5}}{(x^5+3x^2-4x)^2}$$

$$f'(x) = \frac{8 \sqrt{x^3} -22 \sqrt{x^{23}} – 21 \sqrt{x^8}}{5(x^5+3x^2-4x)^2}$$

### EXAMPLE 7

What is the derivative of the given function?

$$f(x) = \frac{\ln{(x)}}{\cos{(x)}}$$

In this case, we have:

$latex u = \ln{(x)}$
$latex v = \cos{(x)}$
$latex f(x) = \frac{u}{v}$

Then, let us derive $latex u$ and $latex v$ individually:

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

$latex v = \cos{(x)}$
$latex v’ = -\sin{(x)}$

After doing this, we can use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\cos{(x)}) \cdot (\frac{1}{x}) – (\ln{(x)}) \cdot (-\sin{(x)})}{(\cos{(x)})^2}$$

Simplifying the obtained expression, we have:

$$f'(x) = \frac{(\frac{\cos{(x)}}{x}) – (-\sin{(x)} \ln{(x)})}{(\cos{(x)})^2}$$

$$f'(x) = \frac{\frac{\cos{(x)}}{x}) + \sin{(x)} \ln{(x)}}{(\cos{(x)})^2}$$

$$f'(x) = \frac{\frac{\cos{(x)}}{x}) + \frac{x \sin{(x)} \ln{(x)}}{x}}{\cos^{2}{(x)}}$$

$$f'(x) = \frac{\cos{(x)} + x \sin{(x)} \ln{(x)} }{x \cos^{2}{(x)}}$$

### EXAMPLE 8

Find the derivative of the given function:

$$f(x) = \frac{x^3}{\sin^{2}{(x)}}$$

$latex u$ is the numerator and $latex v$ is the denominator, so we have

$latex u = x^3$
$latex v = \sin^{2}{(x)}$
$latex f(x) = \frac{u}{v}$

Now, let’s derive $latex u$ and $latex v$ individually and then substitute by the quotient rule formula below:

$latex u = x^3$
$latex u’ = 3x^2$

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

Substituting $latex u$, $latex v$, $latex u’$ and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\sin^{2}{(x)}) \cdot (3x^2) – (x^3) \cdot (2 \sin{(x)} \cos{(x)})}{(\sin^{2}{(x)})^2}$$

Simplifying algebraically and applying some trigonometric identities, we get

$$f'(x) = \frac{(3x^2 \sin^{2}{(x)}) – (2x^3 \sin{(x)} \cos{(x)})}{\sin^{4}{(x)}}$$

$$f'(x) = \frac{3x^2 \sin^{2}{(x)}}{\sin^{4}{(x)}} – \frac{2x^3 \sin{(x)} \cos{(x)}}{\sin^{4}{(x)}}$$

$$f'(x) = \frac{3x^2}{\sin^{2}{(x)}} – \frac{2x^3 \cos{(x)}}{\sin^{3}{(x)}}$$

$$f'(x) = 3x^2 \csc^{2}{(x)} – 2x^3 \cdot \frac{\cos{(x)}}{\sin{(x)}} \cdot \frac{1}{\sin^{2}{(x)}}$$

$$f'(x) = 3x^2 \csc^{2}{(x)} – 2x^3 \cot{(x)} \csc^{2}{(x)}$$

### EXAMPLE 9

What is the derivative of the following function?

$$f(x) = \frac{5x^x}{\cos{(3x)}}$$

We can observe the following:

$latex u = 5x^x$
$latex v = \cos{(3x)}$
$latex f(x) = \frac{u}{v}$

Finding the derivatives of $latex u$ and $latex v$, we have:

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

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

Substituting $latex u$, $latex v$, $latex u’$ and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}f(x) =\frac{(\cos{(3x)}) \cdot (5x^x \ln{(x)} + 5x^x) – (5x^x) \cdot (-3 \sin{(3x)})}{(\cos{(3x)})^2}$$

Simplifying algebraically and applying some trigonometric identities, we get

$$f'(x) =\frac{(5x^x \cos{(3x)} \ln{(x)} + 5x^x \cos{(3x)}) – (-15x^x \sin{(3x)})}{\cos^{2}{(3x)}}$$

$$f'(x) =\frac{5x^x \cos{(3x)} \ln{(x)} + 5x^x \cos{(3x)} + 15x^x \sin{(3x)}}{\cos^{2}{(3x)}}$$

$$f'(x) = \frac{5x^x \cos{(3x)} \ln{(x)}}{\cos^{2}{(3x)}}+ \frac{5x^x \cos{(3x)}}{\cos^{2}{(3x)}} + \frac{15x^x \sin{(3x)}}{\cos^{2}{(3x)}}$$

$$f'(x) = 5x^x \ln{(x)} \cdot \frac{1}{\cos{(3x)}}+ 5x^x \cdot \frac{1}{\cos{(3x)}} + 15x^x \cdot \frac{\sin{(3x)}}{\cos{(3x)}} \cdot \frac{1}{\cos{(3x)}}$$

$$f'(x) = 5x^x \ln{(x)} \sec{(3x)} + 5x^x \sec{(3x)}+ 15x^x \sec{(3x)} \tan{(3x)}$$

### EXAMPLE 10

Find the derivative of the given function:

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

We have $latex x^{e^x}$ as the numerator/dividend and $latex e^{\sin{(x)}}$ as the denominator/divisor. Therefore, we have

$latex u = x^{e^x}$
$latex v = e^{\sin{(x)}}$
$latex f(x) = \frac{u}{v}$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute by the quotient rule formula below:

$latex u = x^{e^x}$
$latex u’ = x^{e^x} e^x \ln{(x)} + \frac{x^{e^x} e^x}{x}$

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

Substituting $latex u$, $latex v$, $latex u’$ and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

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

Simplifying algebraically and applying some trigonometric identities, we get

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

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

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

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

$$f'(x) = \frac{x^{e^x} \left[ (e^x (x \ln{(x)} + 1)) – x \cos{(x)} \right]}{xe^{\sin{(x)}}}$$

## 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

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

Interested in learning more about the quotient rule? Check out these pages: ### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  