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. Other derivative methods such as the chain rule and implicit differentiation have also been used to prove the principles of the quotient rule.

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

## The Quotient Rule and its Formula

### What is the Quotient Rule?

The Quotient Rule is a rule which 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 to 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 given problem to be derived whilst the second function *g(x)* is the divisor or the denominator.

### The Quotient Rule Formula

The quotient rule formula is:

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

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:

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

or

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

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, how about if there are three or more functions to be divided? In this case, instead of adjusting the quotient rule formula, it will be more efficient to just apply the algebraic rules of fractions and then proceed in using both the product and quotient rules.

For instance, if we are given

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

or to be simpler,

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

To derive F(x), we will apply the rules of fraction first

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

Hence, for our given, it will be

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

And from this equation, you can proceed in deriving F(x) by using the quotient rule and then apply product rule to derive the numerator and denominator individually. By doing so, we get

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

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

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

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 to be used to derive a complex multiple division functions.

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## Proofs of The Quotient Rule

### Proof of The Quotient Rule by using Limits

By learning the backbone of Differential Calculus and through the concepts of getting the limits of the slope of a tangent line, we can consider this as the foundation of the product rule.

We can recall that getting the limits of the slope of a tangent line will bring us to the derivative of the equation of the curve. Hence, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$

We can use that formula to write a derivative for the quotient of two functions in terms of limits. Then, by manipulating the equation, we can arrive at the quotient rule. You can explore this step by step in our article for the Proofs of the Quotient Rule

### Proof of The Quotient Rule by using Implicit Differentiation

Unlike the proof of quotient rule using limits, proving the quotient rule by implicit differentiation might not be the backbone of Differential Calculus, but this is actually the shortest method of proving the quotient rule formula considering you know the process of implicit differentiation.

You can look at our article about the Proofs of the Quotient Rule to learn how we can prove the quotient rule using implicit differentiation.

### Proof of The Quotient Rule by using The Product and Chain Rules

The chain rule is one of the derivative formulas that is used to derive functions raised to an exponent whilst the product rule is for the derivative of the product of functions. These two formulas, once learned and understood, can be an easier method to prove the quotient rule.

Learn more about the proof of the quotient rule using the chain rule by visiting our article about the Proofs of the Quotient Rule.

## When to use the Quotient Rule to find derivatives

The quotient rule formula is an efficient tool to use in order to derive given functions like the following:

a. $latex F(x) = \frac{f(x)}{g(x)}$, where *f*(*x*) and *g*(*x*) are the quotient of the given function *F*(*x*).

b. $latex \frac{f}{g}(x) = \frac{f(x)}{g(x)}$, where *f*(*x*) and *g*(*x*) are a quotient of $latex \frac {f}{g}(x)$.

c. $latex f(x) = \frac{u}{v}$, where $latex u$ and $latex v$ are quotient of the given function *f*(*x*).

d. $latex f(x) = \frac{x_1}{x_2}$ where $latex x_1$ and $latex x_2$ are quotient of the given function *f*(*x*).

e. $latex G(x) = \frac{\left( \frac{f(x)}{g(x)} \right)}{h(x)}$, where the fraction $latex \frac{f(x)}{g(x)}$ is the numerator/dividend of the given function *G*(*x*) whilst *h(x)* is the denominator/divisor.

f. $latex H(x) = \frac{f(x)}{\left( \frac{g(x)}{h(x)} \right)}$, where *f(x)* is the numerator/dividend of the given function *G*(*x*) whilst the fraction $latex \frac{g(x)}{h(x)}$ is the denominator/divisor.

g. $latex J(x) = \frac{\left( \frac{f(x)}{g(x)} \right)}{\left( \frac{h(x)}{j(x)} \right)}$, where the fraction $latex \frac{f(x)}{g(x)}$ is the numerator/dividend of the given function *G*(*x*) whilst the fraction $latex \frac{h(x)}{j(x)}$ is the denominator/divisor.

These are the most common examples of functions that can be derived using the quotient rule. Although you may argue that a function can be algebraically divided before it can be derived using the initial and simpler derivative methods (or even a function-specific derivative formula), that is not always the case, even in some basic division/fractional problems that we may encounter.

Thus, we have the quotient rule to make it still possible to derive quotient of functions that are either very difficult to algebraically divide or even impossible to be divided and simplified algebraically.

## How to use the Quotient Rule, a step by step tutorial

We are asked to derive

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

As you can observe, this given function is a fraction but it cannot be algebraically divided nor simplified. But in order to derive this problem, we can use the quotient rule as shown by the following steps:

**Step 1: **It is always recommended to list the formula first if you are still a beginner.

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

Please take note that you may use any form of the quotient rule formula as long as you find it more efficient based on your preference or on the given problem.** Step 2:** Identify if there is a fractional dividend or divisor. If there’s none, proceed to use the quotient rule formula. If there is, apply the rules of fractions to the original form of the given problem before using the quotient rule formula.

** Step 3:** Let’s identify which one is the numerator/dividend and which one is the denominator/divisor. We will denote the numerator/dividend as $latex u$ and denote the denominator/divisor as $latex v$.

Thus, we have

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

$latex v = x$

** Step 4:** If you are a beginner, it is highly recommended that you individually derive each $latex u$ and $latex v$ first before you proceed in applying the quotient rule formula.

Thus, we have

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

$latex u’ = \cos{(x)}$

$latex u’$ used the derivative of trigonometric functions.

$latex v = x$

$latex v’ = 1$

$latex v’$ used the power formula.

** Step 5:** Apply the quotient rule formula now by substituting the $latex u$, $latex u’$, $latex v$, and $latex v’$ in the quotient rule formula.

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

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

** Step 6:** Simplify algebraically and apply the necessary trigonometric identities as well as other rules to the derived equation whenever applicable.

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

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

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

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

## Quotient Rule – Examples with answers

Each of the following examples has its respective detailed solution. It is recommended for you to try to solve the sample problems yourself before looking at the solution so that you can practice and fully master this topic.

**EXAMPLE 1**

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

##### Solution

** Step 1:** List the quotient rule formula for reference:

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

** Step 2:** Identify if there is a fractional dividend or divisor. In this example, we do not have, so we can directly proceed in using the quotient rule formula to derive the problem.

** Step 3:** Identify which one is the numerator/dividend and which one is the denominator/divisor. The numerator/dividend will be denoted as $latex u$ whilst the denominator/divisor will be $latex v$.

Thus, we have

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

$latex v = x^3$

** Step 4:** Derive $latex u$ and $latex v$ individually.

Thus, we have

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

$latex u = x^{\frac{2}{3}}$

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

$latex u’$ used the power formula.

$latex v = x^3$

$latex v’ = 3x^2$

$latex v’$ used the power formula too.

** Step 5:** 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 6:** Simplify algebraically.

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

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

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

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

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

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer.

**The final answer is:**

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

**EXAMPLE 2**

Find the derivative of $latex f(x) = \frac{\cos{(x^3)}}{\sin{(x^3)}}$.

##### Solution

** Step 1:** List the quotient rule formula for reference:

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

** Step 2:** Identify if there is a fractional dividend or divisor. In this example, we do not have, so we can directly proceed in using the quotient rule formula to derive the problem.

** Step 3:** The numerator/dividend will be denoted as $latex u$ whilst the denominator/divisor will be $latex v$.

Thus, we have

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

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

** Step 4:** Derive $latex u$ and $latex v$ individually.

Thus, we have

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

$latex u’ = -3x^2 \sin{(x^3)}$

$latex u’$ used the chain rule formula and derivative of trigonometric functions.

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

$latex v’ = 3x^2 \cos{(x^3)}$

$latex v’$ used the chain rule formula and derivative of trigonometric functions.

** Step 5:** Substitute $latex u$, $latex u’$, $latex v$, and $latex v’$ into the quotient rule formula.

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

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

** Step 6:** Simplify algebraically and apply trigonometric identities if applicable.

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

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

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

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

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

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

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

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer.

**The final answer is:**

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

**EXAMPLE 3**

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

##### Solution

** Step 1:** List the quotient rule formula for reference:

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

** Step 2:** Identify if there is a fractional dividend or divisor. In this example, we do not have, so we can directly proceed in using the quotient rule formula to derive the problem.

** Step 3:** The numerator/dividend will be denoted as $latex u$ whilst the denominator/divisor will be $latex v$.

Thus, we have

$latex u = \ln{(x)}$

$latex v = 5^x$

** Step 4:** Derive $latex u$ and $latex v$ individually.

Thus, we have

$latex u = \ln{(x)}$

$latex u’ = \frac{1}{x}$

$latex v’$ used the derivative of logarithmic functions.

$latex v = 5^x$

$latex v’ = 5^x \ln{(5)}$

$latex u’$ used the derivative of exponential functions.

** Step 5:** Substitute $latex u$, $latex u’$, $latex v$, and $latex v’$ into the quotient rule formula.

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

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

** Step 6:** Simplify algebraically.

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

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

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

** Step 7: **If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer.

**The final answer is:**

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

**EXAMPLE 4**

What is the derivative of $latex f(x) = \frac{\tan^{-1}{(x)}}{\cot^{-1}{(x)}}$?

##### Solution

** Step 1:** List the quotient rule formula for reference:

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

** Step 2:** Identify if there is a fractional dividend or divisor. In this example, we do not have, so we can directly proceed in using the quotient rule formula to derive the problem.

** Step 3:** The numerator/dividend will be denoted as $latex u$ whilst the denominator/divisor will be $latex v$.

Thus, we have

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

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

** Step 4:** Derive $latex u$ and $latex v$ individually.

Thus, we have

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

$latex u’ = \frac{1}{x^2+1}$

$latex v’$ used the derivative of inverse trigonometric functions.

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

$latex v’ = -\frac{1}{x^2+1}$

$latex u’$ used the derivative of inverse trigonometric functions.

** Step 5:** Substitute $latex u$, $latex u’$, $latex v$, and $latex v’$ into the quotient rule formula.

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

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

** Step 6:** Simplify algebraically and apply inverse trigonometric identities if applicable.

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

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

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

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

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer.

**The final answer is:**

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

**EXAMPLE 5**

Derive: $latexf(x) = \frac{\frac{x}{\ln{(x)}}}{\frac{x}{e^x}}$

##### Solution

** Step 1: **List the quotient rule formula for reference:

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

** Step 2: **In this example, both our dividend and divisor are fractions. Therefore, we need to apply the rules of fractions first before we derive it 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 3:** Since we have multiple divisions in our given function, we will label 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 $latex w$. To illustrate, we have

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

By applying the rules of fraction:

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

Thus, we have

$latex s = x$

$latex u = \ln{(x)}$

$latex v = x$

$latex w = e^x$

** Step 4:** Derive $latex s$, $latex u$, $latex v$ and $latex w$ individually.

Thus, we have

$latex s = x$

$latex s’ = 1$

$latex s’$ used the power rule formula.

$latex u = \ln{(x)}$

$latex u’ = \frac{1}{x}$

$latex u’$ used the derivative of logarithmic functions.

$latex v = x$

$latex v’ = 1$

$latex v’$ used the power rule formula.

$latex w = e^x$

$latex w’ = e^x$

$latex w’$ used the derivative of exponential functions.

** Step 5:** Substitute $latex s$, $latex u$, $latex v$ and $latex w$ into the quotient rule formula for functions with multiple divisions:

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

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

** Step 6:** Simplify algebraically:

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

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

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

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

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

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

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer.

**The final answer is:**

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

## Product Rule – Practice problems

Solve the following differentiation problems and test your knowledge on this topic. Use the product rule formula detailed above to solve the exercises. If you have problems with these exercises, you can study the examples solved above.

## See also

Interested in learning more about the quotient rule? Take a look at these pages:

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