Product Rule - Formula, Proof and Examples

The Product Rule is one of the main principles applied in Differential Calculus (or Calculus I). It is commonly used in deriving a function that involves the multiplication operation. The product rule was proven and developed using the backbone of Calculus, which is the limits. Other initial derivative methods like the chain rule have also been used to prove the principles of the product rule.

In this article, we will discuss everything about the product 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 product rule.

CALCULUS

formula for the product rule of derivatives

Relevant for…

Learning about the product rule with examples.

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Contents

CALCULUS

Relevant for…

Learning about the product rule with examples.

See formula

The Product Rule and its Formula

What is the Product Rule?

The Product Rule is a rule which states that a product of at least two functions can be derived by getting the sum of the (a) first function in original form multiplied by the derivative of the second function and (b) second function in original form multiplied by the derivative of the first function. The first function is the first multiplicand of the given problem to be derived whilst the second function is the second multiplicand.

In another statement of the rule, it is when a product of the functions being derived is equal to the first function (or first multiplicand) in its original form multiplied by the derivative of the second function (or the second multiplicand) and then added to the original form of the second function (or the second multiplicand) in its original form multiplied by the derivative of the first function (or the first multiplicand).

The Product Rule Formula

The product rule formula is:

(fg)'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)

where

$u =$ first function $f(x)$ or the first multiplicand
$v =$ first function $g(x)$ or the second multiplicand

Or in other forms, it can be:

$\frac{d}{dx}(F(x)) = f(x) \cdot \frac{d}{dx}(g'(x)) + g(x) \cdot \frac{d}{dx}(f(x))$

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

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

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

The Product Rule for three or more functions

Check out this article to learn about the product rule of three or more functions.

Start now: Explore our additional Mathematics resources

Proofs of the Product Rule

There are three main methods that we can use to prove the product rule, by using limits, by using the chain rule, and by using logarithmic differentiation.

Proof of the Product Rule by using limits

This proof uses the limit definition of a derivative:

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

By using a substitution for $f(x)\cdot g(x)$ , we can write the derivative of a product in terms of limits. Then, we can manipulate the numerator algebraically. If we use the laws of limits, we can solve the limits until we get the product rule.

Take a look at our article about the proofs of the product rule to learn how to prove the product rule step by step. In addition to the proof by using limits, you can also learn about the proofs by using the chain rule and by using logarithmic differentiation.

When to use the Product Rule to find derivatives

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

a. $F(x) = f(x) \cdot g(x)$
where f(x) and g(x) are two multiplicands of the given function F(x).

b. $fg(x) = f(x) \cdot g(x)$
where f(x) and g(x) are two multiplicands and fg(x) is an operation of function.

c. $f(x) = u \cdot v$
where $u$ and $v$ are two multiplicands of the given function f(x).

d. $f(x) = x_1 \cdot x_2$
where $x_1$ and $x_2$ are two multiplicands of the given function f(x).

e. $G(x) = f(x) \cdot g(x) \cdot h(x)$
where f(x), g(x), and h(x) are three multiplicands of the given function G(x)

f. $H(x) = f(x) \cdot g(x) \cdot h(x) \cdot n(x)…..$
where f(x), g(x), h(x), and n(x)… are various multiplicands of the given function H(x)

These are the most common examples of functions that use the product rule for derivation problems. Although you may argue that a function can be algebraically multiplied before it can be derived using the initial and simpler derivative methods (or even a function specific derivative formula), but that is not always the case, even in some basic multiplication problems that we may encounter.

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

How to use the Product Rule, a step by step tutorial

We are asked to derive

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

As you can observe, this given function has two multiplicands but they cannot be algebraically multiplied nor simplified anymore. But in order to derive this problem, we can use the product rule as shown by the following steps:

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

$\frac{d}{dx}(uv) = uv' + vu'$

Please take note that you may use any form of the product rule formula as long as you find it more efficient based on your preference or on the given problem.

Step 2: Identify how many multiplicands you have in the given problem. In this demonstration, our problem has two. The first one is $x^2$ and the second one is $\sin{(x)}$ .

Step 3: In the chosen form of the product rule formula, we will identify the first multiplicand as $u$ and the second multiplicand as $v$ .

Thus, we have

$u = x^2$
$v = \sin{(x)}$

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

Thus, we have

$u' = 2x$
$u'$ used the power formula.

$v' = \cos{(x)}$
$v'$ used the derivative formula for trigonometric functions.

Step 5: Apply the product rule formula now by substituting the $u$ , $u'$ , $v$ , and $v'$ in the product rule formula.

$\frac{d}{dx}(uv) = uv' + vu'$

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

Step 6: Simplify algebraically and apply the necessary trigonometric identities as well as other operations in the new present functions of the derived equation whenever applicable.

$\frac{d}{dx}(uv) = x^2 \cos{(x)} + 2x \sin{(x)}$

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

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

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

Product Rule – Examples with answers

Using the formula detailed above, we can derive various functions that are expressed as products. 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: $f(x) = x^3 (x-5)$

Solution

Step 1: List the product rule formula for reference:

$\frac{d}{dx}(uv) = uv' + vu'$

Please take note that you may use any form of the product rule formula as long as you find it more efficient based on your preference or on the given problem.

Step 2: Identify how many multiplicands you have in the given problem. In this problem, we have two. The first one is $x^3$ and the second one is $(x-5)$ .

Step 3: In the chosen form of the product rule formula of this demonstration, we will mark the first multiplicand as $u$ and the second multiplicand as $v$ .

Thus, we have

$u = x^3$
$v = (x-5)$

Step 4: Derive $u$ and $v$ individually:

$u' = 3x$
$u'$ used the power formula.

$v' = (1-0)$
$v'$ used the power formula, the sum/difference of derivatives, and the derivatives of constants.

Step 5: Apply the product rule formula now by substituting the $u$ , $u'$ , $v$ , and $v'$ in the product rule formula.

$\frac{d}{dx}(uv) = uv' + vu'$

$\frac{d}{dx}(uv) = (x^3) \cdot (1-0) + (x-5) \cdot (3x)$

Step 6: Simplify algebraically:

$\frac{d}{dx}(uv) = x^3 + 3x (x-5)$

$\frac{d}{dx}(uv) = x^3 + 3x^2 - 15x$

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

$f'(x) = x^3 + 3x^2 - 15x$

EXAMPLE 2

Find the derivative of $f(x) = \sin{(x)} \tan{(x)}$ .

Solution

Step 1: List the product rule formula for reference:

$\frac{d}{dx}(uv) = uv' + vu'$

Please take note that you may use any form of the product rule formula as long as you find it more efficient based on your preference or on the given problem.

Step 2: Identify how many multiplicands you have in the given problem. In this problem, we have two. The first one is $\sin{(x)}$ and the second one is $\tan{(x)}$ .

Step 3: In the chosen form of the product rule formula of this demonstration, we will mark the first multiplicand as $u$ and the second multiplicand as $v$ .

Thus, we have

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

Step 4: Derive $u$ and $v$ individually:

$u' = \cos{(x)}$
$u'$ used the derivative formula for trigonometric functions.

$v' = \sec^{2}{(x)}$
$v'$ used the derivative formula for trigonometric functions.

Step 5: Apply the product rule formula now by substituting the $u$ , $u'$ , $v$ , and $v'$ in the product rule formula.

$\frac{d}{dx}(uv) = uv' + vu'$

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

Step 6: Simplify algebraically and since we have a trigonometric function in our derivative, we can also apply some trigonometric identities applicable in our solution:

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

$\frac{d}{dx}(uv) = (\sin{(x)}) (\frac{1}{\cos{(x)}})^2$ $+ (\frac{\sin{(x)}}{\cos{(x)}}) (\cos{(x)})$

$\frac{d}{dx}(uv) = (\sin{(x)}) (\frac{1^{2}}{\cos^{2}{(x)}})$ $+ (\frac{\sin{(x)}}{\cos{(x)}}) (\cos{(x)})$

$\frac{d}{dx}(uv) = (\frac{\sin{(x)}}{\cos{(x)}}) (\frac{1}{\cos{(x)}})$ $+ (\frac{\sin{(x)}}{\cos{(x)}}) (\cos{(x)})$

$\frac{d}{dx}(uv) = \sec{(x)} \tan{(x)} + \sin{(x)}$

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

$f'(x) = \sec{(x)} \tan{(x)} + \sin{(x)}$

EXAMPLE 3

Derive: $x^{2} \sin^{2}{(x)}$

Solution

Step 1: List the product rule formula for reference:

$\frac{d}{dx}(uv) = uv' + vu'$

Step 2: Identify how many multiplicands you have in the given problem. In this problem, we have two. The first one is $x^{2}$ and the second one is $\sin^{2}{(x)}$ .

Step 3: In the chosen form of the product rule formula of this demonstration, we have the first multiplicand as $u$ and the second multiplicand as $v$ .

Thus, we have

$u = x^{2}$
$v = \sin^{2}{(x)}$

Step 4: Derive $u$ and $v$ individually:

$u' = 2x$
$u'$ used the power formula.

$v' = 2 \sin{(x)} \cos{(x)}$
$v'$ used the the chain rule formula followed by the derivative formula for trigonometric functions.

Step 5: Apply the product rule formula now by substituting the $u$ , $u'$ , $v$ , and $v'$ in the product rule formula.

$\frac{d}{dx}(uv) = uv' + vu'$

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

Step 6: Simplify algebraically and since we have a trigonometric function in our derivative, we can also apply some trigonometric identities applicable in our solution:

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

$\frac{d}{dx}(uv) = 2x^{2} \sin{(x)} \cos{(x)} + 2x \sin^{2}{(x)}$

$\frac{d}{dx}(uv) = x^{2} (2 \sin{(x)} \cos{(x)}) + 2x \sin^{2}{(x)}$

$\frac{d}{dx}(uv) = x^{2} \sin{(2x)} + 2x \sin^{2}{(x)}$

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

$\frac{d}{dx}(uv) = x^{2} \sin{(2x)} + 2x \sin^{2}{(x)}$

EXAMPLE 4

What is the derivative of $f(x) = 5x^7 \cot{(x^7)}$ ?

Solution

We start by listing the product rule for reference:

$\frac{d}{dx}(uv) = uv' + vu'$

In this problem, we have two multiplicands in the function f(x). The first multiplicand is $5x^7$ and the other one is $\cot{(x^7)}$ .

We can set $u$ as the first multiplicand and $v$ as the second multiplicand.

Therefore, we have

$u = 5x^7$
$v = \cot{(x^7)}$
$f(x) = uv$

Now, we can use the product rule formula:

$f'(x) = uv' + vu'$

$\frac{d}{dx}f(x) = u \cdot \frac{d}{dx}(v) + v \cdot \frac{d}{dx}(u)$

$\frac{d}{dx}f(x) = 5x^7 \cdot \frac{d}{dx}(\cot{(x^7)}) + \cot{(x^7)} \cdot \frac{d}{dx}(5x^7)$

Note: The derivative of $u$ is found using the power rule formula and the derivative of $v$ is found using the chain rule formula and the derivative formula for the trigonometric function.

By applying the product rule formula along with the other derivative formulas to be used for $\textbf{\emph{u'}}$ and $\textbf{\emph{v'}}$ , we have:

$\frac{d}{dx}f(x) = 5x^7 \cdot (-7x^6 \csc^{2}{(x^7)}) + \cot{(x^7)} \cdot (35x^6)$

Simplifying algebraically, we get

$\frac{d}{dx}f(x) = -35x^{13} \csc^{2}{(x^7)} + 35x^6 \cot{(x^7)}$

And the final answer is:

$f'(x) = 35x^6 \cot{(x^7)} - 35x^{13} \csc^{2}{(x^7)}$

EXAMPLE 5

What is the derivative of $f(x) = x^7 \sin{(\sin^{-1}{(x)})}$ ?

Solution

We list down the product rule formula for our reference:

$\frac{d}{dx}(uv) = uv' + vu'$

Here, the first multiplicand is $x^7$ and the second multiplicand is $\sin{(\sin^{-1}{(x)})}$ .

We will use $u$ for the first multiplicand and $v$ for the second multiplicand.

Therefore, we have

$u = x^7$
$v = \sin{(\sin^{-1}{(x)})}$
$f(x) = uv$

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

$f'(x) = uv' + vu'$

$\frac{d}{dx}f(x) = u \cdot \frac{d}{dx}(v) + v \cdot \frac{d}{dx}(u)$

$\frac{d}{dx}f(x) = x^7 \cdot \frac{d}{dx}(\sin{(\sin^{-1}{(x)})})$ $+ \sin{(\sin^{-1}{(x)})} \cdot \frac{d}{dx}(x^7)$

Note: In this problem, we derivative $u$ using the power rule formula and derive $v$ using the derivative formulas for trigonometric function and inverse trigonometric function.

By applying the product rule formula along with the other derivative formulas to be used for $\textbf{\emph{u'}}$ and $\textbf{\emph{v'}}$ , we have:

$\frac{d}{dx}f(x) = x^7 \cdot (\cos{(\sin^{-1}{(x)})} (\frac{1}{\sqrt{1-x^2}}))$ $+ \sin{(\sin^{-1}{(x)})} \cdot (7x^6)$

Simplifying algebraically and applying trigonometric and inverse trigonometric operations and identities, we get

$\frac{d}{dx}f(x) = x^7 \cdot (1) + 7x^6 \cdot (x)}$

And the final answer is:

$f'(x) = 8x^7$

→ Derivatives Calculator

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.