Product Rule of Derivatives – Examples with Answers

Differentiation problems that involve the product of functions can be solved using the product rule formula. This formula allows us to derive a product of functions, such as but not limited to fg(x) = f(x)g(x).

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

CALCULUS

Relevant for

Exploring examples with answers of the product rule.

See examples

CALCULUS

Relevant for

Exploring examples with answers of the product rule.

See examples

Summary of the product rule

The product rule is a very useful tool for deriving a product of at least two functions. It is a rule that states that the derivative of a product of two functions is equal to the first function f(x) in its original form multiplied by the derivative of the second function g(x) and then added to the original form of the second function g(x) multiplied by the derivative of the first function f(x).

This gives us the formula for the product rule as:

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

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

$$\frac{d}{dx}(uv) = uv’ + vu’$$

where

• $latex u = f(x)$ or the first multiplicand in the given problem
• $latex v = g(x)$ or the second multiplying in 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:

$latex fg(x) = f(x) \cdot g(x)$

or

$latex F(x) = uv$

where $latex f(x)$ or $latex u$ is the first multiplicand while $latex g(x)$ or $latex v$ is the second multiplicand of the given problem.

Product of rule of derivatives – Examples with answers

EXAMPLE 1

Find the derivative of the following function:

$latex f(x) = \sqrt[5]{x^3} \cdot (x^5 + 3x^2 – 4x)$

The first thing to do is to write the product rule formula for our reference:

$$\frac{d}{dx}(uv) = uv’ + vu’$$

We have two multipliers in the given function f(x). The first multiplicand is $latex \sqrt[5]{x^3}$ and the other is $latex (x^5 + 3x^2 – 4x)$.

Therefore, we have

$latex u = \sqrt[5]{x^3}$
$latex v = (x^5 + 3x^2 – 4x)$
$latex f(x) = uv$

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

$latex f'(x) = uv’ + vu’$

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

$$\frac{d}{dx}f(x) = (\sqrt[5]{x^3}) \cdot \frac{d}{dx}(x^5 + 3x^2 – 4x)+ (x^5 + 3x^2 – 4x) \cdot \frac{d}{dx}(\sqrt[5]{x^3})$$

For any radical, it is advisable to rewrite them in fractional exponent form:

$$\frac{d}{dx}f(x) = (x^3)^{\frac{1}{5}} \cdot \frac{d}{dx}(x^5 + 3x^2 – 4x)+ (x^5 + 3x^2 – 4x) \cdot \frac{d}{dx}((x^3)^{\frac{1}{5}})$$

$$\frac{d}{dx}f(x) = (x^3)^{\frac{1}{5}} \cdot (5x^4 + 6x – 4)+ (x^5 + 3x^2 – 4x) \cdot (\frac{1}{5} \cdot (x^3)^{-\frac{4}{5}} \cdot 3x^2)$$

Simplifying algebraically, we obtain

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

$$f'(x) = 5x^{\frac{23}{5}} + 6x^{\frac{8}{5}} – 4x^{\frac{3}{5}}+ \frac{3}{5}x^{\frac{23}{5}} + \frac{9}{5}x^{\frac{8}{5}} – \frac{12}{5}x^{\frac{3}{5}}$$

$$f'(x) = \frac{28}{5}x^{\frac{23}{5}} + \frac{39}{5}x^{\frac{8}{5}} – \frac{32}{5}x^{\frac{3}{5}}$$

$$f'(x) = \frac{28x^{\frac{23}{5}} + 39x^{\frac{8}{5}} – 32x^{\frac{3}{5}}}{5}$$

$$f'(x) = \frac{28\sqrt[5]{x^{23}} + 39\sqrt[5]{x^8} – 32\sqrt[5]{x^3}}{5}$$

EXAMPLE 2

Derive the following function:

$latex f(x) = (5x^5-x^4) \cdot (30x-12x^2)$

Based on the given, we have two multipliers in the given function f(x). The first multiplicand is $latex (5x^5-x^4)$ and the second one is $latex (30x-12x^2)$.

If $latex u$ is the first multiplicand and $latex v$ is the second multiplicand, we have

$latex u = (5x^5-x^4)$
$latex v = (30x-12x^2)$
$latex f(x) = uv$

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

$latex f'(x) = uv’ + vu’$

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

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

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

Simplifying algebraically, we obtain

$$f'(x) = [-120x^6+174x^5-30x^4]+ [-300x^6+798x^5-120x^4]$$

$$f'(x) = -120x^6-300x^6+174x^5+798x^5-30x^4-120x^4$$

$$f'(x) = -420x^6+972x^5-150x^4$$

EXAMPLE 3

What is the derivative of the following function?

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

We have two multiplicands in the given function f(x). The first multiplicand is $latex 6x^3$ and the other is $latex \ln{(x)}$:

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

Using the product rule, we have:

$latex f'(x) = uv’ + vu’$

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

$$\frac{d}{dx}f(x) = 6x^3 \cdot \frac{d}{dx}(\ln{(x)}) + \ln{(x)} \cdot \frac{d}{dx}(6x^3)$$

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

Simplifying, we obtain

$$f'(x) = 6x^2 + 18x^2 \cdot \ln{(x)}$$

$latex f'(x) = 6x^2 + 18x^2\ln{(x)}$

EXAMPLE 4

Find the derivative of:

$latex f(x) = 9x^3 \cdot \sec{(\pi x)}$

We have two multiplicands in the given function f(x). The first multiplicand is $latex 9x^3$ and the other is $latex \sec{(\pi x)}$:

$latex u = 9x^3$
$latex v = \sec{(\pi x)}$
$latex f(x) = uv$

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

$latex f'(x) = uv’ + vu’$

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

$$\frac{d}{dx}f(x) = 9x^3 \cdot \frac{d}{dx}(\sec{(\pi x)})+ \sec{(\pi x)} \cdot \frac{d}{dx}(9x^3)$$

$$\frac{d}{dx}f(x) = 9x^3 \cdot (\pi \sec{(\pi x)} \tan{(\pi x)})+ \sec{(\pi x)} \cdot (27x^2)$$

Simplifying, we obtain

$$f'(x) = 9\pi x^3 \cdot (\sec{(\pi x)} \tan{(\pi x)})+ 27x^2 \cdot \sec{(\pi x)}$$

$$f'(x) = 9\pi x^3 \sec{(\pi x)} \tan{(\pi x)}+ 27x^2 \sec{(\pi x)}$$

EXAMPLE 5

Derive the following function:

$latex f(x) = 5^x \cdot (x+5)^5$

We have two multiplicands in the given function f(x). The first multiplicand is $latex 5^x$ and the other is $latex (x+5)^5$:

$latex u = 5^x$
$latex v = (x+5)^5$
$latex f(x) = uv$

Applying the product rule, we have:

$latex f'(x) = uv’ + vu’$

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

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

$$\frac{d}{dx}f(x) = 5^x \cdot (5 \cdot (x+5)^4 \cdot (1))+ (x+5)^5 \cdot (5^x \cdot \ln{(5)} \cdot (1))$$

When we simplify, we have:

$$f'(x) = 5(x+5)^4 \cdot 5^x + (x+5)^5 \cdot \ln{(x)}$$

$$f'(x) = 5(x+5)^4 \cdot 5^x + (x+5)^5 \ln{(x)}$$

EXAMPLE 6

Find the derivative of the following function

$latex f(x) = 5x^7 \cot{(x^7)}$

Based on the question, we have two multiplicands in the given function f(x). The first multiplicand is $latex 5x^7$ and the other is $latex \cot{(x^7)}$.

Therefore, we have

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

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

$latex f'(x) = uv’ + vu’$

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

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

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

We can simplify as follows:

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

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

EXAMPLE 7

Find the derivative of the given function:

$latex f(x) = x^7 \sin{(\sin^{-1}{(x)})}$

We have two multiplicands in the given function f(x). The first multiplicand is $latex x^7$ and the other is $latex \sin{(\sin^{-1}{(x)})}$.

Therefore, we have

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

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

$latex f'(x) = uv’ + vu’$

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

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

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

Simplifying algebraically, applying identities and trigonometric operations and using trigonometric inverses, we obtain

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

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

EXAMPLE 8

What is the derivative of the following function?

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

We have two multiplicands in the given function f(x). The first multiplicand is $latex 5x^x$ and the other is $latex \cos^{3}(x)$.

Therefore, we have

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

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

$latex f'(x) = uv’ + vu’$

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

$$\frac{d}{dx}f(x) = 5x^x \cdot \frac{d}{dx}(\cos^{3}{(x)}) + \cos^{3}{(x)} \cdot \frac{d}{dx}(5x^x)$$

Note: In this problem, the derivative of $latex u$ uses implicit differentiation, while the derivative of $latex v$ uses the chain rule formula and the derivative formula for the trigonometric function.

$$\frac{d}{dx}f(x) = 5x^x \cdot (-3 \cos^{2}{x} \sin{x})+ \cos^{3}{(x)} \cdot (5x^x(\ln{(x)}+1))$$

Simplifying algebraically, the final answer is:

$$f'(x) = 5x^x(\ln{(x)}+1) \cos^{3}{(x)}– 15x^x \cos^{2}{x} \sin{x}$$

EXAMPLE 9

Find the derivative of the following function

$latex x^{e^x} e^{\sin{(x)}}$

We have two multiplicands in the given function f(x). The first multiplicand is $latex x^{e^x}$ and the other is $latex e^{\sin(x)}$.

Therefore, we have

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

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

$latex f'(x) = uv’ + vu’$

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

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

Note: In this problem, the derivative of $latex u$ will use implicit differentiation, while the derivative of $latex v$ will use derivative formulas for exponential and trigonometric functions.

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

Simplifying algebraically, the final answer is:

$$f'(x) = x^{e^x} e^{sin(x)} \cos{(x)} \hspace{1.15 pt} + \hspace{1.15 pt} x^{e^x} e^{\sin(x)} (e^x \ln{(x)}+\frac{e^x}{x})$$

EXAMPLE 10

Find the derivative of $latex f(x) = (x^3-2x)^3 \cot^{-1}{(x^3-2x)}$.

We have two multiplicands in the given function f(x). The first multiplicand is $latex (x^3-2x)^3$ and the other is $latex \cot^{-1}{(x^3-2x)}$.

Therefore, we have

$latex u = (x^3-2x)^3$
$latex v = \cot^{-1}{(x^3-2x)}$
$latex f(x) = uv$

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

$latex f'(x) = uv’ + vu’$

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

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

Note: In this problem, the derivative of $latex u$ will use the chain rule formula, while the derivative of $latex v$ will use the formula of the derivative for an inverse trigonometric function and the power rule.

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

Simplifying algebraically, the final answer is:

$$f'(x) = (x^3-2x)^2 (9x-6) \cot^{-1}{(x^3-2x)}– \frac{(x^3-2x)^3 (3x^2-2)}{(x^3-2x)^2+1}$$

Product rule of derivatives – Practice problems

Product rule of derivatives quiz
You have completed the quiz!

Find the derivative of the following function and determine the value of $latex F^{\prime}(0)$: $latex F(x) = \sin(x^2+2x)\cos(x)$?

Write the answer in the input box.

$latex F^{\prime}(0)=$