Product Rule - Challenging Examples and Practice Problems

Derivation exercises that involve the product of functions can be solved using the product rule formula. This formula allows us to derive a product of function 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.

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CALCULUS

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Exploring examples with answers of the product rule.

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Summary of The Product Rule

The product rule is a very helpful tool to derive 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 product rule formula as:

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

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

where

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

You can use any of these two forms of the product rule formula depending on your preference.

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

$fg(x) = f(x) \cdot g(x)$
or
$F(x) = uv$

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

Product Rule – Examples with Answers

Using the formula detailed above, we can derive various functions that are written 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

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

Solution

The first thing we always need to do is to list down the product rule formula for our reference:

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

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

Based on the product rule formula, let $u$ be the first multiplicand and $v$ the second multiplicand.

Therefore, we have

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

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

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

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

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

Note: In this sample problem, the derivative of $u$ will use the power rule formula whilst the derivative of $v$ will use 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 2

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

Solution

The first thing we always need to do is to list down the product rule formula for our reference:

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

Based on our given, we have two multiplicands in the given function f(x). The first multiplicand is $x^7$ and the other one is $\sin{(\sin^{-1}{(x)})}$ .

Based on the product rule formula, let $u$ be the first multiplicand and $v$ the second multiplicand.

Therefore, we have

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

After doing this, we can now 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 sample problem, the derivative of $u$ will use the power rule formula whilst the derivative of $v$ will use 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:

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

EXAMPLE 3

Find the derivative of $f(x)=5x^x \cos{3}{x}$

Solution

Based on our given, we have two multiplicands in the given function f(x). The first multiplicand is $5x^x$ and the other one is $\cos^{3}{(x)}}$ .

Based on the product rule formula, let $u$ be the first multiplicand and $v$ the second multiplicand.

Therefore, we have

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

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

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

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

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

Note: In this sample problem, the derivative of $u$ will use implicit differentiation whilst the derivative of $v$ will use the chain rule formula and 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:

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

Simplifying algebraically, the final answer is:

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

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EXAMPLE 4

Find the the derivative of $x^{e^x} e^{\sin{(x)}}$

Solution

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

Based on the product rule formula, let $u$ be the first multiplicand and $v$ the second multiplicand.

Therefore, we have

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

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

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

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

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

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

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

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

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 5

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

Solution

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

Based on the product rule formula, let $u$ be the first multiplicand and $v$ the second multiplicand.

Therefore, we have

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

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

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

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

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

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

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:

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

Simplifying algebraically, the final answer is:

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

EXAMPLE 6

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

Solution

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

Based on the product rule formula, let $u$ be the first multiplicand and $v$ the second multiplicand.

Therefore, we have

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

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

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

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

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

Note: In this sample problem, the derivative of $u$ will use implicit differentiation and the trigonometric function’s derivative formula whilst the derivative of $v$ will only use the derivative formula for the trigonometric function.

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

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

Simplifying algebraically and applying trigonometric operations and identities, the final answer is:

$f'(x) = \sin^{x}{(x)} \hspace{1.15 pt} \cdot \hspace{1.15 pt} [(cos(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} (x \cot{(x)}$ $+\ln{(\sin{(x)})}) \hspace{1.15 pt}) - \sin{(x)}]$

EXAMPLE 7

Find the derivative of $f(x) = x^3 e^x \tan{(x)}$

Solution

Based on our given, we have three multiplicands in the given function f(x). The first multiplicand is $x^3$ , the second one is $e^x$ , and the third one is $\tan{(x)}$ .

BUT, how do we derive a problem that consists of a product of three functions or three multiplicands?

It may seem difficult but it is actually easy.

Let:

$u$ = first multiplicand
$v$ = second multiplicand
$w$ = third multiplicand

Then we have our adjusted product rule formula,

$\frac{d}{dx}(uvw) = [uv \cdot (w)'] + [w \cdot (uv)']$

in which once we expand $(uv)'$ , we have

$\frac{d}{dx}(uvw) = [uv \cdot (w)'] + [w \cdot (uv' + vu')]$

This is the product rule formula we will use for this problem with three multiplicands. Therefore, we have

$u = x^3$
$v = e^x$
$w = \tan{(x)}$
$f(x) = uvw$

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

$f'(x) = [uv \cdot (w)'] + [w \cdot (uv' + vu')]$

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

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

Note: In this sample problem, the derivative of $u$ will use the power rule formula. The derivative of $v$ will use the derivative formula for an exponential function. Then lastly, the derivative of $w$ will use the derivative formula for the trigonometric function.

By applying the adjusted product rule formula shown in this example’s solution formula together with the other derivative formulas to be used for $\textbf{\emph{u'}}$ , $\textbf{\emph{v'}}$ and $\textbf{\emph{w'}}$ ; we have:

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

Simplifying algebraically, the final answer is:

$f'(x) = e^3 \hspace{1.15} [x^3 \sec^{2}{(x)} \hspace{1.15} + \hspace{1.15} (x^3+3x^2) \tan{(2)}]$

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Product Rule – Practice problems

Solve the following derivation 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.

Find the derivative of $f(x)=(9x^3+3x)^4 \sin^2{(x+5)}$

Choose an answer

$(9x^3+3x)^3 \sin{(x+5)} \cos{(x+5)}$
$+ \hspace{1.5 pt} (9x^3+3x)^4 (108x+12) \sin{(x+5)}$

$2(9x^3+3x)^3 (108x^2+12) \sin{(x+5)} \cos{(x+5)}$
$+ \hspace{1.5 pt} (9x^3+3x)^3 \sin^2{(x+5)}$

$2(9x^3+3x)^4 \sin{(x+5)} \cos{(x+5)}$
$+ \hspace{1.5 pt} (9x^3+3x)^3 (108x^2+12) \sin^2{(x+5)}$

$2 \sin^2{(x+5)} \cos{(x+5)}$
$+ \hspace{1.5 pt} (9x^3+3x)^4 (108x^2+12) \sin{(x+5)}$

What is the derivative of $f(x)=\sec^{-1}{(e^{x^2})} \cdot \tan^{-1}{(e^{x^2})}$ ?

Choose an answer

$\frac{2x \hspace{1.15 pt} \tan^{-1}{(e^{x^2})}}{e^{x^2} \cdot \sqrt{1-e^{2x^2}}} \hspace{1.15 pt} + \hspace{1.15 pt} \frac{2x \hspace{1.15 pt} \sec^{-1}{(e^{x^2})}}{1 \hspace{1.15 pt} + \hspace{1.15 pt} e^{2x^2}}$

$\frac{2x \hspace{1.15 pt} \tan^{-1}{(e^{x^2})}}{\sqrt{e^{2x^2}-1}} \hspace{1.15 pt} + \hspace{1.15 pt} \frac{2xe^{x^2} \hspace{1.15 pt} \sec^{-1}{(e^{x^2})}}{1 \hspace{1.15 pt} + \hspace{1.15 pt} e^{2x^2}}$

$\frac{2xe^{x^2} \hspace{1.15 pt} \tan^{-1}{(e^{x^2})}}{e^{2x^2} \cdot \sqrt{e^{2x^2}+1}} \hspace{1.15 pt} + \hspace{1.15 pt} \frac{2xe^{2x^2} \hspace{1.15 pt} \sec^{-1}{(e^{x^2})}}{1 \hspace{1.15 pt} + \hspace{1.15 pt} e^{x^2}}$

None of the above

Derive: $f(x)=(x+2)^{x+2} \cdot \ln{(e^{2^{x}})}$

Choose an answer

$2^x \hspace{1.15 pt} [\ln{(x+2)} \hspace{1.15 pt} + \hspace{1.15 pt} 1}]$

$(x+2)^{x+2} \hspace{1.15 pt} [\ln{(2)} \hspace{1.15 pt} + \hspace{1.15 pt} \ln{(x+2)}}]$

$2^x(x+2)^{x+2} \hspace{1.15 pt} [\ln{(2)} \hspace{1.15 pt} + \hspace{1.15 pt} \ln{(x+2)} \hspace{1.15 pt} + \hspace{1.15 pt} 1}]$

Differentiate $y=10^x \cdot \log{(\sqrt{x^{10}-5x}})}$

Choose an answer

$\frac{10^x (5x^9 - \frac{5}{2}) \log{(e)}}{x^{10} - 5x} \hspace{1.15 pt} + \hspace{1.15 pt} \frac{10^x \ln{(x^{10} - 5x)}}{2}$

$\frac{10^x (5x^9 - \frac{5}{2})}{x^{10} - 5x} \hspace{1.15 pt} + \hspace{1.15 pt} \frac{10^x \log{(x^10 - 5x)}}{2}$

$\frac{10^x (5x^9 - \frac{5}{2})}{\ln{(x^{10}-5x)}} \hspace{1.15 pt} + \hspace{1.15 pt} \frac{10^x \ln{(x^{10} - 5x)}}{x^{10} - 5x}$

None of the above

Find the derivative of $y=x^{\pi} \tan^{-1}{(cos(\pi x))}$

Choose an answer

$\pi x^{\pi} \hspace{1.15 pt} \sin^{-1}{\cos{(\pi x)}} + \frac{\pi x^{\pi} \hspace{1.15 pt} \sin{(\pi x)}}{\sqrt{1 + \cos^{2}{(\pi x)}}}$

$\pi x^{\pi - 1} \hspace{1.15 pt} \sin^{-1}{\cos{(\pi x)}} + \frac{\pi x^{\pi} \hspace{1.15 pt} \sin{(\pi x)}}{\sqrt{1 - \cos^{2}{(\pi x)}}}$

$\pi x^{\pi - 1} \hspace{1.15 pt} \sin^{-1}{\cos{(\pi x)}} - \frac{\pi x^{\pi} \hspace{1.15 pt} \sin{(\pi x)}}{\sqrt{1 - \cos^{2}{(\pi x)}}}$

None of the above

Product Rule – Challenging Examples and Practice Problems

CALCULUS

CALCULUS

Summary of The Product Rule

Product Rule – Examples with Answers

EXAMPLE 1

Solution

EXAMPLE 2

Solution

EXAMPLE 3

Solution

EXAMPLE 4

Solution

EXAMPLE 5

Solution

EXAMPLE 6

Solution

EXAMPLE 7

Solution

Product Rule – Practice problems

Find the derivative of $f(x)=(9x^3+3x)^4 \sin^2{(x+5)}$

What is the derivative of $f(x)=\sec^{-1}{(e^{x^2})} \cdot \tan^{-1}{(e^{x^2})}$ ?

Derive: $f(x)=(x+2)^{x+2} \cdot \ln{(e^{2^{x}})}$

Differentiate $y=10^x \cdot \log{(\sqrt{x^{10}-5x}})}$

Find the derivative of $y=x^{\pi} \tan^{-1}{(cos(\pi x))}$

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Learn mathematics with our additional resources in different topics

Explore our additional Mathematics resources

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Product Rule – Challenging Examples and Practice Problems

CALCULUS

CALCULUS

Summary of The Product Rule

Product Rule – Examples with Answers

EXAMPLE 1

Solution

EXAMPLE 2

Solution

EXAMPLE 3

Solution

EXAMPLE 4

Solution

EXAMPLE 5

Solution

EXAMPLE 6

Solution

EXAMPLE 7

Solution

Product Rule – Practice problems

Find the derivative of

What is the derivative of ?

Derive:

Differentiate

Find the derivative of

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Learn mathematics with our additional resources in different topics

Explore our additional Mathematics resources

Find the derivative of $f(x)=(9x^3+3x)^4 \sin^2{(x+5)}$

What is the derivative of $f(x)=\sec^{-1}{(e^{x^2})} \cdot \tan^{-1}{(e^{x^2})}$ ?

Derive: $f(x)=(x+2)^{x+2} \cdot \ln{(e^{2^{x}})}$

Differentiate $y=10^x \cdot \log{(\sqrt{x^{10}-5x}})}$

Find the derivative of $y=x^{\pi} \tan^{-1}{(cos(\pi x))}$