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.
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:
$latex f(x) = \sqrt[5]{x^3} \cdot (x^5 + 3x^2 – 4x)$
Solution
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
$$f'(x) = x^{\frac{3}{5}} \cdot (5x^4 + 6x – 4)+ (x^5 + 3x^2 – 4x) \cdot (\frac{3}{5}x^{-\frac{2}{5}})$$
$$f'(x) = 5x^{\frac{23}{5}} + 6x^{\frac{8}{5}} – 4x^{\frac{3}{5}}+ \frac{3}{5}x^{\frac{23}{5}} + \frac{9}{5}x^{\frac{8}{5}} – \frac{12}{5}x^{\frac{3}{5}}$$
$$f'(x) = \frac{28}{5}x^{\frac{23}{5}} + \frac{39}{5}x^{\frac{8}{5}} – \frac{32}{5}x^{\frac{3}{5}}$$
And the final answer is:
$$f'(x) = \frac{28x^{\frac{23}{5}} + 39x^{\frac{8}{5}} – 32x^{\frac{3}{5}}}{5}$$
Or using radicals,
$$f'(x) = \frac{28\sqrt[5]{x^{23}} + 39\sqrt[5]{x^8} – 32\sqrt[5]{x^3}}{5}$$
EXAMPLE 2
Derive the following function:
$latex f(x) = (5x^5-x^4) \cdot (30x-12x^2)$
Solution
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
$$f'(x) = [-120x^6+174x^5-30x^4]+ [-300x^6+798x^5-120x^4]$$
$$f'(x) = -120x^6-300x^6+174x^5+798x^5-30x^4-120x^4$$
And the final answer is:
$$ f'(x) = -420x^6+972x^5-150x^4$$
EXAMPLE 3
What is the derivative of the following function?
$latex f(x) = 6x^3 \cdot \ln{(x)}$
Solution
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
$$f'(x) = 6x^2 + 18x^2 \cdot \ln{(x)}$$
And the final answer is:
$latex f'(x) = 6x^2 + 18x^2\ln{(x)}$
EXAMPLE 4
Find the derivative of:
$latex f(x) = 9x^3 \cdot \sec{(\pi x)}$
Solution
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
$$f'(x) = 9\pi x^3 \cdot (\sec{(\pi x)} \tan{(\pi x)})+ 27x^2 \cdot \sec{(\pi x)}$$
And the final answer is:
$$f'(x) = 9\pi x^3 \sec{(\pi x)} \tan{(\pi x)}+ 27x^2 \sec{(\pi x)}$$
EXAMPLE 5
Derive the following function:
$latex f(x) = 5^x \cdot (x+5)^5$
Solution
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:
$$f'(x) = 5(x+5)^4 \cdot 5^x + (x+5)^5 \cdot \ln{(x)}$$
And the final answer is:
$$f'(x) = 5(x+5)^4 \cdot 5^x + (x+5)^5 \ln{(x)}$$
EXAMPLE 6
Find the derivative of the following function
$latex f(x) = 5x^7 \cot{(x^7)}$
Solution
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:
$$\frac{d}{dx}f(x) = -35x^{13} \csc^{2}{(x^7)} + 35x^6 \cot{(x^7)}$$
And the final answer is:
$$f'(x) = 35x^6 \cot{(x^7)} – 35x^{13} \csc^{2}{(x^7)}$$
EXAMPLE 7
Find the derivative of the given function:
$latex f(x) = x^7 \sin{(\sin^{-1}{(x)})}$
Solution
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
$$\frac{d}{dx}f(x) = x^7 \cdot (1) + 7x^6 \cdot (x)$$
And the final answer is:
$latex f'(x) = 8x^7$
EXAMPLE 8
What is the derivative of the following function?
$latex f(x)=5x^x \cos{3}{x}$
Solution
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
$latex x^{e^x} e^{\sin{(x)}}$
Solution
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 $latex f(x) = (x^3-2x)^3 \cot^{-1}{(x^3-2x)}$.
Solution
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
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.
See also
Interested in more information about the product 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.
