Jump to a key chapter
Product rule formula and function notation
There is a formula that you can follow to use the product rule. It is important to try and remember this formula as it is not usually in exam formula booklets. If when u and v are functions of x then the product rule formula is:
\[\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\]
This can also be written in function notation,
If \(f(x) = g(x)h(x)\) then \(f'(x) = g(x)h'(x) + h(x)g'(x)\)
Examples of the product rule
To understand the product rule further, let's look through some examples of using it.
If \(y = 5xe^2\) find \(\frac{dy}{dx}\):
First, you can start by looking at your product rule formula and finding each aspect of the formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
If \(y = 5xe^2\), \(u = 5x\) and \(v = e^2\)
To find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\) you can differentiate your u and v:
\(\frac{du}{dx} = 5\) \(\frac{dv}{dx} = 0\)
Now that you have all aspects of your formula, you can solve it to find \(\frac{dy}{dx}\):\[\frac{dy}{dx} = (5x)(0) + (e^2)(5) = 5e^2\]
You may also be asked to differentiate a trigonometric function using the product rule.
If \(y = (4\sin{x})e^2\) find \(\frac{dy}{dx}\)
As before, you can start by finding each part of the formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
If \(y = (4\sin{x})e^2\), \(u = 4\sin x\) and \(v = e^2\)
Now you can differentiate your u and v to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):
\(\frac{du}{dx} = 4\cos x\) \(\frac{dv}{dx} = 0\)
Finally, you can substitute each part into your formula to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = (4\sin x)(0) + (e^2)(4\cos x)\]
\[\frac{dy}{dx} = 4e^2\cos x\]
You may even be asked to use the product rule in function notation, so let's work through an example.
If \(f(x) = 2x^2(x^2 + 4)\) find \(f'(x)\)
Once again, you can start by breaking down the formula for the product rule in function notation and finding each part.
\(f'(x) = g(x)h'(x) + h(x)g'(x)\)
If \(f(x) = 2x^2(x^2 + 4)\), \(g(x) = 2x^2\) and \(h(x) = x^2 + 4\)
Next, you can differentiate g(x) and h(x) to find the derivatives, g'(x) and h'(x):
\(g' = 4x\) \(h'(x) = 2x\)
Now that you have each part of the formula, you can solve it to find f'(x):
\(f'(x) = (2x^2)(2x) + (x^2 + 4)(4x) = 2(4x^3 + 8x)\)
If \(y = \ln x(x^2)\) find \(\frac{dy}{dx}\)
To begin with, let's look at the product rule formula and find each aspect of it:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
If \(y = \ln x(x^2)\), \(u = \ln x\) and \(v = x^2\)
Let's differentiate the terms to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):
\(\frac{du}{dx} = \frac{1}{x}\) \(\frac{dv}{dx} = 2x\)
Now you can put each part into the formula to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
\[\frac{dy}{dx} = (\ln{x})(2x) + (x^2)(\frac{1}{x}) = 2x \ln{x} + x\]
Product Rule - Key takeaways
The product rule is one of the differentiation rules.
The product rule can be used when differentiating the products of two functions.
When using the product rule, you can either use the formula in y form or in the function notation form.
You may also need to differentiate trigonometric functions using the product rule.
Learn with 0 Product Rule flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
Already have an account? Log in
Frequently Asked Questions about Product Rule
What is the product rule?
The product rule is one of the differentiation rules that is used when you are differentiating the products of two functions.
When do you use the product rule or the chain rule?
You use the product rule when you are dealing with the products of two functions, and you use the chain rule when you are differentiating a composite function, also known as a function of a function.
How do you derive the product rule?
The product rule can be derived from the definition of a derivative, dy/dx f(x) = lim (f(x+h)-f(x))/h
About StudySmarter
StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.
Learn more