Learning Materials
Features
Discover
The quotient rule is a rule used when you are differentiating a quotient function. A quotient function can be described as a function that is being divided by another function.
Millions of flashcards designed to help you ace your studies
Sign up for freeWhat is Vaia?
How does Vaia help me study more efficiently?
Where can I find more explanations like this?
What's smart about Vaia's flashcards?
Can I create my own content on Vaia?
How does spaced repetition work in Vaia flashcards?
What can you do with flashcards in Vaia?
Is Vaia a science-based learning platform?
How do Vaia's smart learning plans support your exam prep?
Can you create your own study sets in Vaia?
What is Vaia?
How does Vaia help me study more efficiently?
Where can I find more explanations like this?
What's smart about Vaia's flashcards?
Can I create my own content on Vaia?
How does spaced repetition work in Vaia flashcards?
What can you do with flashcards in Vaia?
Is Vaia a science-based learning platform?
How do Vaia's smart learning plans support your exam prep?
Can you create your own study sets in Vaia?
Achieve better grades quicker with Premium
Geld-zurück-Garantie, wenn du durch die Prüfung fällst
Did you know that StudySmarter supports you beyond learning?
Review generated flashcards
Start learning or create your own AI flashcards
Team Quotient Rule Teachers
An example of a quotient function is \(y = \frac{3x}{2x + 2}\) or \(y = \frac{x^2}{3x}\).
There is a formula that can be used when using the quotient rule to differentiate It is shown below:
If \(y = \frac{u}{v}\) then\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)
This formula can also be written in function notation:
When \(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)
If \(y = \frac{2x^2}{2x +2}\) find \(\frac{dy}{dx}\)
To begin you can look at the formula and find each part that you need:
\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)
\(u = 2x^2\) \(v = 2x +2\) \(\frac{du}{dx} = 4x\) \(\frac{dv}{dx} = 2\)Next, you can substitute each variable that you have found into the formula:
\(\frac{dy}{dx} = \frac{(2x + 2)(4x) - (2x^2)(2)}{(2x + 2)^2}\)
You're now able to simplify and solve your formula to find :
\(\frac{dy}{dx} = \frac{(8x^2 + 8x) - (4x^2)}{(2x + 2)^2}\)\(\frac{dy}{dx} = \frac{4x^2 + 8x}{(2x + 2)^2}\)
Let's look at an example where a trigonometric function is involved to see how you would solve for \(\frac{dy}{dx}\):
If \(y = \frac{\sin x}{3x + 5}\) find \(\frac{dy}{dx}\).
As before, it is good to start by identifying the formula you would need and breaking it down to find each part of the equation. You know that because there is a fraction involved in the question you can use the quotient rule formula. Let's take a look at the formula and find each part of it:
\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)
\(u = \sin x\) \(v = 3x +5\) \(\frac{du}{dx} = \cos x\) \(\frac{dv}{dx} = 3\)
Now that you have identified each part of the formula you can substitute the parts into the equation to solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{(3x +5)(\cos x) - (\sin x)(3)}{(3x + 5)^2}\)
\(\frac{dy}{dx} = \frac{3x \cos x + 5 \cos x - 3\sin x}{(3x + 5)^2}\)It is useful to also know how to use the quotient rule in terms of its function notation as this may be how it appears within the exam question. Let's remind ourselves of the formula for the function notation before working through some examples!
When \(f'(x) = \frac{g(x)}{h(x)}\)then\(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)
If \(f(x) = \frac{x}{3x^3 + 2}\) find \(f'(x)\).
Once again it is good to start by identifying the formula needed and each part of it. Since there is a quotient involved and the question is written in function form, you know that you need to use the quotient rule in function notation:
\(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)
\(g(x) = x\) \(h(x) = 3x^3 + 2\) \(g'(x) = 1\) \(h'(x) = 9x^2\)
Next, you can substitute each part you have identified into the formula to find \(f'(x)\):
\(f'(x) = \frac{(3x^3 + 2)(1) - (x)(9x^2)}{(3x^3 + 2)^2}\)
\(f'(x) = \frac{(3x^3 + 2) - (9x^3)}{(3x^3 + 2)^2}\)
\(f'(x) = \frac{-6x^3 + 2}{(3x^3 + 2)^2}\)
Let's work through another example.
If \(f(x) = \frac{2x + 2}{\ln x}\) find \(f'(x)\).
You can start by looking at your formula for the function notation of the quotient rule and getting each part of the equation ready to solve:
\(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)
\(g(x) = 2x +2\) \(h(x) = \ln x\) \(g'(x) = 2\) \(h'(x) = \frac{1}{x}\)
Now you can substitute each part into the formula to solve for \(f'(x)\):
\(f'(x) = \frac{(\ln x) (2) - (2x + 2) (\frac{1}{x})}{(\ln x)^2}\) \(f'(x) = \frac{2x \ln x - 2x -2}{x(\ln x)^2}\)
Since functions can be represented visually using graphs, sometimes you may need to solve a question based on the points the function may cross. To do this, you can still simply use the quotient rule formula if it applies, then with some extra steps afterwards you will be able to find the value.
Find the value of \(\frac{dy}{dx}\) for the point (2, 1/3) on the curve where \(y = \frac{x^2}{3x+6}\)
For this type of question you would still start in the same way as before, identify your formula and find each part of it:
\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)
\(u = x^2\) \(v = 3x +6\) \(\frac{du}{du} = 2x\) \(\frac{dv}{dx} = 3\)
Again, just like before you substitute each part into the formula to solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{(3x + 6) (2x) - (x^2)(3)}{(3x +6)^2}\)
\(\frac{dy}{dx} = \frac{6x^2 + 12x - 3x^2}{(3x +6)^2}\)
\(\frac{dy}{dx} = \frac{3x^2 + 12x}{(3x +6)^2}\)
Now, because you are looking to find the value of \(\frac{dy}{dx}\) when the point of the curve is (2, 1/3), you can substitute the x coordinate into the equation above:
\(\frac{dy}{dx} = \frac{3x^2 + 12x}{(3x +6)^2}\)
\(\frac{dy}{dx} = \frac{3(2)^2 + 12(2)}{(3(2) +6)^2}\)
\(\frac{dy}{dx} = \frac{36}{144}\)
The quotient rule is a rule used in differentiation. It is used when you are differentiating a quotient, which is a function that is being divided by another function.
The formula for the quotient rule is \(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\) if \(y = \frac{u}{v}\)
Sign up for free to gain access to all our flashcards.
What is the quotient rule?
The quotient rule is a rule that is used when differentiating.
When do you use the quotient rule?
You can use the quotient rule when you are differentiating a quotient function, this is when a function is being divided by another function.
What is the formula for the quotient rule?
The formula for the quotient rule is dy/dx = (v du/dx - u dv/dx)/v^2 if y = u/v , this can also be written in function notation, when f(x)=g(x)/h(x) then f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2
At StudySmarter, we have created a learning platform that serves millions of students. Meet the people who work hard to deliver fact based content as well as making sure it is verified.
Lily Hulatt is a Digital Content Specialist with over three years of experience in content strategy and curriculum design. She gained her PhD in English Literature from Durham University in 2022, taught in Durham University’s English Studies Department, and has contributed to a number of publications. Lily specialises in English Literature, English Language, History, and Philosophy.
Get to know Lily
Gabriel Freitas is an AI Engineer with a solid experience in software development, machine learning algorithms, and generative AI, including large language models’ (LLMs) applications. Graduated in Electrical Engineering at the University of São Paulo, he is currently pursuing an MSc in Computer Engineering at the University of Campinas, specializing in machine learning topics. Gabriel has a strong background in software engineering and has worked on projects involving computer vision, embedded AI, and LLM applications.
Get to know Gabriel
Vaia 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 moreSave explanations to your personalised space and access them anytime, anywhere!
Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of Vaia.
Already have an account? Log in
Sign up to highlight and take notes. It’s 100% free.
Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App 
