Quotient Rule

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. 

Quotient Rule Quotient Rule

Create learning materials about Quotient Rule with our free learning app!

  • Instand access to millions of learning materials
  • Flashcards, notes, mock-exams and more
  • Everything you need to ace your exams
Create a free account
Contents
Table of contents

    An example of a quotient function is \(y = \frac{3x}{2x + 2}\) or \(y = \frac{x^2}{3x}\).

    Quotient rule formula

    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\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)

    Examples using the quotient rule formula

    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 \(\frac{dy}{dx}\):

    \(\frac{dy}{dx} = \frac{(3x +5)(\cos x) - (\sin x)(3)}{(3x + 5)^2}\)

    \(\frac{dy}{dx} = \frac{3x \cos x + 5 \cos x - 3\sin x}{(3x + 5)^2}\)

    Examples using the function notation

    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 \(f'(x)\):

    \(f'(x) = \frac{(\ln x) (2) - (2x + 2) (\frac{1}{x})}{(\ln x)^2}\) \(f'(x) = \frac{2x \ln x - 2x -2}{x(\ln x)^2}\)

    How to solve problems using the quotient rule

    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:

    \(\frac{dy}{dx} = \frac{3x^2 + 12x}{(3x +6)^2}\)

    \(\frac{dy}{dx} = \frac{3(2)^2 + 12(2)}{(3(2) +6)^2}\)

    \(\frac{dy}{dx} = \frac{36}{144}\)

    Quotient Rule - Key takeaways

    • 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 \(y = \frac{u}{v}\)

    • The formula can also be written in function notation when \(f(x) = \frac{g(x)}{h(x)}\) then\(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)
    Frequently Asked Questions about Quotient Rule

    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

    Discover learning materials with the free StudySmarter app

    Sign up for free
    1
    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
    StudySmarter Editorial Team

    Team Math Teachers

    • 6 minutes reading time
    • Checked by StudySmarter Editorial Team
    Save Explanation

    Study anywhere. Anytime.Across all devices.

    Sign-up for free

    Sign up to highlight and take notes. It’s 100% free.

    Join over 22 million students in learning with our StudySmarter App

    The first learning app that truly has everything you need to ace your exams in one place

    • Flashcards & Quizzes
    • AI Study Assistant
    • Study Planner
    • Mock-Exams
    • Smart Note-Taking
    Join over 22 million students in learning with our StudySmarter App

    Get unlimited access with a free StudySmarter account.

    • Instant access to millions of learning materials.
    • Flashcards, notes, mock-exams, AI tools and more.
    • Everything you need to ace your exams.
    Second Popup Banner