Quotient Rule: Formula, Proof, Definition, Examples

Quotient Rule is a method for finding the derivative of a function that is the quotient of two other functions. It is a method used for differentiating problems where one function is divided by another. We use the quotient rule when we have to find the derivative of a function of the form: f(x)/g(x).

Let's learn about the Quotient Rule in Calculus, its formula, and derivation, with the help of solved examples.

Quotient Rule Definition

Quotient rule is the rule of differentiation of those functions that are given in the form of fractions, where both the numerator and the denominator are individual functions. The Quotient Rule is a fundamental technique in calculus for finding the derivative of a function that is the quotient (ratio) of two differentiable functions. It provides a method to differentiate expressions where one function is divided by another.

Suppose we are given a function f(x) = g(x)/h(x) then the differentiation of f(x), f'(x) is found as,

f'(x) = [g(x) × h'(x) - h(x) × g'(x)] / [h(x)]²

Quotient Rule Formula

The quotient rule formula is the formula used to find the differentiation of the function which is expressed as the quotient function. Below is the formula of quotient rule is,

d/dx [u(x)/v(x)] = [v(x) × u'(x) - u(x) × v'(x)] / [v(x)]²

Where,

• u(x) is the first function which is a differentiable function, 
• u'(x) is the derivative of function u(x), 
• v(x) is the second function which is a differentiable function, and
• v'(x) is the derivative of the function v(x).

Quotient Rule Proof

We can derive the quotient rule using the following methods:

• Using Chain Rule
• Using Implicit Differentiation
• Using Derivative and Limit Properties

Now let's learn about them in detail.

Derivation of Quotient Rule Using Chain Rule

To Prove: H'(x) = d/dx [f(x)/g(x)] = [f(x) × g'(x) - g(x) × h'(x)] / [g(x)]²

Given: H(x) = f(x)/g(x)

Proof:

H(x) = f(x)/g(x)
⇒ H(x) = f(x).g(x)^-1

Using Product Rule,
H'(x) = f(x). d/dx [g(x)^-1] + g(x)^-1. f'(x)

Applying the power rule,
H'(x) = f(x). (-1)[g(x)^-2.g'(x)] + g(x)^-1. f'(x)
⇒ H'(x) = - [f(x).g'(x)] / g(x)² + f'(x) / g(x)

H'(x) = [-f(x).g'(x)] + f'(x).g(x)] / g²(x)

Thus, the quotient rule is proved.

➣ Read More: Chain Rule

Derivation of Quotient Rule Using Implicit Differentiation

Let's take a differentiable function f(x), such that f(x) = u(x)/v(x).

u(x) = f(x).v(x)

using the product rule,
u'(x) = f'(x)⋅v(x) + f(x)v'(x)

Now solving for f'(x)
f'(x) = [u'(x) - f(x)v'(x)] / v(x)

Substituting the value of f(x) as, f(x) = u(x)/v(x)
f'(x) = {u'(x) - u(x)/v(x).[v'(x)]}/v(x)

f'(x) = {u'(x)v(x) - u(x).v'(x)} / v²(x)

Thus, the quotient rule is proved.

➣ Read More: Implicit Differentiation

Derivation of Quotient Rule Using Derivative and Limit Properties

Let's take a differentiable function f(x) such that f(x) = u(x)/v(x),

We know that, f'(x) = lim_h→0 [f(x+h) - f(x)] / h

Substituting the value of f(x) = u(x)/v(x)
f'(x) = lim_h→0 [u(x+h)/v(x+h) - u(x)/v(x)] / h
f'(x) = lim_h→0 [u(x+h).v(x) - u(x).v(x+h)] / h.v(x).v(x+h)

Distributing the limit,

f'(x) = {lim_h→0 [u(x+h).v(x) - u(x).v(x+h)] / h}.{lim_h→0 1/v(x).v(x+h)}
⇒ f'(x) = {lim_h→0 [u(x+h).v(x) - u(x).v(x+h) + u(x)v(x) - u(x)v(x)] / h}.{ 1/v(x).v(x)}
⇒ f'(x) = {lim_h→0 [u(x+h).v(x) - u(x).v(x)] / h} {lim_h→0  [u(x)v(x+h) - u(x)v(x)] / h}.{ 1/v²(x)}
⇒ f'(x) = v(x){lim_h→0 [u(x+h) - u(x)] / h} -u(x) {lim_h→0  [-v(x+h) + v(x)] / h}.{ 1/v²(x)}

f'(x) = [v(x).u'(x) - u(x).v'(x)] / v²(x)

Which is the required quotient rule.

➣ Read More

• Properties of Limits
• Rules of Derivatives

How to Use Quotient Rule in Differentiation?

To apply the quotient rule, we follow the following steps,

Step 1: Write the individual functions as u(x) and v(x).

Step 2: Find the derivative of the individual function u(x) and v(x), i.e. find u'(x) and v'(x). Now apply the quotient rule formula,

f'(x) = [u(x)/v(x)]' = [u'(x) × v(x) - u(x) × v'(x)] / [v(x)]²

Step 3: Simplify the above equation and it gives the differentiation of f(x).

We can understand this concept with the help of an example.

Example: Find f'(x) if f(x) = 2x³/(x+2)

Given, f(x) = 2x³/(x + 2)

Comparing with f(x) = u(x)/v(x), we get

• u(x) = 2x³
• v(x) = (x + 2)

Now Differentiating u(x) and v(x)

• u'(x) = 6x²
• v'(x) = 1

Using Quotient rule,

f'(x) = [v(x)u'(x) - u(x)v'(x)]/[v(x)]²
⇒ f'(x) = [(x+2)•6x² - 2x³•1]/(x + 2)²
⇒ f'(x) = (6x³ + 12x² - 2x³)/(x + 1)²
⇒ f'(x) = (4x³ + 12x²​​​​)/(x + 1)²

Product and Quotient Rule

The product rule of differentiation is used to find the differentiation of a function when the function is given as product of two function.

Product rule of differentiation states that , if P(x) = f(x).g(x)

P'(x) = f(x).g'(x) + f'(x).g(x)

Whereas the quotient rule of differentiation is used to differentiate a function that is represented as, division of two functions, i.e. f(x) = p(x)/q(x).

Then the derivation of f(x) using the quotient rule is calculated as,

f'(x) = {q(x).p'(x) - p(x).q'(x)}/q²(x)

Must Read

• Product Rule in Calculus
• Chain Rule
• Differentiation and Integration Formula
• Logarithmic Differentiation
• Fundamentals of Calculus
• Application of Derivatives

Solved Examples on Quotient Rule

Let's solve some sample questions on the Quotient Rule.

Example 1: Differentiate \bold{y=\frac{x^3-x+2}{x^2+5}}   .

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

y'=\dfrac {d}{dx}[\dfrac{x^3-5+2}{x^2+5}]

⇒ y'= \dfrac{[d/dx(x^3-x+2)(x^2+5)-(x^3-x+2)d/dx(x^2+5)]}{[x^2+5]^2}

⇒ y'= \dfrac{[(3x^2-1)(x^2+5)-(x^3-x+2)(2x)]}{[x^2+5]^2}\\=\dfrac{(3x^4+15x^2-x^2-5)-(2x^4-2x^2+4x)}{[x^2+5]^2}

⇒ y'= \dfrac{x^4+16x^2-4x-5}{[x^2+5]^2}

Example 2: Differentiate, f(x) = tan x.

Solution:

tan x is written as sinx/cosx, i.e.

tan x = (sin x) / (cos x)

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

f'(x)=\dfrac{(d/dx(sinx))(cosx)-(d/dx(cosx))(sinx)}{cos^2x}

⇒ f'(x)= \dfrac{cosx.cosx-(-sinx)(sinx)}{cos^x}

⇒ f'(x)=\dfrac{cos^2x+sin^2x}{cos^2x}

⇒ f'(x)=\dfrac{1}{cos^2x}

Example 3: Differentiate, f(x)= e^x/x²

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

f'(x)=[\dfrac{d/dx(e^x)(x^2)-d/dx(x^2)(e^x)}{x^4}]

⇒ f'(x)=\dfrac{e^x.x^2-2xe^x}{x^4}

Example 4: Differentiate, y=\frac{cosx}{x^2}

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

y'=\dfrac{d/dx(cosx)(x^2)-d/dx(x^2)(cosx)}{x^4}

⇒ y'=\dfrac{-sinx(x^2)-(2x)(cosx)}{x^4}

⇒ y'=\dfrac{-(x^2)sinx-(2xcosx)}{x^4}

Example 5: Differentiate, f(p) = p+5/p+7

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

f'(p)=d/dx[\dfrac{p+5}{p+7}]

⇒ f'(p)=[\dfrac{d/dx(p+5)(p+7)-d/dx(p+7)(p+5)}{(p+7)^2}]

⇒ f'(p)=[\dfrac{p+7-p-5}{(p+7)^2}]

⇒ f'(p)=[\dfrac{2}{(p+7)^2}]

Practice Problems

Here are a few practice problems on the Quotient Rule for you to solve.

Question 1: Find the derivative of f(x) = (x² + 3)/(sin x)

Question 2: Find the derivative of f(x) = (2x² + 3x + 5)/(x + 3)

Question 3: Find the derivative of f(x) = (x + 3)/(ln x)

Question 4: Find the derivative of f(x) = (x.sin x)/(x²)

Question 5: Differentiate f(x)= \frac{Sin(x)}{x^2}

Question 6: Find the derivative of f(x) = \frac{Cos(x)}{1+Sin(x)}

Question 7: Use the quotient rule to differentiate f(x) = \frac{x^2+3x+2}{x-4}

Question 8: Compute the derivative of f(x)= \frac{2x+3}{4x^2-5}

Question 9: Differentiate f(x)= \frac{x^3-2x}{√x}

Question 10: Find the derivative of f(x) = \frac{\tan{x}}{x}