Quotient Rule

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, a function of the form:f(x)/g(x).

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.

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

Derivation of Quotient Rule Using Chain Rule

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

Thus, the quotient rule is proved.

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.

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.

How to Use Quotient Rule in Differentiation?

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 + 2)²
⇒ f'(x) = (4x³ + 12x²​​​​)/(x + 2)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)

Solved Examples on Quotient Rule

Example 1: Differentiate .

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

⇒ 

⇒ 

⇒ 

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,

⇒ 

⇒ 

⇒ 

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

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

Differentiate:

Substitute and simplify:

⇒ a

Example 4: Differentiate, 

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

⇒ 

⇒ 

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

Solution:

Both Numerator and Denominator functions are differentiable.

Applying Quotient Rule,

⇒ 

⇒ 

⇒ 

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

Question 6: Find the derivative of

Question 7: Use the quotient rule to differentiate

Question 8: Compute the derivative of

Question 9: Differentiate

Question 10: Find the derivative of

Related Reads

• Product Rule in Calculus
• Chain Rule
• Differentiation and Integration Formula
• Logarithmic Differentiation
• Fundamentals of Calculus
• Application of Derivatives
• Implicit Differentiation
• Properties of Limits
• Rules of Derivatives