In Calculus, the quotient rule is used to find the derivative of a function that is written as a fraction or ratio. It is also called the division rule of differentiation because it deals with dividing one function by another. The rule can be applied when a function is in the form f(x)/g(x), where both f(x) and g(x) are differentiable, and g(x) ≠ 0. To use the quotient rule correctly, it helps to understand the product rule and the concept of limits in differentiation. The same idea extends in integration as the integral quotient rule or division rule of integration, which helps in integrating functions given as the ratio of two expressions.

In this maths article, we shall read about quotient rule in differentiation, its formula and derivation. The proof of the quotient rule can be performed using derivatives and limit properties, implicit differentiation, and also chain rule. We shall also solve some examples for better understanding of the concept.

Quotient Rule Differentiation

In Calculus, quotient rule is used to define the value of differentiation of a function that is given in the form of the quotient obtained on dividing two differentiable functions.

In other words, we can define the quotient rule as the derivative of a quotient is equal to the ratio of the result obtained on subtracting the numerator times the derivative of the denominator from the denominator times the derivative of numerator to the square of the denominator.

Quotient Rule Formula

As mentioned above, let us put these words in the form of an equation to get the quotient rule formula.

If f(x) = u(x)/v(x) is the given function, we can find the derivative of f(x) using the formula:

\(f^{\prime}\left(x\right)=\left[\frac{u\left(x\right)}{v\left(x\right)}\right]^{\prime}=\frac{\left[v\left(x\right)\times u^{\prime}\left(x\right)-u\left(x\right)\times v^{\prime}\left(x\right)\right]}{\left[v\left(x\right)\right]^2}\)

Here,

f(x) is the given function of the form u(x)/v(x) for which the derivative is to be calculated.

u(x) is a differentiable function that makes the numerator of f(x).

v(x) is a differentiable function that makes the denominator of f(x).

u’(x) is the derivative of u(x).

v’(x) is the derivative of v(x).

Derivation of Quotient Rule Formula

The quotient rule formula can be derived in three main ways: using the properties of limits and derivatives, through implicit differentiation, or by applying the chain rule. Each method provides a different approach to understanding how the derivative of a function written as a ratio is calculated. We will explain all three methods in detail.

Proof of Quotient Rule Formula Using Properties of Limits and Derivatives

Let us prove the quotient rule formula using the definition of limits or derivatives.

Let the given function be f(x) = u(x)/v(x)

\(f^{\prime}\left(x\right)=\lim_{h\to0}\ \frac{\left[f\left(x+h\right)-f\left(x\right)\right]}{h}\)

\(f^{\prime}\left(x\right)=\lim_{h\to0}\ \frac{\left[\frac{u\left(x+h\right)}{v\left(x+h\right)}-\frac{u\left(x\right)}{v\left(x\right)}\right]}{h}\)

\(f^{\prime}\left(x\right)=\lim_{h\to0}\ \frac{\left[u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)\right]}{h.v\left(x\right).v\left(x+h\right)}\)

\(f^{\prime}\left(x\right)=\left(\lim_{h\to0}\ \frac{\left[u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)\right]}{h}\right)\left(\lim_{h\to0}\ \frac{1}{v\left(x\right).v\left(x+h\right)}\right)\)

\(f^{\prime}\left(x\right)=\left(\lim_{h\to0}\ \frac{\left[u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x\right)+u\left(x\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)\right]}{h}\right)\left(\frac{1}{\left[v\left(x\right)\right]^2}\right)\)

\(f^{\prime}\left(x\right)=\left(\left(\lim_{h\to0}\ \frac{\left[u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x\right)\right]}{h}\right)-\left(\lim_{h\to0}\ \frac{\left[u\left(x\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)\right]}{h}\right)\right)\left(\frac{1}{\left[v\left(x\right)\right]^2}\right)\)

\(f^{\prime}\left(x\right)=\left(\left(v\left(x\right)\lim_{h\to0}\ \frac{\left[u\left(x+h\right)-u\left(x\right)\right]}{h}\right)-u\left(x\right)\left(\lim_{h\to0}\ \frac{\left[v\left(x\right)-v\left(x+h\right)\right]}{h}\right)\right)\left(\frac{1}{\left[v\left(x\right)\right]^2}\right)\)

\(f^{\prime}\left(x\right)=\frac{\left[v\left(x\right)u^{\prime}\left(x\right)-u\left(x\right)v^{\prime}\left(x\right)\right]}{\left[v\left(x\right)\right]^2}\)

Hence Proved

Proof of Quotient Rule Formula Using Implicit Differentiation

In order to prove the quotient rule formula using implicit differentiation formula, let us first take a differentiable function f(x)=u(x)/v(x).

So, we can write u(x) = f(x). v(x)

By using product rule, we can write:

u’(x)=f’(x).v(x)+f(x).v’(x)

On solving f’(x), we get:

\(f^{\prime}(x)=\frac{[u^{\prime}(x)-f(x)v^{\prime}(x)]}{v\left(x\right)}\)

Substituting the value of f(x),

\(f^{\prime}(x)=\frac{[u^{\prime}(x)-\frac{u\left(x\right)}{v\left(x\right)}v^{\prime}(x)]}{v\left(x\right)}\)

\(f^{\prime}(x)=\frac{[u^{\prime}(x).v\left(x\right)-u\left(x\right).v^{\prime}(x)]}{\left[v\left(x\right)\right]^2}\)

Proof of Quotient Rule Formula Using Chain Rule

We can also use chain rule to derive quotient rule.

For this, let f(x) is a differentiable function of the form f(x) = u(x)/v(x).

\(f\left(x\right)=\ u\left(x\right)v^{-1}\left(x\right)\)

Using product rule:

\(f^{\prime}\left(x\right)=\ u^{\prime}\left(x\right)v^{-1}\left(x\right)+u\left(x\right).\left(\frac{d}{dx}\left(v^{-1}\left(x\right)\right)\right)\)

Using the power rule to solve the derivative in the second term:

\(f^{\prime}\left(x\right)=\ u^{\prime}\left(x\right)v^{-1}\left(x\right)+u\left(x\right).\left(-1\right)\left[v\left(x\right)\right]^{-2}v^{\prime}\left(x\right)\)

\(f^{\prime}\left(x\right)=\ \frac{u^{\prime}\left(x\right)}{v\left(x\right)}-\frac{u\left(x\right)v^{\prime}\left(x\right)}{\left[v\left(x\right)\right]^2}\)

\(f^{\prime}\left(x\right)=\ \frac{u^{\prime}\left(x\right)v\left(x\right)-u\left(x\right)v^{\prime}\left(x\right)}{\left[v\left(x\right)\right]^2}\)

Steps to Apply Quotient Rule

We have already read that if we have to apply quotient rule on f(x) = u(x

/v(x), then u(x) and v(x) both should be differentiable.

Here are the steps that can be applied in order to find the derivative of the differentiable function f(x).

Step 1: Make a note of the value of u(x) and v(x).

Step 2: First find the values of u’(x)and v’(x) and then put the values in the formula of quotient rule, i.e.

\(f^{\prime}\left(x\right)=\ \frac{u^{\prime}\left(x\right)v\left(x\right)-u\left(x\right)v^{\prime}\left(x\right)}{\left[v\left(x\right)\right]^2}\).

Quotient Rule Integration

Similar to the quotient rule for differentiation, the integration quotient rule is also used to integrate a function given in the form of numerator and denominator. This rule is also named as anti-derivative quotient or division rule.

The formula for quotient rule for integration is taken from integration by parts formula, that is:

\(\int_{ }^{ }u.dv=u.v-\int_{ }^{ }v.du\)

Let us suppose that

F(x)=F(u)/F(v) is a divisible function.

Differentiating F(x), we get:

\(F^{\prime}\left(x\right)=\frac{F^{\prime}\left(u\right)F\left(v\right)-F^{\prime}\left(v\right)F\left(u\right)}{\left[F\left(v\right)\right]^2}\)

Now, on integrating F’(x) we get:

\(\frac{F\left(u\right)}{F\left(v\right)}+C=\int_{ }^{ }\frac{F^{\prime}\left(u\right)F\left(v\right)-F^{\prime}\left(v\right)F\left(u\right)}{\left[F\left(v\right)\right]^2}dx\)

How to Use the Quotient Rule in Differentiation

The quotient rule helps differentiate functions written as a fraction of two functions. Find the derivatives of the numerator and denominator, then apply the formula f'(x) = [u'v − uv'] / v² to get the result.

Step 1: Write the numerator and denominator as separate functions, u(x) and v(x).

Step 2: Find the derivatives of both functions, u'(x) and v'(x). Then apply the quotient rule formula:

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

Step 3: Simplify the resulting expression to get the derivative.

Example: Find f'(x) if f(x) = (x² + 1) / (x − 3)

Here, u(x) = x² + 1 and v(x) = x − 3

Differentiating:
u'(x) = 2x, v'(x) = 1

Applying the quotient rule:
f'(x) = [(x − 3)(2x) − (x² + 1)(1)] / (x − 3)²
= [2x² − 6x − x² − 1] / (x − 3)²
= [x² − 6x − 1] / (x − 3)²

This gives the derivative of the function.

Product and Quotient Rule

The product rule is applied when a function is the multiplication of two functions. If P(x) = f(x) · g(x), its derivative is P'(x) = f(x)·g'(x) + f'(x)·g(x), helping calculate derivatives efficiently.

P'(x) = f(x) · g'(x) + f'(x) · g(x)

The quotient rule is used when a function is the division of two functions. If f(x) = p(x) / q(x), then its derivative is:

f'(x) = [q(x) · p'(x) - p(x) · q'(x)] / [q(x)]²

These rules help differentiate functions efficiently when they are multiplied or divided.

Quotient Rule Solved Examples

Question 1: Differentiate \(\frac{\log x}{x}\) using quotient rule?

Answer: Let f(x) = log x and g(x) = x

Then, \(\frac{d}{dx}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}=\frac{g\left(x\right)f^{\prime}\left(x\right)-f\left(x\right)g^{\prime}\left(x\right)}{\left(g\left(x\right)\right)^2}\)

Substituting the values of f(x), g(x), f’(x), and g’(x) in the above equation, we get

\(\frac{d}{dx}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}=\frac{x.\ \frac{1}{x}-\log\ x.1}{x^2}\)

=\(\frac{1-\log x}{x^2}\)

This is the required answer.

Question 2: Find the derivative of tan x using quotient rule?

Answer: We know that tan x = sin x/ cos x

Let f(x) = sin x and g(x) = cos x

Then, \(\frac{d}{dx}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}=\frac{g\left(x\right).f^{\prime}\left(x\right)-f\left(x\right).g^{\prime}\left(x\right)}{\left(g\left(x\right)\right)^2}\)

Substituting the values of f(x), g(x), f’(x), and g’(x) in the above equation, we get

\(\frac{d}{dx}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}=\frac{\cos x.\cos\ x-\sin x.\left(-\sin\ x\right)}{\left(\cos^2x\right)}\)

=\(\frac{\cos ^2x+\sin ^2x}{\left(\cos ^2x\right)}\)

=\(\frac{1}{\cos ^2x}\)

=\(\sec ^2x\)

This is the required answer.

We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.