AOS3 Topic 6: Quotient Rule

In Calculus, the Quotient Rule is similar to the Product Rule. The Quotient Rule is stated as follows:

The derivative of the quotient of two functions is given by the ratio of the denominator function times the derivative of the numerator function, minus the numerator function times the derivative of the denominator function, to the square of the denominator function.

In short, the Quotient Rule is a method for differentiating the division of functions or quotients. This is also known as the quotient rule differentiation in mathematics.

Quotient Rule Formula

Let \( F(x) = \frac{f(x)}{g(x)} \), where \( g(x) \neq 0 \). If \( f'(x) \) and \( g'(x) \) exist, then:

\[ F'(x) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2} \]

If \( F(x) = \frac{x^3 + 2x}{x^5 + 2} \), then \( F \) can be considered as a quotient of two functions \( f \) and \( g \), where \( f(x) = x^3 + 2x \) and \( g(x) = x^5 + 2 \). The Quotient Rule gives:

\[ F'(x) = \frac{(x^5 + 2) \cdot (3x^2 + 2) - (x^3 + 2x) \cdot 5x^4}{(x^5 + 2)^2} \]

Expanding this, we get:

\[ F'(x) = \frac{3x^7 + 6x^2 + 2x^5 + 4 - 5x^7 - 10x^5}{(x^5 + 2)^2} \]

\[ F'(x) = \frac{-2x^7 - 8x^5 + 6x^2 + 4}{(x^5 + 2)^2} \]

The Quotient Rule (Function Notation)

Let \( F(x) = \frac{f(x)}{g(x)} \), where \( g(x) \neq 0 \). If \( f'(x) \) and \( g'(x) \) exist, then:

\[ F'(x) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2} \]

Proof

The quotient rule can be proved from first principles, but instead, we will use the product rule and the chain rule.

We can write \( F(x) = f(x) \cdot h(x) \), where \( h(x) = \frac{1}{g(x)} \). Using the chain rule, we have:

\[ h'(x) = -\frac{1}{[g(x)]^2} \cdot g'(x) \]

Therefore, using the product rule, we obtain:

\[ F'(x) = f(x) \cdot h'(x) + h(x) \cdot f'(x) \]

Substitute \( h'(x) \) and simplify:

\[ F'(x) = f(x) \cdot \left(-\frac{1}{[g(x)]^2} \cdot g'(x)\right) + \frac{1}{g(x)} \cdot f'(x) \]

\[ = \frac{-f(x) \cdot g'(x) + g(x) \cdot f'(x)}{[g(x)]^2} \]

\[ = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2} \]

The Quotient Rule (Leibniz Notation)

If \( y = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \) and \( v \neq 0 \), then:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

Using the Quotient Rule to Find the Derivative of \( \tan \theta \)

Let \( y = \tan \theta \). We write \( y = \frac{\sin \theta}{\cos \theta} \) and apply the quotient rule to find the derivative:

\[ \frac{dy}{d\theta} = \frac{\cos \theta \cdot \cos \theta - \sin \theta \cdot (-\sin \theta)}{(\cos \theta)^2} \]

\[ = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} \]

\[ = \frac{1}{\cos^2 \theta} \quad \text{(by the Pythagorean identity)} \]

\[ = \sec^2 \theta \]

Example 1

Find the Derivative of \( \tan x \) using the Quotient Rule.

Solution:

We know, \( \tan x = \frac{\sin x}{\cos x} \).

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

Then, \( f'(x) = \cos x \) and \( g'(x) = -\sin x \).

Using the quotient rule:

\[ \frac{d}{dx}(\tan x) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{(\cos x)^2} \]


Simplifying the numerator, we have: \[ \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we get: \[ \frac{1}{\cos^2 x} \]

Recognizing that \( \frac{1}{\cos^2 x} = \sec^2 x \), we conclude: \[ \frac{d}{dx}(\tan x) = \sec^2 x \]

Example 2

Find the derivative of the following function with respect to \( x \):

\[ y = \frac{(x+3)^4}{\sqrt{x^2+5}} \]

olution:

We will use the quotient rule to differentiate this function.

The quotient rule states:

\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2} \]

Let \( f(x) = (x+3)^4 \) and \( g(x) = \sqrt{x^2+5} \).

First, we find the derivatives of \( f(x) \) and \( g(x) \):

\[ f'(x) = 4(x+3)^3 \]

\[ g(x) = (x^2+5)^{1/2} \]

\[ g'(x) = \frac{1}{2}(x^2+5)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+5}} \]

Now, apply the quotient rule:

\[ \frac{dy}{dx} = \frac{\sqrt{x^2+5} \cdot 4(x+3)^3 - (x+3)^4 \cdot \frac{x}{\sqrt{x^2+5}}}{(x^2+5)} \]

We can combine the terms in the numerator:

\[ \frac{dy}{dx} = \frac{(x+3)^3 \left( 4\sqrt{x^2 + 5} - \frac{x(x+3)}{\sqrt{x^2 + 5}} \right)}{x^2 + 5} \]

Simplify the expression inside the parentheses:

\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot \frac{4(x^2 + 5) - x(x+3)}{\sqrt{x^2 + 5}}}{x^2 + 5} \]

Expanding the terms:

\[ 4(x^2 + 5) = 4x^2 + 20 \]

\[ x(x+3) = x^2 + 3x \]

So, we have:

\[ 4x^2 + 20 - (x^2 + 3x) = 3x^2 - 3x + 20 \]

Substituting back:

\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot \frac{3x^2 - 3x + 20}{\sqrt{x^2 + 5}}}{x^2 + 5} \]

Simplify the fraction:

\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot (3x^2 - 3x + 20)}{(x^2 + 5)^{3/2}} \]

Final Answer:

The derivative of the function is:

\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot (3x^2 - 3x + 20)}{(x^2 + 5)^{3/2}} \]

Example 3

Find the derivative of \( f(x) = \frac{x + \cos(x)}{\tan(x)} \) using the quotient rule.

Solution:

We are given the function \( f(x) = \frac{x + \cos(x)}{\tan(x)} \). We need to find the derivative of \( f(x) \) with respect to \( x \) using the quotient rule.

The quotient rule states:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]

For our function, let \( u(x) = x + \cos(x) \) and \( v(x) = \tan(x) \). We need to find the derivatives of \( u(x) \) and \( v(x) \):

\[ u(x) = x + \cos(x), \quad u'(x) = 1 - \sin(x) \]

\[ v(x) = \tan(x), \quad v'(x) = \sec^2(x) \]

Now, substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula:

\[ f'(x) = \frac{\tan(x) \cdot (1 - \sin(x)) - (x + \cos(x)) \cdot \sec^2(x)}{[\tan(x)]^2} \]

Simplify the expression:

\[ f'(x) = \frac{\tan(x) - \tan(x)\sin(x) - x\sec^2(x) - \cos(x)\sec^2(x)}{[\tan(x)]^2} \]

Example 4

Differentiate the following with respect to \( x \):

\[ \frac{e^x}{e^{2x} + 1} \quad \text{(a)} \]

Solution:

Let \( y = \frac{e^x}{e^{2x} + 1} \).

Applying the quotient rule:

\[ \frac{dy}{dx} = \frac{(e^{2x} + 1)e^x - e^x \cdot 2e^{2x}}{(e^{2x} + 1)^2} \]

Simplifying the expression:

\[ \frac{dy}{dx} = \frac{e^{3x} + e^x - 2e^{3x}}{(e^{2x} + 1)^2} = \frac{e^x - e^{3x}}{(e^{2x} + 1)^2} \]

Example 5

Find \( f'(x) \) if \( f(x) = \frac{\ln x}{x} \).

Solution

To find the derivative of \( f(x) = \frac{\ln x}{x} \), we'll use the quotient rule.

Given:

\[ f(x) = \frac{\ln x}{x} \]

Here, \( g(x) = \ln x \) and \( h(x) = x \).

Step 1: Find \( g'(x) \) and \( h'(x) \)

- \( g(x) = \ln x \), so \( g'(x) = \frac{1}{x} \).

- \( h(x) = x \), so \( h'(x) = 1 \).

Step 2: Apply the Quotient Rule

\[ f'(x) = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} \]

Step 3: Simplify

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

Exercise &&1&& (&&1&& Question)

Find the derivative of \( \cot(x) \) using the quotient rule.

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Exercise &&2&& (&&1&& Question)

Differentiate with respect to \( x \):\[\frac{\sin(x)}{x + 1} \quad \text{(b)}, \quad x \neq -1\]

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Exercise &&3&& (&&1&& Question)

Find \( f'(x) \) if \( f(x) = \frac{2x}{\cos x} \).

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Exercise &&4&& (&&1&& Question)

Find the derivative of the following function with respect to \( x \):

\[ f(x) = \frac{x^2 + 1}{x^2} \]

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