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
\[ \frac{d}{dx}(\tan x) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{(\cos x)^2} \]
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
\[ y = \frac{(x+3)^4}{\sqrt{x^2+5}} \]
\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2} \]
\[ f'(x) = 4(x+3)^3 \]
\[ g(x) = (x^2+5)^{1/2} \]
\[ g'(x) = \frac{1}{2}(x^2+5)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+5}} \]
\[ \frac{dy}{dx} = \frac{\sqrt{x^2+5} \cdot 4(x+3)^3 - (x+3)^4 \cdot \frac{x}{\sqrt{x^2+5}}}{(x^2+5)} \]
\[ \frac{dy}{dx} = \frac{(x+3)^3 \left( 4\sqrt{x^2 + 5} - \frac{x(x+3)}{\sqrt{x^2 + 5}} \right)}{x^2 + 5} \]
\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot \frac{4(x^2 + 5) - x(x+3)}{\sqrt{x^2 + 5}}}{x^2 + 5} \]
\[ 4(x^2 + 5) = 4x^2 + 20 \]
\[ x(x+3) = x^2 + 3x \]
\[ 4x^2 + 20 - (x^2 + 3x) = 3x^2 - 3x + 20 \]
\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot \frac{3x^2 - 3x + 20}{\sqrt{x^2 + 5}}}{x^2 + 5} \]
\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot (3x^2 - 3x + 20)}{(x^2 + 5)^{3/2}} \]
\[ \frac{dy}{dx} = \frac{(x+3)^3 \cdot (3x^2 - 3x + 20)}{(x^2 + 5)^{3/2}} \]
Example 3
\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]
\[ u(x) = x + \cos(x), \quad u'(x) = 1 - \sin(x) \]
\[ v(x) = \tan(x), \quad v'(x) = \sec^2(x) \]
\[ f'(x) = \frac{\tan(x) \cdot (1 - \sin(x)) - (x + \cos(x)) \cdot \sec^2(x)}{[\tan(x)]^2} \]
\[ f'(x) = \frac{\tan(x) - \tan(x)\sin(x) - x\sec^2(x) - \cos(x)\sec^2(x)}{[\tan(x)]^2} \]
Example 4
\[ \frac{e^x}{e^{2x} + 1} \quad \text{(a)} \]
\[ \frac{dy}{dx} = \frac{(e^{2x} + 1)e^x - e^x \cdot 2e^{2x}}{(e^{2x} + 1)^2} \]
\[ \frac{dy}{dx} = \frac{e^{3x} + e^x - 2e^{3x}}{(e^{2x} + 1)^2} = \frac{e^x - e^{3x}}{(e^{2x} + 1)^2} \]
Example 5
\[ f(x) = \frac{\ln x}{x} \]
\[ f'(x) = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} \]
\[ f'(x) = \frac{1 - \ln x}{x^2} \]