The Quotient Rule Revision Notes for VCE SSCE Mathematical Methods

The Quotient Rule

The quotient rule is a technique for differentiating functions that are written as a fraction (quotient) of two other functions. When you have a function divided by another function, the quotient rule allows you to find the derivative efficiently.

Understanding the quotient rule

When we have a function $F(x) = \frac{f(x)}{g(x)}$ , where $g(x) \neq 0$ , we cannot simply differentiate the numerator and denominator separately. Instead, we need to use the quotient rule.

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The quotient rule states that if $F(x) = \frac{f(x)}{g(x)}$ , where $g(x) \neq 0$ , and if $f'(x)$ and $g'(x)$ exist, then:

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

A helpful way to remember this is: "denominator times derivative of numerator minus numerator times derivative of denominator, all over the denominator squared".

Key components

To apply the quotient rule, you need to:

Identify the numerator function f(x) and its derivative f'(x)
Identify the denominator function g(x) and its derivative g'(x)
Apply the formula carefully, maintaining the correct order of terms

infoNote

The order of terms in the quotient rule is crucial. Many students make mistakes by reversing the order in the numerator. Always remember: it's denominator times derivative of numerator minus numerator times derivative of denominator.

Proof of the quotient rule

The quotient rule can be proved using the product rule and chain rule together. Here's how it works:

We can rewrite $F(x) = \frac{f(x)}{g(x)}$ as $F(x) = f(x) \cdot h(x)$ , where $h(x) = [g(x)]^{-1}$ .

Using the chain rule to find $h'(x)$ :

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

Now, applying the product rule:

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

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

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

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

This confirms the quotient rule formula.

infoNote

While you won't typically need to prove the quotient rule in exams, understanding how it's derived from the product and chain rules can help deepen your understanding of differentiation techniques and how different rules relate to each other.

Leibniz notation

The quotient rule can also be expressed using 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\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

This is the same rule, just written with different notation. Use whichever notation you find most comfortable.

Worked example: polynomial quotient

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Worked Example: Differentiating a Polynomial Quotient

Find the derivative of $\frac{x - 2}{x^2 + 4x + 1}$ with respect to $x$ .

Solution:

Let $y = \frac{x - 2}{x^2 + 4x + 1}$ .

Step 1: Identify the numerator and denominator:

Numerator: $u = x - 2$ , so $\frac{du}{dx} = 1$
Denominator: $v = x^2 + 4x + 1$ , so $\frac{dv}{dx} = 2x + 4$

Step 2: Apply the quotient rule:

\frac{dy}{dx} = \frac{(x^2 + 4x + 1)(1) - (x - 2)(2x + 4)}{(x^2 + 4x + 1)^2}

Step 3: Expand the numerator:

= \frac{x^2 + 4x + 1 - (2x^2 - 8)}{(x^2 + 4x + 1)^2}

= \frac{x^2 + 4x + 1 - 2x^2 + 8}{(x^2 + 4x + 1)^2}

Step 4: Simplify:

= \frac{-x^2 + 4x + 9}{(x^2 + 4x + 1)^2}

Worked example: exponential and trigonometric functions

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Worked Example: Exponential Function

Part a: Differentiate $\frac{e^x}{e^{2x} + 1}$ with respect to $x$ .

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

Step 1: Identify the components:

Numerator: $u = e^x$ , so $\frac{du}{dx} = e^x$
Denominator: $v = e^{2x} + 1$ , so $\frac{dv}{dx} = 2e^{2x}$

Step 2: Apply the quotient rule:

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

Step 3: Simplify:

= \frac{e^{3x} + e^x - 2e^{3x}}{(e^{2x} + 1)^2}

= \frac{e^x - e^{3x}}{(e^{2x} + 1)^2}

Step 4: Factorise the numerator:

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

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Worked Example: Trigonometric Function

Part b: Differentiate $\frac{\sin x}{x + 1}$ with respect to $x$ , where $x \neq -1$ .

Let $y = \frac{\sin x}{x + 1}$ for $x \neq -1$ .

Step 1: Identify the components:

Numerator: $u = \sin x$ , so $\frac{du}{dx} = \cos x$
Denominator: $v = x + 1$ , so $\frac{dv}{dx} = 1$

Step 2: Apply the quotient rule:

\frac{dy}{dx} = \frac{(x + 1)\cos x - \sin x \cdot 1}{(x + 1)^2}

Step 3: Simplify:

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

Application: the derivative of tan θ

The quotient rule can be used to find the derivative of tan θ, which is an important result in trigonometry.

We can write $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and apply the quotient rule:

\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}

Using the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$ :

= \frac{1}{\cos^2 \theta}

= \sec^2 \theta

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Therefore, $\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta$ .

This is an important result that you should remember for working with trigonometric functions.

Exam tips

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Tips for Success:

Always identify your numerator and denominator functions clearly before applying the rule
Remember the order: denominator × derivative of numerator - numerator × derivative of denominator
Don't forget to square the denominator in your final answer
Simplify your answer where possible, but be careful with algebraic manipulation
Check that your denominator is non-zero (the quotient rule requires $g(x) \neq 0$ )

bookmarkSummary

Key Points to Remember:

The quotient rule is used for differentiating functions written as $\frac{f(x)}{g(x)}$ where $g(x) \neq 0$
The formula is: $F'(x) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}$
The quotient rule can be proved using the product rule and chain rule together
Always maintain the correct order: denominator times derivative of numerator minus numerator times derivative of denominator, all divided by the square of the denominator
The derivative of $\tan \theta$ is $\sec^2 \theta$ , which can be found using the quotient rule

The Quotient Rule (VCE SSCE Mathematical Methods): Revision Notes