Quotient Rule (HSC SSCE Mathematics Advanced): Revision Notes

Quotient Rule

Introduction to the quotient rule

The quotient rule is a differentiation technique used to find the derivative of functions that are expressed as one function divided by another. When you have a function that is the ratio of two differentiable functions, the quotient rule provides a systematic way to calculate its derivative.

This rule is particularly useful when dealing with rational functions, where simplifying before differentiating would be complicated or impossible. Understanding the quotient rule is essential for solving problems involving rates of change in fractional expressions and finding tangents to curves defined by quotient functions.

The quotient rule formula

If you have a function $y = \frac{u}{v}$, where both $u$ and $v$ are functions of $x$, then the derivative of $y$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Alternatively, using function notation, if $f(x) = \frac{g(x)}{h(x)}$, where $g(x)$ is not equal to $0$, then:

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

Understanding the components

Let's break down what each part of the quotient rule formula represents:

• $\frac{dy}{dx}$ is the derivative of y with respect to x (what we're trying to find)
• $v$ is the function in the denominator (the bottom of the fraction)
• $\frac{du}{dx}$ is the derivative of u (the numerator) with respect to $x$
• $u$ is the function in the numerator (the top of the fraction)
• $\frac{dv}{dx}$ is the derivative of v (the denominator) with respect to $x$
• $v^2$ is the square of the denominator function

The formula tells us to multiply the denominator by the derivative of the numerator, then subtract the product of the numerator and the derivative of the denominator, and finally divide everything by the square of the denominator.

Memory aid for the quotient rule

Worked example 1: Basic differentiation

Alternative method

For this particular function, you could also simplify before differentiating by dividing each term in the numerator by $x$:

$$y = \frac{x^2}{x} + \frac{6x}{x} + \frac{5}{x} = x + 6 + 5x^{-1}$$ $$\frac{dy}{dx} = 1 + 0 - 5x^{-2} = 1 - \frac{5}{x^2} = \frac{x^2 - 5}{x^2}$$

Worked example 2: Finding a tangent equation

Key learning objectives

After studying this topic, you should be able to:

Remember!