Average and instantaneous rates of change, the definition of the derivative as a limit $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, and the use of this definition to differentiate from first principles

What this dot point is asking

VCAA wants you to distinguish average rate of change (the slope between two points) from instantaneous rate of change (the slope at a single point), express the instantaneous rate as a limit, and apply the limit definition to differentiate simple polynomials from first principles. This question appears on Paper 1 most years.

Average versus instantaneous rate of change

The average rate of change of $f$ between $x = a$ and $x = b$ is the slope of the secant line through the two points:

The instantaneous rate of change at $x = a$ is the slope of the tangent line at that point:

This is the slope of secant lines becoming the slope of the tangent as the two points come together.

The limit definition of the derivative

For a function $f$, the derivative at $x$ is

provided this limit exists. When it does, $f$ is differentiable at $x$.

An equivalent form (using the alternate variable $a$ for the fixed point):

The four-step Paper 1 method

Differentiation from first principles is a routine procedure.

1. Write the limit definition. Always start by writing the formula. Markers want to see it.
2. Compute $f(x + h)$. Substitute $x + h$ everywhere $x$ appears in the rule for $f$.
3. Simplify the difference quotient $\frac{f(x + h) - f(x)}{h}$. Subtract, then factor out $h$ and cancel.
4. Take the limit as $h \to 0$. Substitute $h = 0$ into the simplified expression.

Worked example: a quadratic

Differentiate $f(x) = x^2 + 3x$ from first principles.

Step 1

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.

Step 2

$f(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h$.

Step 3

$f(x + h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h) - (x^2 + 3x) = 2xh + h^2 + 3h = h(2x + h + 3)$.

Divide by $h$: $\frac{f(x + h) - f(x)}{h} = 2x + h + 3$.

Step 4. $f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3$.

Worked example: a cubic

Differentiate $f(x) = x^3$ from first principles.

$f(x + h) = (x + h)^3 = x^3 + 3 x^2 h + 3 x h^2 + h^3$.

$f(x + h) - f(x) = 3 x^2 h + 3 x h^2 + h^3 = h(3 x^2 + 3 x h + h^2)$.

$\frac{f(x + h) - f(x)}{h} = 3 x^2 + 3 x h + h^2$.

$f'(x) = \lim_{h \to 0} (3 x^2 + 3 x h + h^2) = 3 x^2$.

This confirms the standard power-rule result $\frac{d}{dx}(x^3) = 3 x^2$.

When the limit fails

The limit $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ may not exist. Common causes:

• Corner in the graph (e.g. $f(x) = |x|$ at $x = 0$): the left and right limits differ.
• Vertical tangent (e.g. $f(x) = x^{1/3}$ at $x = 0$): the limit is infinite.
• Discontinuity in $f$: the limit cannot exist at a jump or a hole.

A function that is differentiable at $x$ must be continuous at $x$. The converse is false: $|x|$ is continuous everywhere but not differentiable at $0$.

Examples in context

Example 1. Average versus instantaneous speed. A drone's height is $f(t) = t^2$ metres at time $t$ seconds. Its average speed between $t = 2$ and $t = 4$ is $\frac{f(4) - f(2)}{4 - 2} = \frac{16 - 4}{2} = 6$ m/s. From first principles, $f'(t) = \lim_{h \to 0}\frac{(t + h)^2 - t^2}{h} = \lim_{h \to 0}(2t + h) = 2t$, so the instantaneous speed at $t = 3$ (the midpoint) is $f'(3) = 6$ m/s, matching the average over the symmetric interval.

Example 2. Confirming a rule. To check the derivative of $f(x) = 3x^2 - x$, apply first principles: $f(x + h) - f(x) = 3(x + h)^2 - (x + h) - 3x^2 + x = 6xh + 3h^2 - h$. Dividing by $h$ gives $6x + 3h - 1$, and the limit as $h \to 0$ is $f'(x) = 6x - 1$, agreeing with term-by-term differentiation.

Try this

Q1. Differentiate $f(x) = x^2 - 5x$ from first principles. [3 marks]

• Cue. Numerator $2xh + h^2 - 5h = h(2x + h - 5)$; divide and take limit: $f'(x) = 2x - 5$.

Q2. Find the average rate of change of $f(x) = x^3$ between $x = 1$ and $x = 2$. [2 marks]

• Cue. $\frac{8 - 1}{2 - 1} = 7$.

Q3. Differentiate $f(x) = 4x^2 + 2$ from first principles, then state $f'(1)$. [3+1 marks]

• Cue. $\frac{8xh + 4h^2}{h} = 8x + 4h \to 8x$, so $f'(x) = 8x$ and $f'(1) = 8$.