Define the derivative of a function as a limit and use first principles to find the derivative of a polynomial function

What this dot point is asking

QCAA wants you to define the derivative as the limit of an average rate of change and apply the first-principles definition to differentiate polynomial functions. This is the foundation for every shortcut rule (power rule, sum rule, constant multiple) that follows.

Average and instantaneous rates of change

The average rate of change of $f$ between $x = a$ and $x = a + h$ is the slope of the secant line:

The instantaneous rate of change at $x = a$ is the limit of this average as $h \to 0$. This is the slope of the tangent line at $a$ and is called the derivative.

The derivative as a limit

The derivative of $f$ at $x$ is:

provided the limit exists. Other notations: $\dfrac{dy}{dx}$, $\dfrac{d}{dx} f(x)$, $Df(x)$.

First-principles procedure

Four standard steps:

1. Write $f(x + h)$ by substituting $x + h$ into the function.
2. Compute $f(x + h) - f(x)$ and simplify.
3. Divide by $h$ and simplify so that no $h$ appears in the denominator.
4. Take the limit as $h \to 0$ by direct substitution.

The key algebraic move is to factor $h$ out of every term in the numerator of step 2; then it cancels with the denominator in step 3, leaving a polynomial in $x$ and $h$. Step 4 then sends $h$ to zero.

The basic results

For $f(x) = c$: $f'(x) = 0$ (constant function has zero slope everywhere).

For $f(x) = x$: $f'(x) = 1$.

For $f(x) = x^2$: $f'(x) = 2x$.

For $f(x) = x^n$ with $n$ a positive integer: $f'(x) = n x^{n-1}$ (the power rule, proved by binomial expansion in first principles).

Worked example

Find $f'(x)$ from first principles for $f(x) = x^3$.

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

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

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

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

This matches the power rule.

Common traps

Cancelling $h$ before factoring. If $h$ appears as a sum like $f(x+h) - f(x) = 2xh + h^2$, you must factor $h$ first to cancel with the denominator.

Forgetting the limit. The derivative is the limit, not the difference quotient. Writing $f'(x) = 2x + h$ and stopping loses the final mark.

Substituting $h = 0$ in the difference quotient before simplifying. This gives $0/0$, which is undefined. Simplify first, then take the limit.

Treating $(x+h)^2$ as $x^2 + h^2$. It is $x^2 + 2xh + h^2$. Expand all binomials carefully.

How this appears in IA1 and EA

IA1. A four-mark first-principles question on a polynomial up to degree $3$. The procedure is the marker's focus, not the final answer.

EA Paper 1. Multiple choice or short response on identifying the difference quotient or the derivative of a simple polynomial.

EA Paper 2. Used as the launching pad for the power rule and combined-rule applications in Year 12 Unit 3.

In one sentence

The derivative is defined as $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$, found by expanding $f(x+h)$, subtracting $f(x)$, factoring $h$ out of the numerator, cancelling with the $h$ in the denominator, and substituting $h = 0$; this proves the power rule $\frac{d}{dx} x^n = n x^{n-1}$ for polynomial inputs.