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The derivative from first principles (QCE Math Methods Unit 2)

Syllabus dot point

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

7 min read

Quick answer

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

Jump to a section
  1. What this dot point is asking
  2. Average and instantaneous rates of change
  3. The derivative as a limit
  4. First-principles procedure
  5. The basic results
  6. Why the limit is necessary
  7. The geometric meaning
  8. How this appears in assessment
  9. Exam-style questions

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 ff between x=ax = a and x=a+hx = a + h is the slope of the secant line:

f(a+h)−f(a)h\frac{f(a+h) - f(a)}{h}

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

The derivative as a limit

The derivative of ff at xx is:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

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

First-principles procedure

Four standard steps:

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

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

The basic results

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

For f(x)=xf(x) = x: f′(x)=1f'(x) = 1.

For f(x)=x2f(x) = x^2: f′(x)=2xf'(x) = 2x.

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

First principles for a cubic

Find f′(x)f'(x) from first principles for f(x)=x3f(x) = x^3.

Step 1: Expand f(x+h)f(x + h)

The first-principles definition requires us to evaluate the function at x+hx + h. For a cubic, this means expanding the binomial (x+h)3(x + h)^3 using the binomial theorem or by successive multiplication:

f(x+h)=(x+h)3=x3+3x2h+3xh2+h3.f(x + h) = (x + h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3.

Step 2: Form the difference f(x+h)−f(x)f(x + h) - f(x) and factor out hh

Subtracting f(x)=x3f(x) = x^3 causes the x3x^3 terms to cancel, leaving only terms that contain hh. This is the key algebraic step: every remaining term has hh as a factor, so we can factor it out to remove the hh from the denominator in the next step:

f(x+h)−f(x)=3x2h+3xh2+h3=h(3x2+3xh+h2).f(x + h) - f(x) = 3x^2 h + 3x h^2 + h^3 = h(3x^2 + 3xh + h^2).

Step 3: Divide by hh and simplify

Dividing the difference by hh cancels the common factor, leaving an expression with no hh in the denominator:

f(x+h)−f(x)h=3x2+3xh+h2.\frac{f(x + h) - f(x)}{h} = 3x^2 + 3xh + h^2.

Step 4: Take the limit as h→0h \to 0

Now that hh no longer appears in the denominator, we can substitute h=0h = 0 directly. The terms 3xh3xh and h2h^2 both vanish:

f′(x)=lim⁡h→0(3x2+3xh+h2)=3x2.f'(x) = \lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2.

This confirms the power rule ddxx3=3x2\dfrac{d}{dx}x^3 = 3x^2.

Final answer: f′(x)=3x2f'(x) = 3x^2.

Common traps

Cancelling hh before factoring. If hh appears as a sum like f(x+h)−f(x)=2xh+h2f(x+h) - f(x) = 2xh + h^2, you must factor hh first to cancel with the denominator.

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

Substituting h=0h = 0 in the difference quotient before simplifying. This gives 0/00/0, which is undefined. Simplify first, then take the limit.

Treating (x+h)2(x+h)^2 as x2+h2x^2 + h^2. It is x2+2xh+h2x^2 + 2xh + h^2. Expand all binomials carefully.

Why the limit is necessary

The difference quotient f(x+h)−f(x)h\dfrac{f(x + h) - f(x)}{h} is the gradient of the secant line joining two points on the curve. As hh shrinks, the second point slides toward the first and the secant rotates toward the tangent. You cannot simply set h=0h = 0, because that gives 00\dfrac{0}{0}, which is undefined; the limit is what makes the idea rigorous. Algebraically, simplifying first removes the hh in the denominator so that substituting h=0h = 0 is then legitimate, which is the whole reason the four-step procedure works.

The geometric meaning

The value f′(a)f'(a) is the gradient of the tangent line to y=f(x)y = f(x) at x=ax = a, and equivalently the instantaneous rate of change of ff there. A positive derivative means the function is increasing, a negative derivative means decreasing, and a zero derivative marks a stationary point. This geometric reading is what connects first principles to the later study of stationary points, optimisation and curve sketching, where the derivative is the central tool.

How this appears in assessment

In IA1 a four-mark first-principles question on a polynomial up to degree three is standard, and the marker's focus is the procedure (difference quotient, factor, cancel, limit) as much as the final answer. In the external assessment, Paper 1 may ask you to identify the difference quotient or differentiate a simple polynomial from first principles, while Paper 2 builds on the result to introduce the power rule and combined-rule applications in Unit 3. Showing every step, and not collapsing the limit prematurely, is what secures full marks.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Paper 1 (technique). Use first principles to determine the derivative of f(x)=3x2−5xf(x) = 3x^2 - 5x.
Show worked answer →

f(x+h)=3(x+h)2−5(x+h)=3x2+6xh+3h2−5x−5hf(x + h) = 3(x + h)^2 - 5(x + h) = 3x^2 + 6xh + 3h^2 - 5x - 5h.

f(x+h)−f(x)=6xh+3h2−5hf(x + h) - f(x) = 6xh + 3h^2 - 5h. Divide by hh: 6xh+3h2−5hh=6x+3h−5\dfrac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5.

Take the limit as h→0h \to 0: f′(x)=6x−5f'(x) = 6x - 5.

Markers reward the explicit difference quotient, simplifying before taking the limit, and the limit step at the end.

Practice5 marks
Paper 2 (complex familiar). For f(x)=x2−4xf(x) = x^2 - 4x, (a) use first principles to determine f′(x)f'(x), and (b) hence determine the equation of the tangent to y=f(x)y = f(x) at the point where x=1x = 1.
Show worked answer →

(a) f(x+h)=(x+h)2−4(x+h)=x2+2xh+h2−4x−4hf(x + h) = (x + h)^2 - 4(x + h) = x^2 + 2xh + h^2 - 4x - 4h. Then f(x+h)−f(x)=2xh+h2−4hf(x + h) - f(x) = 2xh + h^2 - 4h, so the difference quotient is 2x+h−42x + h - 4, and f′(x)=lim⁡h→0(2x+h−4)=2x−4f'(x) = \lim_{h \to 0}(2x + h - 4) = 2x - 4.

(b) At x=1x = 1: gradient f′(1)=2(1)−4=−2f'(1) = 2(1) - 4 = -2, and f(1)=1−4=−3f(1) = 1 - 4 = -3. Tangent: y−(−3)=−2(x−1)y - (-3) = -2(x - 1), so y=−2x−1y = -2x - 1.

Markers reward the first-principles derivative, evaluating the gradient and point, and the tangent equation.

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