Introduction to Calculus — Worked Solutions (Preliminary Maths Advanced)
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Worked examples for Preliminary Maths Advanced introduction to calculus. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Maths Advanced — Introduction to Calculus. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Differentiation from first principles
Question
Use first principles to find the derivative of $f(x) = x^2 + 3x$.
Solution
Apply the definition $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$.
$f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$.
Subtract $f(x)$: $f(x+h) - f(x) = 2xh + h^2 + 3h$.
Divide by $h$: $2x + h + 3$. Let $h \to 0$: $f'(x) = 2x + 3$.
Write the limit at every line until you cancel the $h$ — examiners want to see it, not just the final answer.
First principles means going back to the definition of the derivative as a limit: $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$. The idea is the gradient of a chord as the two points slide together.
Start by expanding $f(x+h)$: $(x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$.
Subtracting the original $f(x) = x^2 + 3x$ cancels the $x^2$ and $3x$, leaving $2xh + h^2 + 3h$. Notice every term has a factor of $h$ — that's what lets us divide safely.
Dividing by $h$ gives $2x + h + 3$. Finally we let $h \to 0$, and the lone $h$ disappears, so $f'(x) = 2x + 3$. That cancellation of $h$ is the whole point: it removes the $0/0$ problem.
Definition: $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$.
- $f(x+h) = x^2 + 2xh + h^2 + 3x + 3h$
- $f(x+h) - f(x) = 2xh + h^2 + 3h$
- $\div h$: $2x + h + 3$
- $h \to 0$: $f'(x) = 2x + 3$
Where the marks go
- 1 mark: States the first-principles limit and expands $f(x+h)$
- 1 mark: Simplifies the difference quotient to $2x + h + 3$
- 1 mark: Takes the limit to get $f'(x) = 2x + 3$
Key idea
First principles uses $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$; expand, cancel the common factor of $h$, then let $h \to 0$.
Example 2 — Differentiation rules
Question
For $y = 2x^3 - 5x^2 + 4$, find $\dfrac{dy}{dx}$ and hence the gradient of the tangent at $x = 2$.
Solution
Differentiate term by term with the power rule, $\frac{d}{dx}(ax^n) = nax^{n-1}$.
$\frac{dy}{dx} = 6x^2 - 10x + 0 = 6x^2 - 10x$.
Now substitute $x = 2$: $6(4) - 10(2) = 24 - 20 = 4$.
Gradient $= 4$. The constant $4$ differentiates to $0$ — don't carry it through.
The gradient function is the derivative, so differentiate first using the power rule: multiply by the index, then drop the index by one.
Term by term: $2x^3 \to 6x^2$, $-5x^2 \to -10x$, and the constant $4 \to 0$ (a flat line has no slope). So $\frac{dy}{dx} = 6x^2 - 10x$.
The gradient at a point means we evaluate this derivative there. At $x = 2$: $6(2)^2 - 10(2) = 6 \times 4 - 20 = 24 - 20 = 4$.
So the tangent's gradient at $x = 2$ is $4$. The two-step pattern — differentiate to get the gradient function, then substitute the $x$-value — comes up constantly.
Power rule term by term.
- $\frac{dy}{dx} = 6x^2 - 10x$
Evaluate at $x = 2$.
- $6(2)^2 - 10(2)$
- $= 24 - 20$
- $= 4$
Gradient $= 4$.
Where the marks go
- 1 mark: Differentiates the cubic and quadratic terms correctly
- 1 mark: Constant differentiates to $0$, giving $\frac{dy}{dx} = 6x^2 - 10x$
- 1 mark: Substitutes $x = 2$ into the derivative
- 1 mark: Correct gradient $4$
Key idea
The derivative is the gradient function; evaluate it at a particular $x$ to get the gradient of the tangent there.
Frequently asked questions
Step-by-step solutions to Introduction to Calculus questions in Preliminary Maths Advanced, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Advanced, using the methods and notation expected in exams and assessments.