The Definition of a Derivative
Differentiation measures the rate of change of a curve at any given point. While shortcut rules like the power rule ($d/dx[x^n] = nx^{n-1}$) are used for speed, A-Level boards (Edexcel, AQA, OCR) expect you to prove these formulas using **First Principles**.
The derivative $f'(x)$ is mathematically defined as the limit of the gradient of a secant line between points $(x, f(x))$ and $(x+h, f(x+h))$ as the width $h$ approaches 0:
Step-by-Step Proof Methodology
To differentiate any function $f(x)$ from first principles, follow these four distinct steps:
State the limit formula
Always write down: $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$ to secure initial method marks.
Expand $f(x+h)$ algebraically
Substitute $x+h$ into your function and expand the brackets fully (e.g. using Binomial Expansion for higher powers).
Subtract $f(x)$ and divide by $h$
Subtract the original function. All terms without an $h$ factor must cancel out. Factorise $h$ out of the remaining terms and cancel it with the denominator.
Evaluate the limit as $h \rightarrow 0$
Set all remaining $h$ terms to 0 to yield the final derivative function.
Worked A-Level Exam Proof
Worked Exam Example
Question: Prove, from first principles, that the derivative of $x^3$ is $3x^2$.
Solution:
- Identify function: $f(x) = x^3$.
Therefore: $f(x+h) = (x+h)^3$. - Expand $(x+h)^3$ using binomial expansion: $$f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3$$
- Substitute into the limit formula: $$f'(x) = \lim_{h \rightarrow 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}$$
- Simplify the numerator (notice the $x^3$ terms cancel): $$f'(x) = \lim_{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$
- Divide through by $h$: $$f'(x) = \lim_{h \rightarrow 0} (3x^2 + 3xh + h^2)$$
- Apply the limit $h \rightarrow 0$ (terms containing $h$ become zero): $$f'(x) = 3x^2 + 3x(0) + (0)^2 = 3x^2$$
Result: $f'(x) = 3x^2$ (as required).
Frequently Asked Questions
If you substitute $h = 0$ at the start, the formula yields $\frac{f(x) - f(x)}{0} = \frac{0}{0}$. This is an indeterminate form (division by zero is undefined). We must simplify the algebra first to cancel $h$ from the denominator before taking the limit.
Yes. The proof uses the compound angle identity $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$, and evaluates standard small-angle approximation limits: $\lim_{h \rightarrow 0} \frac{\sin(h)}{h} = 1$ and $\lim_{h \rightarrow 0} \frac{\cos(h)-1}{h} = 0$, leading to $d/dx[\sin(x)] = \cos(x)$.
Secure an A* in A-Level Maths
Learn calculus proofs, mechanics vectors, and trigonometric identities 1-on-1 with elite tutors. Text or WhatsApp us today.