Slide a second point towards the first: Pick a point on a curve. Pick a second point a little distance h to the right. The straight line joining them (a chord) has gradient
(rise)/(run) = (f(x+h) − f(x))/h.
Now let the second point slide back towards the first: h shrinks towards 0, and the chord swings round until it becomes the tangent. Its gradient is the derivative f'(x).
Why we can't just set h = 0: If you put h = 0 straight away you get 0/0, which is meaningless. The trick is to simplify the fraction first (the h's cancel), and only then let h → 0 in what's left. The limit tells us the value the gradient approaches, even though we never actually divide by zero.
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Expand, cancel the lone h, then let h → 0: The method is always the same four moves:
1. Write f(x+h) and expand it.
2. Subtract f(x) and form (f(x+h) − f(x))/h.
3. Cancel the single h on the bottom (every top term has an h to give it).
4. Let h → 0 — every leftover term still containing h disappears.
IB-style question — differentiate x² from first principles
A student must show, using the limit definition, that the derivative of f(x) = x² is 2x.
Differentiate f(x) = x² from first principles.
Step by step
- Write down the definition.
- Expand the top: (x+h)² = x² + 2xh + h². The x² terms cancel.
- Every top term has an h, so cancel the h on the bottom.
- Now let h → 0: the leftover h vanishes.
Final answer
f'(x) = 2x — matching the power rule (bring down the 2, drop the power by 1).
IB-style question — differentiate x³ from first principles
Show, from first principles, that the derivative of g(x) = x³ is 3x².
Step by step
- Set up the definition.
- Expand (x+h)³ = x³ + 3x²h + 3xh² + h³; the x³ terms cancel.
- Cancel one h from every term.
- Let h → 0: the two terms with an h vanish.
Final answer
g'(x) = 3x² — again confirming the power rule.