Back to the path
f'

Introduction to the derivative

Level 6 · Ages 16–17

Quick recall· Let's wake up what you already know before we start.

What is the slope of the line through (1, 3) and (3, 7)?

The whole lesson

Read it again any time, including while you work through the exercises.

On a straight line, the slope is the same everywhere: it is the gradient. But on a curve, the slope keeps changing! Think of a mountain road: sometimes it climbs steeply, sometimes it is almost flat, sometimes it goes down. How can we measure the slope of a curve at one precise point? That is exactly the job of the derivative.

First idea: between two points of a curve, we can compute an average slope, called the rate of change: (f(b) - f(a)) ÷ (b - a). It is the slope of the line joining the two points, like the average speed of a journey.

Second idea, the stroke of genius: bring the two points closer. We compute the rate of change between a and a + h, then let h tend to 0. If this rate approaches a fixed number, that number is the derivative of f at a, written f'(a). It is the instantaneous slope, like the speed shown on the dashboard at one precise moment.

Geometrically, f'(a) is the slope of the tangent line: the line that skims the curve at the point with x-coordinate a, matching its direction. Its equation is y = f'(a)(x - a) + f(a). A useful special case: if the tangent is horizontal, then f'(a) = 0, which often happens at the top of a bump or the bottom of a dip.

For the function f(x) = x², we will accept the result computed in the example: f'(x) = 2x. At the point with x-coordinate 3, the slope of the curve is therefore 2 × 3 = 6.

The worked example, step by step

Compute the derivative of f(x) = x² at a = 2.

  1. t(h) = ((2 + h)² - 2²) ÷ h

    I write the rate of change between 2 and 2 + h: change in f divided by change in x.

  2. (4 + 4h + h² - 4) ÷ h = (4h + h²) ÷ h

    I expand (2 + h)² = 4 + 4h + h², and the 4s cancel.

  3. t(h) = 4 + h

    I simplify by h. The rate of change equals 4 + h.

  4. h → 0 : t(h) → 4

    As h approaches 0, the rate approaches 4: so f'(2) = 4. Generalising this computation gives f'(x) = 2x.

Result: f'(2) = 4: at the point with x-coordinate 2, the curve climbs with slope 4.

Give the equation of the tangent to the curve of f(x) = x² at the point with x-coordinate 2.

  1. f(2) = 4 et f'(2) = 4

    I need two ingredients: the value of f and the slope of the tangent at x-coordinate 2.

  2. y = f'(2)(x - 2) + f(2) = 4(x - 2) + 4

    I apply the tangent formula y = f'(a)(x - a) + f(a).

  3. y = 4x - 8 + 4 = 4x - 4

    I expand and simplify. Check: for x = 2, y = 4, the tangent does pass through the point (2, 4).

Result: The tangent has equation y = 4x - 4.