Differentiation from First Principles · Hook

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Ever wondered how a speedometer knows your speed at one exact moment?

Tunde's danfo covers 60 m between two bus stops in 5 seconds — an average of 12 m/s. But the needle shows the speed right NOW. Calculus finds that 'right now' rate by making the time gap smaller and smaller.

Revision notes

📒 Differentiation from First Principles: lesson notes

Find the gradient of a curve at any point by shrinking a chord until it becomes a tangent.

Tunde's danfo covers 60 m between two bus stops in 5 seconds — an average of 12 m/s. But the needle shows the speed right NOW. Calculus finds that 'right now' rate by making the time gap smaller and smaller.

The gradient of a curve

A straight line has the same gradient everywhere, but a curve gets steeper or flatter as you move along it.

The gradient of a curve at a point P is the gradient of the tangent — the line that just touches the curve at P.

To find it, draw a chord from P to a nearby point Q on the curve, then slide Q closer and closer to P.

🔑 Key idea: Gradient of a curve at a point = gradient of the tangent at that point.

Check it with numbers

On y = x2, take P at x = 3 (y = 9). With Q at x = 3.1, y = 9.61, so the chord gradient is 0.61 ÷ 0.1 = 6.1.

Move Q to x = 3.01: y = 9.0601, so the gradient is 0.0601 ÷ 0.01 = 6.01.

The gradients get closer and closer to 6, which is 2 × 3. That limit is the gradient of the tangent at x = 3.

🔑 Key idea: As the chord shrinks, its gradient approaches the tangent's gradient.

The δx method (first principles)

Let x increase by a tiny amount δx, so y increases by δy. For y = x2: y + δy = (x + δx)2 = x2 + 2xδx + (δx)2.

Subtract y = x2 to get δy = 2xδx + (δx)2. Divide by δx: δy/δx = 2x + δx — the gradient of the chord.

Now let δx → 0. The δx term vanishes, so dy/dx = 2x. At x = 3 that gives 6, just as the numbers showed.

🔑 Key idea: dy/dx is the limit of δy/δx as δx → 0.

Worked example

Using first principles, find dy/dx for y = x2 − 8x + 6.

  1. Step 1: Write y + δy

    y + δy = (x + δx)2 − 8(x + δx) + 6

    = x2 + 2xδx + (δx)2 − 8x − 8δx + 6

  2. Step 2: Subtract y

    δy = 2xδx + (δx)2 − 8δx

    Why? The x2, x and constant terms are in both, so they cancel.

  3. Step 3: Divide by δx

    δy/δx = 2x + δx − 8

    Why? This is the gradient of the chord.

  4. Step 4: Let δx → 0

    dy/dx = 2x − 8

    Why? As the chord shrinks to a point, it becomes the tangent.

Answer: 2x − 8

Summary

  • The gradient of a curve at a point is the gradient of the tangent there.
  • A chord PQ gets closer to the tangent as Q slides towards P.
  • Steps: find y + δy, subtract y, divide by δx, then let δx → 0.
  • Constant terms always cancel when you subtract.
  • From first principles, y = ax2 + bx + c gives dy/dx = 2ax + b.

Ready to check you've got it? Scroll up and work through the lesson: the interactive visuals, practice questions and mastery check are all there.