Maths › Differentiation › The derivative from first principles

The derivative from first principles

A curve has no single gradient, but each point on it does, and catching that number takes a genuinely new idea: measure the slope of a chord, then let the chord shrink. What survives the shrinking is the derivative, and everything else in calculus is built on top of this one move.

Year 12-13EDEXCEL 9MA0 7.1

Builds on Straight lines.

IN THIS TOPIC

  • Explain the derivative as the limit of chord gradients, and use f'(x) and dy/dx notation.
  • Differentiate small powers of x from first principles.
  • Sketch a gradient function from a curve, and meet the second derivative.

WHAT YOU PROBABLY THINK

The gradient of a curve at a point is the gradient of the line to a nearby point.

The shrinking chord

A straight line has one gradient; a curve changes slope constantly. To measure the slope at a point, start with what can be measured: the gradient of a chord from (x, f(x)) to a nearby point (x + h, f(x + h)), which is the change in f over the change in x. That chord's gradient is not the answer, which is where the opening lie falls short. It is an approximation whose error shrinks with h, and the answer is what it approaches.

Chords on y equals x squared from the point 1 comma 1: with h equal to 1 the chord slope is 3, with h a half it is 2.5, and as h shrinks the chords settle onto the tangent of slope exactly 2chord h = 1: slope 3chord h = ½: slope 2.5tangent: slope 2shrink h and the chord slides onto the tangent
FIG. 1Chords from (1, 1) on y = x²: h = 1 gives slope 3, h = ½ gives 2.5, and the shrinking family settles onto the tangent of slope exactly 2.
f'(x) = limh → 0 f(x + h) − f(x)h

This limit is the derivative, written f'(x) or dy/dx, and it is the gradient of the tangent at the point. Computing it directly is called differentiating from first principles.

WORKED EXAMPLE

x² from first principles

Prove from first principles that the derivative of f(x) = x2 is 2x.

Form the chord gradient: [(x + h)2 − x2]/h = (2xh + h2)/h.

For h ≠ 0, divide through: the chord gradient is 2x + h.

As h → 0 the surviving part is 2x, so f'(x) = 2x. ∎

The h must be cancelled before the limit is taken; substituting h = 0 into the original fraction gives 0/0, which is precisely the trap the algebra exists to avoid.

YOUR TURN

A cubic from first principles

Differentiate f(x) = x3 from first principles, before opening the working.

Show the working

Expand: (x + h)3 = x3 + 3x2h + 3xh2 + h3.

The chord gradient is (3x2h + 3xh2 + h3)/h = 3x2 + 3xh + h2.

As h → 0, both terms carrying h vanish, leaving f'(x) = 3x2. ∎

Every surviving term of the expansion had exactly one h to cancel; the terms with more vanish in the limit. That pattern is the power rule being born.

The gradient is a function

Reading the slope at every point of a curve produces a new function, the gradient function, and the two graphs speak to each other. Where the curve climbs, its gradient function is positive; where it is momentarily flat, the gradient function crosses zero; where it falls, negative.

The curve y equals x cubed minus 3x above its own gradient function 3 x squared minus 3: the curve's hilltop and valley at x equal to plus and minus 1 sit exactly over the points where the gradient function crosses zeroy = x³ − 3xits gradient: 3x² − 3flat on top exactly where the gradient crosses zero
FIG. 2y = x³ − 3x drawn above its own gradient function 3x² − 3. The hilltop and valley of the curve sit exactly over the zeros of the gradient.

Differentiating the gradient function gives the second derivative, written f''(x) or d2y/dx2, the rate of change of the gradient itself. It measures how the slope is turning, and the next two lessons put both derivatives to work.

TRY IT UNSEEN

First principles with two terms

Differentiate f(x) = x2 − 4x from first principles.

Show the working

The chord gradient is [(x + h)2 − 4(x + h) − x2 + 4x]/h.

The numerator expands to 2xh + h2 − 4h, so the chord gradient is 2x + h − 4.

As h → 0, f'(x) = 2x − 4. ∎

The −4x term contributed the constant −4, its own gradient, and the limit treated each term independently. Differentiation will always respect sums this way.

THE EXAM BIT

  • First-principles proofs are marked line by line: chord gradient stated, expansion shown, h cancelled, limit taken. Skipping the cancellation forfeits the middle marks.
  • Write the limit statement as well as the algebra: “as h → 0, the gradient → 2x” is the concluding mark.
  • f'(x), dy/dx and “the gradient function” name the same object; expect any of the three in a question stem.
  • When sketching y = f'(x) from y = f(x), plot the zeros first, at the flat points of the curve, then fix signs from rising and falling.
  • The chord gradient before the limit, 2x + h, is exact for the chord; only the limit turns it into the tangent's gradient.

CHECK YOURSELF

From first principles, find the gradient of y = 3x2 at the point (2, 12).

Show a hint

Chord from x = 2, expand, cancel h, then let h → 0.

Show the answer

The chord gradient is [3(2 + h)2 − 12]/h = (12h + 3h2)/h = 12 + 3h.

As h → 0 the gradient approaches 12.

The general derivative 6x, evaluated at x = 2, agrees, and having both routes is exactly how to check first-principles work.

The derivative is the limit of chord gradients as the chord shrinks to a point.

Cancel h first, then let it vanish; what survives is the gradient function.

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.

  • Explain the derivative as the limit of chord gradients, and use f'(x) and dy/dx notation.
  • Differentiate small powers of x from first principles.
  • Sketch a gradient function from a curve, and meet the second derivative.

No animated video for this topic yet; these notes stand alone.