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The derivative from first principles

The derivative is the limit of a difference quotient and gives the gradient of a tangent. Differentiate small positive integer powers of x from first principles, and relate the signs of f' and f'' to the behaviour of the original function.

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 the graph of a curve.
Read the second derivative as the rate of change of the gradient.

COMMON MISCONCEPTION

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

That line is a chord, and its gradient only approximates the curve's. The gradient at the point is the limit of the chord gradients as h tends to zero. On y = x² the chord from x has gradient 2x + h, which is exact for that chord; only the limit turns it into the tangent gradient 2x.

The shrinking chord

A straight line has a single gradient; a curve does not. Start with the gradient of a chord joining (x, f(x)) to a neighbouring point (x + h, f(x + h)), which is the change in y divided by the change in x. That chord gradient approximates the gradient of the curve at x, and the error decreases as h decreases, so the gradient of the curve is the limit of the chord gradients as h tends to zero.

The parabola y equals x squared with two dashed chords leaving the point 1 comma 1, of computed slopes 3 and 2.5 for h equal to 1 and one half, and the solid amber tangent of slope 2 that the chords approach as h shrinks to nothing. — 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) = \text{lim}_{h → 0} \frac{f(x + h) − f(x)}{h}

IN THE FORMULAE BOOKLET

That limit is the derivative, written f'(x) or dy/dx, and it gives the gradient of the tangent at the point. It is printed in the formulae booklet, under Differentiation in the A level Mathematics section, so the statement can be quoted rather than recalled. Computing it directly from the limit is called differentiating from first principles. Some questions write δx in place of h; the quantity and the proof are the same.

The proofs below are for small positive integer powers of x. The derivatives of sin x and cos x are proved the same way, using the compound angle formulae to expand the chord, in Differentiating trig, exponentials and logs.

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, and the chord gradient is 2x + h.

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

Cancel the h before taking the limit. Substituting h = 0 into the original fraction gives 0/0, which is why the numerator is expanded and the factor of h divided out first.

GUIDED PRACTICE

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 containing h tend to zero, leaving f'(x) = 3x². ∎

The surviving term of the expansion is the one with exactly one factor of h, which the division cancels; terms with higher powers of h tend to zero in the limit.

The gradient is a function

Evaluating the derivative across the domain gives the gradient function. Where f'(x) > 0, f is increasing; where f'(x) < 0, it is decreasing; and f'(x) = 0 identifies stationary points. At x = 0, f(x) = x³ has a stationary point of inflection because f'(x) = 3x² touches zero without changing sign.

Two aligned computed graphs: the cubic y equals x cubed minus 3x on top, and beneath it the gradient function 3 x squared minus 3, with dashed verticals showing the cubic's maximum and minimum sitting directly above the gradient's two zero crossings. — FIG. 2y = x³ − 3x drawn above its own gradient function 3x² − 3. The maximum and minimum of the curve lie directly above the zeros of the gradient function.

To sketch y = f'(x) from y = f(x), mark the zeros of f' first, directly below the stationary points of f. Then fix the sign of f' on each interval between them, and only then consider the steepness of the sketch.

INDEPENDENT PRACTICE

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 −4, its own gradient, and the limit acted on each term separately. The derivative of a sum is the sum of the derivatives.

Differentiating twice

Differentiating the gradient function gives the second derivative, written f''(x) or d²y/dx². It is the rate of change of the gradient. On an interval where the curve is convex the gradient is increasing, so f'' is positive there, and that is the basis of the second-derivative test for classifying stationary points.

The cubic y equals x cubed minus 3 x, its gradient 3 x squared minus 3 and its second derivative 6 x, stacked so the sign of each lines up with the shape of the one above — FIG. 3The same cubic and its first and second derivatives, drawn one above the other on a shared x-axis. Reading down a dashed line, the curve has a stationary point where the gradient row crosses its axis, and changes between convex and concave where the second-derivative row crosses.

ASSESSMENT FOCUS

A complete first-principles proof has four stages: the chord gradient stated, the numerator expanded, the factor of h divided out, and the limit taken.
Write the limit statement as well as the algebra. “As h → 0, the gradient → 2x” is the conclusion of the proof.
f'(x), dy/dx and “the gradient function” all name the same object, and a question stem may use any of the three.
The chord gradient before the limit, 2x + h, is exact for that chord. Only the limit turns it into a tangent 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, gives the same value, which checks the first-principles working.

The derivative is the limit of the chord gradient as h tends to zero, and it gives the gradient of the tangent.

Divide the factor of h out first, then take the limit; the remaining expression is the gradient function.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Or read them with their worked answers on the derivative from first principles questions page.

CHECK YOUR PROGRESS

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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 the graph of a curve.
Read the second derivative as the rate of change of the gradient.

Open the full revision checklist to see every objective in the course in one place.

The derivative from first principles

The shrinking chord

The gradient is a function

Differentiating twice

Question workbook

Question workbook

Answer book