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Differentiation from First Principles

Q: Why do we need a limit at all — why not just use a small h?

A small but non-zero h still gives the gradient of a CHORD, which is only an approximation to the gradient at a single point. The limit h 0 is what makes the chord become the exact tangent.

Q: Why does dividing by h before letting h 0 matter so much?

Substituting h=0 directly into f(x+h)-f(x)h gives 00, which is undefined. You must simplify algebraically first so the h in the denominator cancels, THEN take the limit.

The gradient of a straight line is constant and simple to find. The gradient of a curve changes at every point — so what does "the gradient at a single point" even mean? First principles answers this by zooming in so far that a curve looks straight, and measuring the gradient of that infinitesimally short chord.

30 min Video by TLMaths Differentiation

The big picture

Every differentiation rule you will ever use — the power rule, the chain rule, all of it — is really just a shortcut for a single underlying idea: the gradient of a curve at a point is the LIMIT of the gradient of a chord as the two points defining that chord merge into one. This lesson builds that idea rigorously, from a chord of finite length $h$ down to $h \to 0$ , so that every later shortcut has a solid foundation you could always fall back on and re-derive.

A-Level Maths G1-11: Differentiation From First PrinciplesWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Understand the gradient of a curve at a point as a limit of chord gradients
Use the definition $f \overset{ˊ}{(} x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$
Differentiate simple polynomials from first principles
Recognise why this process becomes impractical for complicated functions (motivating the rules ahead)

From a chord to a tangent

Pick two points on a curve $y = f (x)$ : $(x, f (x))$ and a nearby point $(x + h, f (x + h))$ , a small horizontal distance $h$ apart. The gradient of the CHORD joining them is $\frac{f ( x + h ) - f ( x )}{h}$ — an ordinary "change in $y$ over change in $x$ " calculation.

As $h$ gets smaller and smaller, the second point slides closer to the first, and the chord rotates to hug the curve more and more closely. In the limit as $h \to 0$ , the chord becomes the $tangent$ at that exact point — and its gradient is the gradient of the curve there.

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

The formal definition of the derivative.

Tip — This limit process is the entire content of "differentiation" — every rule you meet later is a proven shortcut for evaluating this exact limit for a particular type of function.

Differentiating $x^2$ from first principles

Substituting $f (x) = x^{2}$ into the definition and expanding shows exactly how the algebra collapses to a clean result as $h \to 0$ .

f (x + h) = (x + h)^{2} = x^{2} + 2 x h + h^{2}

f (x + h) - f (x) = 2 x h + h^{2}

3Divide by

h

\frac{2 x h + h ^{2}}{h} = 2 x + h

4Let

h \to 0

2 x + h \to 2 x

Answer

f^{'} (x) = 2 x

Why this works

Every term with an $h$ still attached (like $h^{2} / h = h$ ) vanishes as $h \to 0$ — this is why the algebra is deliberately set up to factor out an $h$ from the numerator before dividing.

A general polynomial term

The same process on $f (x) = x^{3}$ (and any other power) always produces the same pattern of cancellation, which is exactly what justifies the power rule you will use for the rest of the course.

f (x + h) = (x + h)^{3} + 2 (x + h) = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} + 2 x + 2 h

f (x + h) - f (x) = 3 x^{2} h + 3 x h^{2} + h^{3} + 2 h

3Divide by

h

3 x^{2} + 3 x h + h^{2} + 2

4Let

h \to 0

3 x^{2} + 2

Answer

f^{'} (x) = 3 x^{2} + 2

Tip — Notice the pattern: differentiating $x^{n}$ from first principles always leaves $n x^{n - 1}$ once every term containing a surviving $h$ vanishes — this is the power rule, proved rather than just stated.

Think like an examiner

Always write out the full definition $lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ before substituting — examiners award method marks for showing this structure.
Expand $f (x + h)$ FULLY (e.g. binomial expansion for powers) before subtracting $f (x)$ — do not skip straight to cancelling.
State clearly that you are letting $h \to 0$ as your final step, showing which terms vanish and why.

Common misconceptions

Substituting

h = 0

before dividing by

h

You must divide by

h

FIRST (after the

h

cancels out of every term), then let

h \to 0

— substituting too early gives

\frac{0}{0}

, which is undefined.

Forgetting to expand

f (x + h)

completely, e.g. leaving

(x + h)^{2}

unexpanded.

Every bracket must be expanded fully so the

f (x)

terms cancel and a common factor of

h

can be taken out.

Thinking first principles only works for polynomials.

The definition works for ANY differentiable function (it is how

sin x

cos x

and others are proved too) — polynomials are just the most algebraically manageable starting point.

Definition of the derivative

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

Gradient of a curve at a point.

Stretch yourself

Use first principles to show that the derivative of $f (x) = x^{3}$ is $3 x^{2}$ .

Hint — Expand $(x + h)^{3}$ using the binomial expansion, subtract $x^{3}$ , divide by $h$ , then let $h \to 0$ .

Questions students ask

Why do we need a limit at all — why not just use a small

h

A small but non-zero

h

still gives the gradient of a CHORD, which is only an approximation to the gradient at a single point. The limit

h \to 0

is what makes the chord become the exact tangent.

Why does dividing by

h

before letting

h \to 0

matter so much?

Substituting

h = 0

directly into

\frac{f ( x + h ) - f ( x )}{h}

gives

\frac{0}{0}

, which is undefined. You must simplify algebraically first so the

h

in the denominator cancels, THEN take the limit.

Do I need to use first principles for every derivative in the course?

No — first principles proves the rules (like the power rule) that you then use as shortcuts. AQA only expects first-principles working for simple polynomials and, later, for proving the derivatives of

sin x

and

cos x

Key takeaways

The derivative $f^{'} (x)$ is defined as $lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ .
This is the limit of a chord gradient as the chord shrinks to a single point, becoming the tangent.
Expand $f (x + h)$ fully, subtract $f (x)$ , divide by $h$ , then let $h \to 0$ .
This process proves the power rule and every other differentiation rule that follows.

How this fits the course

Build on

Straight-Line Graphs

Differentiating xⁿ

Leads to

Differentiating xⁿ

Test yourself

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Differentiation from First Principles

30 min Video by TLMaths Differentiation