Differentiation From First Principles (So It Finally Makes Sense)

Quick answer: Differentiating from first principles means finding a gradient using the limit definition f'(x) = lim(h->0) [f(x+h) - f(x)] / h, instead of the power-rule shortcut. You expand f(x+h), subtract f(x), divide every term by h, then let h -> 0. For f(x) = x^2 this gives 2x; for f(x) = x^3 it gives 3x^2.

When I first met "differentiate from first principles" in A-Level Pure, I could already turn x^2 into 2x in my sleep using the power rule, so it felt like being told to walk to school when I owned a bike. It took me a while to realise first principles isn't the slow way to differentiate — it's the proof of why the fast way works. And the moment a board can put "prove" in front of something, it becomes a predictable few marks. This shows up in Edexcel, AQA and OCR A-Level, in Cambridge A-Level (9709), and as the definition of the derivative in AP Calculus.

The idea: a gradient of a chord that keeps shrinking

• A straight line has one gradient everywhere: rise over run. A curve doesn't — its steepness changes as you move along it.
• Pick a point on the curve, (x, f(x)). Now pick a second point a short horizontal distance h to the right: (x + h, f(x + h)).
• The straight line joining those two points is a chord. Its gradient is the change in y over the change in x: [f(x+h) - f(x)] / h.
• That chord is only an approximation to the steepness at your point. But as you slide the second point closer — as h shrinks towards 0 — the chord swings round until it rests on the curve as the tangent. The gradient of that tangent is the gradient of the curve at x.

"First principles" just means building the gradient from that shrinking chord, using nothing but the definition.

The definition you must be able to write from memory

f'(x) = lim(h->0) [f(x+h) - f(x)] / h

Some boards and textbooks use δx (delta x) instead of h, so you may also see dy/dx = lim(δx->0) [f(x+δx) - f(x)] / δx. Same idea, different letter. Whatever symbol your board uses, three parts must appear or you lose marks:

• the word lim (or the words "as h -> 0"),
• the gradient-of-a-chord fraction with f(x+h) on top,
• h on the bottom.

Miss the limit notation and, in most mark schemes, you drop a whole mark even if your algebra is flawless. More on that below.

Worked example 1: differentiate f(x) = x^2 from first principles

1. Write the definition: f'(x) = lim(h->0) [f(x+h) - f(x)] / h.
2. Work out f(x + h) by replacing every x with (x + h): f(x + h) = (x + h)^2 = x^2 + 2xh + h^2.
3. Subtract f(x): f(x+h) - f(x) = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2.
4. Divide every term by h: (2xh + h^2) / h = 2x + h.
5. Take the limit as h -> 0: the "+ h" vanishes, so f'(x) = 2x.

That is the power rule — bring the power down, drop it by one — proved for x^2, not assumed.

Worked example 2: differentiate f(x) = x^3 from first principles

Same five moves; the only harder bit is the expansion.

1. Write the definition: f'(x) = lim(h->0) [f(x+h) - f(x)] / h.
2. Expand f(x + h) = (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3.
3. Subtract f(x): f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3. (the x^3 terms cancel)
4. Divide every term by h: (3x^2h + 3xh^2 + h^3) / h = 3x^2 + 3xh + h^2.
5. Let h -> 0: the 3xh and h^2 terms both disappear, leaving f'(x) = 3x^2.

Notice the pattern: at step 4, every term you are left with still carries an h except the one that survives. That surviving term is your answer. If nothing survives, or everything survives, you have slipped in the expansion.

The "show that" version — and where the marks live

Boards rarely just say "differentiate x^2." They say "Prove from first principles that the derivative of x^2 is 2x," or "Using differentiation from first principles, show that..." It is the same five steps, but now it is a proof, so your presentation is graded. From a typical 5-mark scheme, the marks are handed out roughly like this:

1. Writing the gradient-of-a-chord expression [f(x+h) - f(x)] / h with the correct f(x+h) — 1 mark.
2. Expanding the bracket correctly — 1 mark.
3. Simplifying and dividing through by h — 1 mark.
4. Stating the limit as h -> 0 — the notation itself, not just the final answer — 1 mark.
5. Reaching the correct derivative — 1 mark.

The three mistakes that cost real marks every year:

• Setting h = 0 too early. If you put h = 0 before you divide, step 3 becomes 0/0, which is meaningless. Divide by h first, then let h -> 0. Never both at once.
• Dropping the limit line. Cancelling the h and jumping straight to "= 2x" loses the notation mark. Write the "as h -> 0" line explicitly — it is a free mark sitting right there.
• Expansion errors on the cubic. (x + h)^3 has four terms, not two. Forgetting the middle 3x^2h and 3xh^2 is the single most common way to throw away the method marks.

If the algebra of the expansion feels shaky, drill it on its own — our step-by-step math solver will expand (x + h)^2 and (x + h)^3 line by line, so you can check your own working against it. And if the why still feels fuzzy — why a shrinking chord becomes a tangent — the free explainer tool will walk it through at whatever level you ask.

Test yourself

1. Differentiate f(x) = x^2 + 3x from first principles.
2. Differentiate f(x) = 2x^2 from first principles.
3. In the definition f'(x) = lim(h->0) [f(x+h) - f(x)] / h, why can't you simply substitute h = 0 straight away?

Quick answers:

1. f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h. Subtract f(x): 2xh + h^2 + 3h. Divide by h: 2x + h + 3. Let h -> 0: f'(x) = 2x + 3.
2. f(x+h) = 2(x+h)^2 = 2x^2 + 4xh + 2h^2. Subtract f(x): 4xh + 2h^2. Divide by h: 4x + 2h. Let h -> 0: f'(x) = 4x.
3. Because the fraction becomes 0/0, which is undefined. The point of the limit is to simplify first — cancel the h on the bottom — and only then let h approach 0.

Want these marked properly, with the notation checked? Drop your working into the math solver, or turn the whole method into revision cards with the flashcards tool.

FAQ

Do I really need first principles if I know the power rule?

For the differentiating itself, no — everywhere else in the exam you will use the power rule. But boards set first principles as its own question precisely because it proves the rule, and you cannot answer "prove from first principles" with the shortcut. It is a reliable few marks once you have drilled it.

Is it h or δx?

Both stand for the same small step in x. Edexcel and AQA usually write h; you will also meet δx (delta x) in Cambridge and in older textbooks. Use whichever your board's mark scheme uses, and stay consistent within a single answer.

Does AP Calculus test this?

Yes — AP calls it the definition of the derivative, f'(x) = lim(h->0) [f(x+h) - f(x)] / h, with an alternate form f'(a) = lim(x->a) [f(x) - f(a)] / (x - a). AP questions can ask you to set up or use the definition directly, so the same expand-simplify-limit routine applies.

What is the hardest function they will ask from first principles?

At A-Level it is usually a cubic like x^3, or a quadratic with a coefficient — the algebra is the challenge, not the idea. Cambridge and AP can push to things like 1/x, but the method never changes: form the chord, simplify, take the limit.

In short: differentiation from first principles is just the gradient of a chord that you shrink to nothing. Expand f(x+h), subtract f(x), divide by h, then let h -> 0 — and always write that limit line, because in the "show that" version it is a whole mark on its own.