IB Maths AA HLTopic 5 — CalculusPaper 1 & 2HL~10 min read
First Principles Differentiation
“From first principles” means deriving f′(x) directly from the limit definition — no power rule, no shortcuts. It’s a Paper-1 favourite “show that” question because it tests whether you really understand WHY the derivative exists. The technique is mechanical: substitute, expand, simplify, cancel h, take the limit. The challenge is staying sharp on the algebra and keeping limh → 0 written until the last line.
📘 What you need to know
The definition: f′(x) = limh → 0f(x+h) − f(x)h — given in the formula booklet, you don’t need to memorise it.
The geometric meaning: f(x+h) − f(x)h is the SLOPE of the secant joining (x, f(x)) to (x+h, f(x+h)). As h → 0, the secant becomes the tangent.
Why “limit” and not “set h = 0″: at h = 0 the quotient is 0/0 (undefined). The limit captures the value the quotient APPROACHES.
Step 1 — substitute: write f(x+h) explicitly by replacing every x in the formula with (x+h).
Step 2 — expand: for polynomials, multiply out; for (x+h)n use the binomial theorem; for surds, rationalise with the conjugate; for fractions, find a common denominator.
Step 3 — simplify: subtract f(x) — every term WITHOUT h cancels. Constants disappear. The remaining numerator should have h as a common factor.
Step 4 — cancel: factor h out of the numerator and cancel with the h in the denominator (valid because h ≠ 0 inside the limit).
Step 5 — take the limit: set h = 0 in the remaining expression. The final answer must NOT contain h.
Marking tip: keep limh → 0 written on EVERY line until the final concluding step — examiners deduct marks for omitting it.
The definition and what it means
Limit definition of the derivativef′(x) = limh → 0f(x+h) − f(x)h
Geometrically, this is the slope of the secant line connecting two nearby points on the curve, in the limit as those two points merge into one — the tangent.
The first-principles formula computes the slope of the secant AB. As B slides toward A (i.e., as h → 0), the secant becomes the tangent at A — whose slope is f′(x).
Why “lim” matters: setting h = 0 directly gives 0/0, which is undefined. The limit lets us say what value the quotient approaches WITHOUT actually putting h = 0. The algebraic trick (cancelling the h) is what makes the limit computable.
🧭 Recipe — differentiate from first principles
Substitute into the formula: write f′(x) = limh → 0 [f(x+h) − f(x)] / h with the specific f from the question. Keep limh → 0 attached.
Expand f(x+h): polynomials — multiply out; (x+h)n — binomial theorem; surds — multiply by the conjugate; reciprocals — common denominator.
Subtract f(x) in the numerator. All the terms without an h should cancel — constants disappear.
Factor h out of the numerator, then cancel it against the h in the denominator. The expression left should still depend on h, but no longer be 0/0.
Take the limit by setting h = 0 in what’s left. State the final derivative as the concluding line — this is where limh → 0 finally drops off.
Worked examples
WE 1
Quadratic with linear and constant terms
Prove, from first principles, that the derivative of f(x) = 4x² + 5x − 1 is 8x + 5.
Step 1 — substitute into the limit definitionf'(x) = lim h→0 [f(x+h) − f(x)] / h = lim h→0 [(4(x+h)² + 5(x+h) − 1) − (4x² + 5x − 1)] / hStep 2 — expand f(x+h)4(x+h)² = 4(x² + 2xh + h²) = 4x² + 8xh + 4h²5(x+h) = 5x + 5h⇒ f(x+h) = 4x² + 8xh + 4h² + 5x + 5h − 1Step 3 — subtract f(x); constants and “no-h” terms cancelf(x+h) − f(x) = 8xh + 4h² + 5hStep 4 — factor out h, cancel with denominator= h(8x + 4h + 5)÷ h: f'(x) = lim h→0 (8x + 4h + 5)Step 5 — let h → 0= 8x + 0 + 5 = 8x + 5 ✓∴ f'(x) = 8x + 5notice the constant “−1” never appears in f(x+h) − f(x): every term WITHOUT an h cancels. The technique always sweeps constants away.
WE 2
Cubic with two terms
Prove, from first principles, that the derivative of f(x) = 2x³ − 3x² is 6x² − 6x.
Step 1 — set upf'(x) = lim h→0 [2(x+h)³ − 3(x+h)² − (2x³ − 3x²)] / hStep 2 — expand using (x+h)³ = x³ + 3x²h + 3xh² + h³ and (x+h)² = x² + 2xh + h²2(x+h)³ = 2x³ + 6x²h + 6xh² + 2h³3(x+h)² = 3x² + 6xh + 3h²f(x+h) = 2x³ + 6x²h + 6xh² + 2h³ − 3x² − 6xh − 3h²Step 3 — subtract f(x); 2x³ and −3x² cancelf(x+h) − f(x) = 6x²h + 6xh² + 2h³ − 6xh − 3h²Step 4 — factor h, cancel= h(6x² + 6xh + 2h² − 6x − 3h)÷ h: f'(x) = lim h→0 (6x² + 6xh + 2h² − 6x − 3h)Step 5 — let h → 0= 6x² + 0 + 0 − 6x − 0 = 6x² − 6x ✓∴ f'(x) = 6x² − 6xevery term that survives the limit has NO h in it. Every h-term contributes 0 once h → 0 — that’s how the surviving terms get isolated.
WE 3
Reciprocal function — common denominator technique
Prove, from first principles, that the derivative of f(x) = 1/x is −1/x².
Step 1 — set upf'(x) = lim h→0 [1/(x+h) − 1/x] / hStep 2 — combine the numerator over a common denominator1/(x+h) − 1/x = [x − (x+h)] / [x(x+h)] = −h / [x(x+h)]Step 3 — divide by h (i.e. multiply by 1/h)f'(x) = lim h→0 [−h / (x(x+h))] · (1/h) = lim h→0 −1 / [x(x+h)]Step 4 — let h → 0= −1 / [x(x+0)] = −1 / x² ✓∴ f'(x) = −1/x²no polynomial expansion needed — the key move is creating a common denominator. The h in the numerator (after combining fractions) cancels the h in the outer denominator.
Prove, from first principles, that the derivative of f(x) = √x is 1/(2√x).
Step 1 — set upf'(x) = lim h→0 [√(x+h) − √x] / hStep 2 — multiply num and den by the conjugate (√(x+h) + √x)num: (√(x+h) − √x)(√(x+h) + √x) = (x+h) − x = hden: h · (√(x+h) + √x)Step 3 — simplifyf'(x) = lim h→0 h / [h · (√(x+h) + √x)]Step 4 — cancel h = lim h→0 1 / [√(x+h) + √x]Step 5 — let h → 0= 1 / [√x + √x] = 1 / (2√x) ✓∴ f'(x) = 1/(2√x)conjugate trick: (a − b)(a + b) = a² − b² turns “√(x+h) − √x” into “(x+h) − x = h” in the numerator. Every surd problem in first principles uses this move.
WE 5
Quartic — binomial theorem on (x+h)⁴
Prove, from first principles, that the derivative of f(x) = x⁴ is 4x³.
Step 1 — set upf'(x) = lim h→0 [(x+h)⁴ − x⁴] / hStep 2 — binomial expansion of (x+h)⁴(x+h)⁴ = x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴Step 3 — subtract x⁴(x+h)⁴ − x⁴ = 4x³h + 6x²h² + 4xh³ + h⁴Step 4 — factor h, cancel with denominator= h(4x³ + 6x²h + 4xh² + h³)÷ h: f'(x) = lim h→0 (4x³ + 6x²h + 4xh² + h³)Step 5 — let h → 0= 4x³ + 0 + 0 + 0 = 4x³ ✓∴ f'(x) = 4x³binomial coefficients (1, 4, 6, 4, 1) for n = 4. After subtracting x⁴ and factoring h, only the 4x³ term survives the limit — exactly the second binomial coefficient times x³.
WE 6
First principles at a specific point
The function f is defined by f(x) = x³ + 1. Use first principles to find the gradient of y = f(x) at the point where x = 2.
Step 1 — set up with x = 2 directlyf'(2) = lim h→0 [f(2+h) − f(2)] / hStep 2 — compute f(2) and expand f(2+h)f(2) = 2³ + 1 = 9f(2+h) = (2+h)³ + 1 = 8 + 12h + 6h² + h³ + 1 = 9 + 12h + 6h² + h³Step 3 — subtract; the constant 9 cancelsf(2+h) − f(2) = 12h + 6h² + h³Step 4 — factor h, cancel= h(12 + 6h + h²)÷ h: f'(2) = lim h→0 (12 + 6h + h²)Step 5 — let h → 0= 12 + 0 + 0 = 12f'(2) = 12 (gradient at x = 2 is 12)you can plug the number in right at step 1 — no need to find the general f'(x) first. The “specific point” version often runs faster because the numerator simplifies sooner.
💡 Top tips
Keep limh → 0 on EVERY line until the very last step — examiners deduct marks for omitting it, and dropping it early signals you’ve set h = 0 too soon.
Pick the right expansion tool: binomial for (x+h)n; conjugate for surds; common denominator for reciprocals. The form of f(x) tells you which.
Expect every term without h to cancel. If they don’t, you’ve made an algebra slip — recheck the expansion before going further.
The final answer must have NO h. If you’ve finished and your “derivative” still has h in it, you haven’t taken the limit yet.
Use the formula booklet — the first-principles formula is given. No memorisation needed for the form itself; just for the techniques.
⚠ Common mistakes
Setting h = 0 before cancelling — gives 0/0, which is undefined. You must algebraically remove the h in the denominator FIRST.
Algebra error in (x+h)n — using (x+h)² = x² + h² (missing the 2xh) is the classic trap. Always write out the full binomial expansion.
Forgetting to subtract f(x) cleanly — leaving an extra f(x) term, or not distributing the minus sign across all of it.
Dropping limh → 0 too early — even if your algebra is correct, the formal presentation loses marks if the limit symbol vanishes before the conclusion.
Cancelling h before it’s been factored out: you must show h as a common factor in the numerator BEFORE you cancel — examiners want to see the factoring step.
Up next: Differentiating Inverse Functions. If y = f−1(x), then x = f(y), and the chain rule gives dy/dx = 1 / (dx/dy). This is the cleanest way to differentiate things you can’t easily isolate — including the classic proof that d/dx[ln x] = 1/x via the inverse of ex.
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