IB Maths AI HLDifferentiationPaper 1 & 2~9 min read
Gradients, Tangents & Normals
Now you put the derivative to work. Substitute an x-value into f′(x) and you get the gradient at that point — which is the gradient of the tangent. From a point and a gradient you can build the straight-line equation of the tangent, and flipping the gradient to −1/f′ gives the normal, the perpendicular line through the same point.
📘 What you need to know
Gradient at a point: substitute x into the derivative — f′(x1) is the gradient at x = x1.
Tangent: the straight line that touches the curve at a point; its gradient is f′(x1).
Normal: the straight line through the point that is perpendicular to the tangent.
Tangent equation: y − y1 = f′(x1)(x − x1).
Normal gradient: −1f′(x1) (negative reciprocal of the tangent gradient).
GDC: can evaluate a derivative at a point with ddx( … )|x=…, but won’t give the derivative function.
Not in the booklet: derive the tangent/normal equations from the straight-line equation that is given.
Finding the gradient at a point
The gradient of a curve at a point is just the gradient of its tangent there. To get a number, substitute the x-value into the derivative function.
Gradient at x = x₁
gradient = f′(x1)
Differentiate first, then substitute — not the other way round ✗
For example, if f(x) = x2 + 3x − 4 then f′(x) = 2x + 3, so the gradient at x = 1 is f′(1) = 5, and at x = −2 it’s f′(−2) = −1.
GDC shortcut: your calculator can evaluate the derivative at a point directly — type ddx( f(x) )|x=value. Handy for awkward values like x = −2.5, but it won’t hand you the algebraic f′(x).
Tangent vs normal
At a point on a curve, the tangent grazes the curve (same direction), while the normal cuts straight across it at a right angle. Their gradients are negative reciprocals.
Tangent & normal at a point
Tangent (red) and normal (green) meet the curve at the same point at right angles.
tangent gradientm = f′(x1)Straight from the derivative — same steepness as the curve at that point.
normal gradient−1mNegative reciprocal of the tangent gradient (perpendicular lines).
🤔 Why “negative reciprocal”?
Perpendicular straight lines always have gradients that multiply to −1. So if the tangent has gradient m, the normal must have gradient −1/m (their product is m × (−1/m) = −1). A flat tangent (m = 0) therefore has a vertical normal — that’s the one case where −1/m breaks down.
Building the equations
You aren’t given the tangent or normal formulas — but you are given the straight-line equation y − y1 = m(x − x1). Substitute the right gradient and point and you’re done.
Tangent & normal lines
tangent: y − y1 = f′(x1)(x − x1) | normal: y − y1 = −1f′(x1)(x − x1)
Derived from the booklet’s straight-line equation ✓
🧭 Recipe — tangent or normal at x = x1
Differentiate to get f′(x) (rewrite roots/fractions as powers first).
Find the point: substitute x1 into f(x) for y1.
Find the gradient: substitute x1 into f′(x) for m. For a normal, use −1/m.
Substitute into y − y1 = m(x − x1) and rearrange to the requested form.
🧠 “Tangent Touches, Normal kNocks across”
The tangent touches the curve (same gradient); the normal cuts across at 90°. Flip and negate the tangent gradient to get the normal’s.
Worked examples
WE 1–3 use the SME function f(x) = x3 + 6x2 + 5x − 12; WE 4–5 use f(x) = 2x4 + 3x2 (with x ≠ 0).
WE 1
Find f′(x) for f(x) = x3 + 6x2 + 5x − 12
Differentiate term by term.
3x² + 2·6x¹ + 5x⁰ + 0f′(x) = 3x² + 12x + 5
WE 2
Show the gradient at x = 1 is 20
Substitute x = 1 into the derivative.
f′(1) = 3(1)² + 12(1) + 5= 3 + 12 + 5f′(1) = 20 ✓
WE 3
Find the gradient at x = −2.5
An awkward value — substitute (or use the GDC’s derivative-at-a-point).
Use the same point (1, 5); the normal gradient is the negative reciprocal of 2.
normal gradient = −1/f′(1) = −1/2y − 5 = −½(x − 1) → y = −½x + 11/2×2: 2y = −x + 11normal: x + 2y − 11 = 0
💡 Top tips
Differentiate, then substitute — the gradient at a point is f′(x1), never f(x1).
Get the y-coordinate from the original f(x), not the derivative.
Normal gradient = −1/m — flip and change sign.
Use y − y1 = m(x − x1) — usually quicker than y = mx + c.
Match the requested form — clear fractions to get integer coefficients for ax + by + d = 0.
GDC checks: evaluate the derivative at the point to confirm your gradient before building the line.
⚠ Common mistakes
Substituting before differentiating — putting the x-value into f(x) instead of f′(x).
Using the tangent gradient for the normal — forgetting the −1/m flip.
Reading y1 off the derivative — the point’s height comes from f(x1).
Leaving fractions when integer form is requested.
Sign slips when rearranging into ax + by + d = 0.
Next up — Increasing & Decreasing Functions. You now read off a single gradient value at a point; the next step is to read its sign across a whole interval. Where f′(x) > 0 the curve climbs (increasing), where f′(x) < 0 it falls (decreasing), and where f′(x) = 0 it’s momentarily flat — which sets up everything about stationary points to come.
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