IB Maths AI SLDifferentiationPaper 1 & 2tangent & normal~6 min read
Gradients, Tangents & Normals
The derivative gives the gradient of a curve at any point — and a gradient plus a point is all you need to pin down a straight line. This note uses that to find two key lines at any point on a curve: the tangent, which just touches the curve, and the normal, which cuts across it at a right angle.
📘 What you need to know
The gradient of a curve at a point is found by substituting that x-value into the derivativef′(x).
The tangent at a point touches the curve there; its gradient is f′(x1).
The normal at a point is perpendicular to the tangent; its gradient is −1/f′(x1).
Perpendicular gradients multiply to −1 — so flip the tangent gradient and change its sign.
With a gradient m and a point (x1, y1), the line is y − y1 = m(x − x1).
Find the point (x1, y1) first — substitute x1 into f(x) for the y-coordinate.
Finding the gradient at a point
The derivativef′(x) is the gradient function — so to find the gradient of a curve at a particular point, substitute that point’s x-value into f′(x).
For example, if f(x) = x2 + 3x − 4 then f′(x) = 2x + 3, so the gradient at x = 1 is f′(1) = 5. Your GDC won’t give you the derivative function, but it can evaluate the derivative at a point directly.
The equation of a tangent
The tangent at a point on a curve is the straight line that just touches the curve there without crossing it. Its gradient is the gradient of the curve at that point — that is, f′(x1).
A straight line is fixed by a gradient and a point, so once you have both, substitute them into the equation of a line.
Equation of a tangenty − y1 = f′(x1)(x − x1)
the gradient of the tangent at (x1, y1) is f′(x1) — slot the gradient and the point into the straight-line equation
The equation of a normal
The normal at a point is the straight line through that point that is perpendicular to the tangent. Since perpendicular gradients multiply to −1, the normal’s gradient is −1 divided by the tangent’s gradient.
Equation of a normaly − y1 = −1f′(x1)(x − x1)
the normal is perpendicular to the tangent — its gradient is −1 ÷ f′(x1)
The tangent touches the curve at P; the normal cuts through P at right angles to it. The tangent’s gradient is f′(x1); the normal’s is −1/f′(x1).
The point (x1, y1) is the same for both lines — only the gradient changes.
🧠Recipe — tangent or normal at a point
Differentiate the function to get f′(x).
Find the point: substitute x1 into f(x) to get the y-coordinate y1.
Find the gradient at x1: substitute x1 into f′(x) — this is the tangent gradient.
For a normal, take the perpendicular gradient: −1 ÷ (tangent gradient).
Substitute the gradient and point into y − y1 = m(x − x1), then rearrange to the form asked for.
Worked examples
WE 1
Gradient at a point
Find the gradient of the curve f(x) = x2 − 5x + 2 at the point where x = 4.
the gradient at a point = f′(x) evaluated theref′(x) = 2x − 5f′(4) = 2(4) − 5 = 8 − 5gradient at x = 4 is 3differentiate first, then substitute the x-value — the gradient is simply f′(4).
WE 2
Where is the gradient zero?
The curve y = x3 − 12x. Find the x-values where the gradient of the curve is 0.
gradient 0 means dy/dx = 0dy/dx = 3x2 − 123x2 − 12 = 0 ⇒ x2 = 4x = 2 or x = −2setting the derivative equal to 0 finds where the curve is flat — here at two separate points.
WE 3
Equation of a tangent
Find the equation of the tangent to the curve y = x2 + 2x at the point where x = 1, in the form y = mx + c.
point: at x = 1, y = 1 + 2 = 3 ⇒ (1, 3)gradient: dy/dx = 2x + 2, at x = 1 ⇒ m = 4y − 3 = 4(x − 1)y = 4x − 1find the point on the curve first (substitute into the function), then the gradient (substitute into the derivative).
WE 4
Equation of a normal
Find the equation of the normal to the curve y = x2 − 3x + 5 at the point where x = 2, in the form y = mx + c.
point: at x = 2, y = 4 − 6 + 5 = 3 ⇒ (2, 3)tangent gradient: dy/dx = 2x − 3, at x = 2 ⇒ 1normal gradient = −1 ÷ 1 = −1y − 3 = −1(x − 2)y = −x + 5the normal uses the perpendicular gradient — flip the tangent gradient and change its sign.
WE 5
Tangent with a negative power
Find the equation of the tangent to f(x) = x + 4/x at the point where x = 1.
rewrite: f(x) = x + 4x−1point: at x = 1, f(1) = 1 + 4 = 5 ⇒ (1, 5)f′(x) = 1 − 4x−2 = 1 − 4/x2f′(1) = 1 − 4 = −3y − 5 = −3(x − 1)y = −3x + 8rewrite the fraction as a negative power before differentiating, then carry on as usual.
WE 6
Full question: tangent and normal
For the curve y = x3 − 4x + 1, at the point where x = 2: (a) find the gradient; (b) find the equation of the tangent in the form y = mx + c; (c) find the equation of the normal in the form ax + by + d = 0.
point: at x = 2, y = 8 − 8 + 1 = 1 ⇒ (2, 1)(a) dy/dx = 3x2 − 4, at x = 2 ⇒ 12 − 4gradient = 8(b) tangent: y − 1 = 8(x − 2)y = 8x − 15(c) normal gradient = −1/8y − 1 = (−1/8)(x − 2) ⇒ ×8: 8y − 8 = −x + 2(a) 8 · (b) y = 8x − 15 · (c) x + 8y − 10 = 0for the integer form, multiply through to clear the fraction, then collect every term on one side.
💡 Top tips
The gradient of a curve at a point is f′(x) evaluated there — differentiate, then substitute.
Find the y-coordinate of the point from the originalf(x), not from the derivative.
The normal gradient is −1 ÷ (tangent gradient): flip it and switch the sign.
y − y1 = m(x − x1) is usually quicker than y = mx + c — the point and gradient slot straight in.
For a form like ax + by + d = 0, multiply through to clear fractions, then move every term to one side.
âš Common mistakes
Using f(x) for the gradient instead of the derivative f′(x).
Getting the y-coordinate from f′(x) — the point’s y-value comes from the original f(x).
Using the tangent gradient for the normal — the normal needs the perpendicular gradient.
Getting the perpendicular gradient wrong: the negative reciprocal of 2 is −1⁄2, not −2.
Stopping at y − y1 = m(x − x1) when a specific form is asked for — always rearrange.
Next up: Increasing & Decreasing Functions — using the sign of f′(x) to describe how a curve behaves. For tangents and normals the method never changes: differentiate, find the point, find the gradient, then substitute into y − y1 = m(x − x1). The only twist for a normal is the perpendicular gradient, −1/f′(x1).
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