Find equations of tangents and normals
Use the derivative at a point to find the tangent gradient, the perpendicular normal gradient and both line equations.
Section 14 · Calculus
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How do you find the tangent gradient to y = f(x) at x = a?
What you will learn
- Find the gradient of a curve at a specified point
- Write a tangent equation using point-gradient form
- Find the normal gradient as the negative reciprocal when the tangent is not horizontal
The derivative gives the tangent gradient
At x = a, calculate y = f(a) and m = f′(a). The tangent passes through (a, f(a)) with gradient m, so y − f(a) = m(x − a).
A normal is perpendicular to the tangent. If m is nonzero, its gradient is −1/m. If the tangent is horizontal, the normal is vertical and has equation x = a.
Keep the point and gradient attached to the correct line
The tangent and normal share the same point on the curve, but their gradients are different. Substitute the point into each point-gradient equation before expanding.
A tangent touches the curve locally; a normal crosses it at right angles at the specified point. Do not use a secant gradient from two different curve points.
Pause and explain
What is the tangent point on y = x² when x = 3?
Worked example
Find the tangent and normal to y = x² at x = 2.
Show the worked solution
- The point is (2, 4) and the derivative is dy/dx = 2x, so the tangent gradient is 4.
- Tangent: y − 4 = 4(x − 2), giving y = 4x − 4.
- Normal gradient is −1/4; use the same point in point-gradient form.
Answer Tangent y = 4x − 4; normal y − 4 = −¼(x − 2).
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the tangent equation to y = x³ − 2x at x = 1.
Give me a hint
Find the point and derivative gradient before using point-gradient form.
Compare my reasoning
- At x = 1, y = 1 − 2 = −1, so the point is (1, −1).
- dy/dx = 3x² − 2; at x = 1 the gradient is 1.
- Use y + 1 = 1(x − 1).
Tangent y = x − 2.
Look for these in your work
- Curve point found
- Derivative evaluated
- Line equation simplified
Find the normal equation to y = √x at x = 4.
Give me a hint
Differentiate x^(1/2), find the tangent gradient, then take the negative reciprocal.
Compare my reasoning
- The point is (4, 2); f′(x) = 1/(2√x), so the tangent gradient is 1/4.
- The normal gradient is −4.
- Use y − 2 = −4(x − 4).
Normal y − 2 = −4(x − 4), or y = −4x + 18.
Look for these in your work
- Point on curve checked
- Reciprocal sign correct
- Normal passes through the point
At x = 1, find the tangent and normal to y = ln x.
Give me a hint
Use the logarithm derivative and note the tangent gradient is 1 at this point.
Compare my reasoning
- The point is (1, 0) and f′(1) = 1.
- The tangent is y = x − 1.
- The normal gradient is −1, so its equation is y = −x + 1.
Tangent y = x − 1; normal y = −x + 1.
Look for these in your work
- Point and gradient correct
- Tangent constructed
- Perpendicular normal constructed
Common mistakes
- Using f(a) as the gradient instead of f′(a)
- Using the tangent gradient for the normal
- Taking a reciprocal without changing the sign
- Forgetting the vertical-normal case when the tangent is horizontal
What can you explain now?
If a tangent gradient is 3, what is the perpendicular normal gradient?
Compare with the explanation
−1/3
Perpendicular non-vertical line gradients multiply to −1.
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Make it stick
How does the derivative give the tangent equation, and how do you find the normal?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.