Differentiation
Cambridge IGCSE Mathematics (0580) · Unit 2: Algebra and graphs · 9 flashcards
Differentiation is topic 2.9 in the Cambridge IGCSE Mathematics (0580) syllabus , positioned in Unit 2 — Algebra and graphs , alongside Algebraic notation and manipulation, Equations and Inequalities. In one line: A stationary point is where the gradient of the curve is zero (dy/dx = 0). To find it, differentiate the function, set the derivative equal to zero, and solve for x. Substitute x into the original equation to find the y-coordinate.
This topic is examined across Paper 1 (Core) or Paper 2 (Extended) — non-calculator — and Paper 3 (Core) or Paper 4 (Extended) — calculator. It is a Supplement (Extended-tier) topic, so it appears only on the Extended-tier papers.
The deck below contains 9 flashcards — 1 definition, 3 key concepts and 2 application cards — covering the precise wording mark schemes reward. Use the definition card to lock down command-word answers (define, state), then move on to the concept and application cards to handle explain, describe and compare questions.
A stationary point, and how do you find it
A stationary point is where the gradient of the curve is zero (dy/dx = 0). To find it, differentiate the function, set the derivative equal to zero, and solve for x. Substitute x into the original equation to find the y-coordinate.
Questions this Differentiation deck will help you answer
- › What is the derivative of a function f(x), and what does it represent geometrically?
- › Explain how to find the equation of the tangent to the curve y = x^2 at the point (2, 4).
- › How do you determine whether a stationary point is a maximum or minimum point?
- › A particle moves along a line such that its displacement s (in meters) from a fixed point O at time t (in seconds) is given by s = t^3 - 6t^2 + 9t. Find the times when the particle is instantaneously at rest.
- › Explain the meaning of 'rate of change' in the context of differentiation.
What is the derivative of a function f(x), and what does it represent geometrically?
The derivative, denoted as f'(x) or dy/dx, represents the instantaneous rate of change of the function. Geometrically, it represents the gradient (slope) of the tangent line to the curve of f(x) at a specific point.
Differentiate: y = 3x^4 - 2x^2 + 5x - 7
Using the power rule (dy/dx(x^n) = nx^(n-1)), dy/dx = 12x^3 - 4x + 5. The derivative gives the gradient function of the original equation.
Explain how to find the equation of the tangent to the curve y = x^2 at the point (2, 4).
First, find the derivative: dy/dx = 2x. At x = 2, the gradient is 2(2) = 4. Then use the point-gradient form: y - 4 = 4(x - 2) giving y = 4x - 4.
What is a stationary point, and how do you find it?
A stationary point is where the gradient of the curve is zero (dy/dx = 0). To find it, differentiate the function, set the derivative equal to zero, and solve for x. Substitute x into the original equation to find the y-coordinate.
How do you determine whether a stationary point is a maximum or minimum point?
Use the second derivative test. If the second derivative (d^2y/dx^2) is positive at the stationary point, it's a minimum. If it's negative, it's a maximum. If it's zero, the test is inconclusive.
Find the coordinates of the stationary points of the curve y = x^3 - 3x
dy/dx = 3x^2 - 3. Set dy/dx = 0: 3x^2 - 3 = 0 => x = ±1. When x = 1, y = -2; when x = -1, y = 2. Stationary points are (1, -2) and (-1, 2).
A particle moves along a line such that its displacement s (in meters) from a fixed point O at time t (in seconds) is given by s = t^3 - 6t^2 + 9t. Find the times when the particle is instantaneously at rest.
The particle is at rest when its velocity is zero. Velocity, v = ds/dt = 3t^2 - 12t + 9. Setting v = 0: 3t^2 - 12t + 9 = 0 => t = 1 and t = 3 seconds.
Explain the meaning of 'rate of change' in the context of differentiation.
Rate of change describes how one variable changes in relation to another. In differentiation, dy/dx represents the rate of change of y with respect to x.
Find the maximum value of the function f(x) = -x^2 + 4x + 3.
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f''(x) = -2, which is negative, so x=2 is a maximum. f(2) = -4 + 8 + 3 = 7. Maximum value is 7.
Key Questions: Differentiation
What is a stationary point, and how do you find it?
A stationary point is where the gradient of the curve is zero (dy/dx = 0). To find it, differentiate the function, set the derivative equal to zero, and solve for x. Substitute x into the original equation to find the y-coordinate.
More topics in Unit 2 — Algebra and graphs
Differentiation sits alongside these Mathematics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
Cambridge syllabus keywords to use in your answers
These are the official Cambridge 0580 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Key terms covered in this Differentiation deck
Every term below is defined in the flashcards above. Use the list as a quick recall test before your exam — if you can't define one of these in your own words, flip back to that card.
Related Mathematics guides
Long-read articles that go beyond the deck — cover the whole subject's common mistakes, high-yield content and revision pacing.
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