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Paper 2
Calculus: Differentiation
Both
The derivative gives the gradient of the tangent.
Key Facts
- Differentiate x^n → nx^(n-1)
- Constant terms differentiate to 0.
- Stationary points: dy/dx = 0
- Tangent gradient = dy/dx
- Normal is perpendicular to tangent.
Key Equations
dy/dx = nx^(n-1)
Normal gradient = -1/(dy/dx)
Topics Covered
Gradient Function
What you need to know
- •Differentiate x^n to get nx^(n-1).
- •The derivative gives the gradient at any point.
- •dy/dx represents the rate of change.
Exam Tips
- Write dy/dx clearly in your working.
Tangents and Normals
What you need to know
- •Gradient of tangent = dy/dx at that point.
- •Gradient of normal = -1/(dy/dx).
- •Use y - y₁ = m(x - x₁) for equation of line.
Exam Tips
- Find gradient first, then substitute coordinates.
Stationary Points
What you need to know
- •Stationary points have dy/dx = 0.
- •Solve dy/dx = 0 to find x-coordinates.
- •Substitute back to find y-coordinates.
Exam Tips
- Check whether it's a maximum or minimum using second derivative or gradient changes.
Key Terms
Derivative
Rate of change of a function, written dy/dx.
Tangent
Line that touches a curve at one point with same gradient.
Normal
Line perpendicular to the tangent at a point.
Stationary point
Point where gradient is zero (maximum, minimum, or inflection).
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Common Exam Questions
Differentiate y = 3x^4 - 2x^2 + 5.
2 markseasyPaper 2
Model Answer
dy/dx = 12x^3 - 4x
What examiners want to see
- ✓Apply power rule to each term.
Find the gradient of y = x^3 - 4x at x = 2.
3 marksmediumPaper 2
Model Answer
dy/dx = 3x^2 - 4. At x=2: 3(4) - 4 = 8.
What examiners want to see
- ✓Differentiate first.
- ✓Substitute x value.
Find stationary points of y = x^3 - 3x + 1.
4 markshardPaper 2
Model Answer
dy/dx = 3x^2 - 3 = 0. x^2 = 1, so x = ±1. Points are (1, -1) and (-1, 3).
What examiners want to see
- ✓Set dy/dx = 0.
- ✓Solve for x.
- ✓Find y-coordinates.
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