Mathematics (9709)
y = 3x⁴ → dy/dx = 12x³ y = 5 → dy/dx = 0 (constant) y = x⁻² → dy/dx = −2x⁻³
Maximum: d²y/dx² < 0 (curve concave down) Minimum: d²y/dx² > 0 (curve concave up) If d²y/dx² = 0, check nature using a sign table
Maximum/minimum problems (optimisation) Rate of change problems Finding equations of tangents and normals
Topic 3 of 4Cambridge A Levels
Differentiation
First and second derivatives, tangents, normals and stationary points
Differentiation finds the rate of change of a function — the gradient at any point.
Basic rules: If y = xⁿ, then dy/dx = nxⁿ⁻¹.
Tangent at point (a, f(a)): gradient = f'(a). Equation: y − f(a) = f'(a)(x − a).
Normal is perpendicular: gradient = −1/f'(a).
Stationary points (where dy/dx = 0):
Applications:
Key Points to Remember
- 1If y = xⁿ, dy/dx = nxⁿ⁻¹
- 2Stationary points where dy/dx = 0
- 3Second derivative test: negative = max, positive = min
- 4Tangent gradient = f'(a); normal gradient = -1/f'(a)
Pakistan Example
Maximising Profit — Calculus in Pakistani Business
A Lahore textile factory's profit function is P(x) = −2x² + 200x − 3000 where x is production in thousands. Setting dP/dx = 0: −4x + 200 = 0, x = 50. Since d²P/dx² = −4 < 0, this is a maximum. Optimal production is 50,000 units for maximum profit of Rs. 2,000 per unit.