Free Differentiation questions for AS Level Mathematics (9709), part of Pure Mathematics 2. Each question has the answer and the marking points an examiner would look for.
Generate a full Differentiation worksheet →All Pure Mathematics 2 questionsA curve has equation y = ln xx2 for x > 0.
Find the exact coordinates of the stationary point.
Determine the nature of the stationary point.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | (e1/2, 12e) quotient rule M1 1 − 2ln xx3 A1 ln x = 12 M1 (e1/2, 12e) A1 | 4 |
| (b) | maximum: gradient changes from positive to negative valid method (sign of derivative either side or second derivative) M1 maximum A1 | 2 |
The parametric equations of a curve are x = t2 + 3, y = t3 − 9t.
Find dydx in terms of t and the coordinates of the stationary points.
| Part | Answer and marking points | Marks |
|---|---|---|
| 3t2 − 92t; (6, ±6√3) dx/dt and dy/dt M1 3t2 − 92t A1 t = ±√3 M1 x = 6 A1 y = ∓6√3 A1 | 5 |
Find the gradient of the curve y = x2e3x at the point where x = 1. Give your answer correct to 3 significant figures.
| Part | Answer and marking points | Marks |
|---|---|---|
| 100 Product rule attempted M1 2xe3x + 3x2e3x A1 Substitute x = 1 M1 100 A1 | 4 |
A curve has equation y = ln xx3 for x > 0.
Find the exact coordinates of the stationary point.
Determine the nature of the stationary point.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | (e1/3, 13e) quotient rule M1 1 − 3ln xx4 A1 ln x = 13 M1 (e1/3, 13e) A1 | 4 |
| (b) | maximum: gradient changes from positive to negative valid method (sign of derivative either side or second derivative) M1 maximum A1 | 2 |
The parametric equations of a curve are x = t2 + 4, y = t3 − 12t.
Find dydx in terms of t and the coordinates of the stationary points.
| Part | Answer and marking points | Marks |
|---|---|---|
| 3t2 − 122t; (8, ±8√4) dx/dt and dy/dt M1 3t2 − 122t A1 t = ±√4 M1 x = 8 A1 y = ∓8√4 A1 | 5 |
Find the gradient of the curve y = x2e4x at the point where x = 2. Give your answer correct to 3 significant figures.
| Part | Answer and marking points | Marks |
|---|---|---|
| 59600 Product rule attempted M1 2xe4x + 4x2e4x A1 Substitute x = 2 M1 59600 A1 | 4 |
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