Pure: Differentiation Depth: Fluency and Exam Drill - Worksheets, Questions and Revision

20 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Pure Mathematics

P14D Pure: Differentiation Depth: Fluency and Exam Drill

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

y = (3x - 2)⁵. Find dy/dx.

(Total for Question 1 is 1 mark)

y = e^6x. Find dy/dx.

(Total for Question 2 is 1 mark)

y = (x² + 1)³. Find dy/dx.

(Total for Question 3 is 1 mark)

y = e^x². Find dy/dx.

(Total for Question 4 is 1 mark)

y = ln(4x), for x > 0. Find dy/dx.

(Total for Question 5 is 1 mark)

y = 1/x², for x not equal to 0. Find dy/dx.

(Total for Question 6 is 1 mark)

y = √x, for x > 0. Find dy/dx.

(Total for Question 7 is 1 mark)

y = x³ - 4x. Find the value of dy/dx when x = 2.

(Total for Question 8 is 1 mark)

A curve has parametric equations x = t², y = 3t. State dx/dt and dy/dt.

(Total for Question 9 is 1 mark)

A curve satisfies x² + y² = 25. Differentiate implicitly with respect to x, giving your answer in the form ... = 0.

(Total for Question 10 is 1 mark)

y = (5x + 2)⁴. Find dy/dx, and hence find the exact value of dy/dx when x = 0.

(Total for Question 11 is 2 marks)

y = x e^4x. Show that dy/dx can be written as e^4x(1 + 4x).

(Total for Question 12 is 2 marks)

y = (3x - 1)/(x + 2). Find dy/dx, giving your answer as a single simplified fraction.

(Total for Question 13 is 2 marks)

A curve has parametric equations x = t³, y = 2t², for t in R. Find dy/dx in terms of t.

(Total for Question 14 is 2 marks)

A curve has equation y = (4x² - 3)³.

(a)Find dy/dx, giving your answer in fully factorised form.(2)

(b)Hence find the exact value of dy/dx when x = 1.(2)

(Total for Question 15 is 4 marks)

A curve has equation y = x³ e^-2x.

(a)Show that dy/dx can be written as x² e^-2x(3 - 2x).(2)

(b)Hence find the exact value(s) of x for which the tangent to the curve is horizontal.(2)

(Total for Question 16 is 4 marks)

A curve has equation y = (5x + 1)/(x² + 3).

(a)Find dy/dx, giving your answer as a single simplified fraction.(2)

(b)Hence find the exact x-coordinates of the points where dy/dx = 0, giving your answers in the form (a ± √b)/c.(2)

(Total for Question 17 is 4 marks)

A curve C has equation x² + y² + 6x - 4y = 12. Show that dy/dx = -(x+3)/(y-2).

(Total for Question 18 is 3 marks)

A curve has parametric equations x = 2t - 1, y = t² + 3t, for t in R. Find dy/dx in terms of t, and hence find the value of dy/dx when t = 2.

(Total for Question 19 is 3 marks)

A cylindrical water tank has a fixed radius of 4 m. Water is pumped into the tank at a constant rate of 6 cubic metres per minute. At time t minutes, the depth of water in the tank is h metres.

(a)Show that dh/dt = 3/(8 π).(2)

(b)Find the time taken, in minutes to 3 significant figures, for the depth to increase from 0 m to 2.5 m.(2)

(Total for Question 20 is 4 marks)

Mark scheme · P14D Pure: Differentiation Depth: Fluency and Exam Drill

Question 1

B1 15(3x-2)⁴ cao
Answer: 15(3x-2)⁴

Question 2

B1 6e^6x cao
Answer: 6e^6x

Question 3

B1 6x(x²+1)² cao
Answer: 6x(x²+1)²

Question 4

B1 2x e^x² cao
Answer: 2x e^x²

Question 5

B1 1/x cao
Answer: 1/x

Question 6

B1 -2/x³ cao
Answer: -2/x³

Question 7

B1 1/(2sqrt(x)) cao
Answer: 1/(2sqrt(x))

Question 8

B1 8 cao
Answer: 8

Question 9

B1 dx/dt = 2t, dy/dt = 3, both required
Answer: dx/dt = 2t, dy/dt = 3

Question 10

B1 2x + 2y(dy/dx) = 0 oe
Answer: 2x + 2y(dy/dx) = 0

Question 11

M1 dy/dx = 20(5x+2)³
A1 at x=0: 160 cao
Answer: dy/dx = 20(5x+2)³; at x=0, dy/dx = 160

Question 12

M1 uses the product rule: dy/dx = e^4x + 4x e^4x
A1 cso: factorises to e^4x(1+4x)
Answer: dy/dx = e^4x(1+4x)

Question 13

M1 uses the quotient rule
A1 dy/dx = 7/(x+2)² cao
Answer: dy/dx = 7/(x+2)²

Question 14

M1 finds dy/dt = 4t and dx/dt = 3t²
A1 dy/dx = 4/(3t) cao
Answer: dy/dx = 4/(3t)

Question 15

(a) M1 applies the chain rule: dy/dx = 3(4x²-3)² x 8x
(a) A1 24x(4x²-3)² cao
(a) Answer: dy/dx = 24x(4x²-3)²
(b) M1 substitutes x=1 into the expression from (a)
(b) A1 24 cao
(b) Answer: 24

Question 16

(a) M1 uses the product rule: dy/dx = 3x² e^-2x - 2x³ e^-2x
(a) A1 cso: factorises to x² e^-2x(3-2x)
(a) Answer: dy/dx = x² e^-2x(3-2x)
(b) M1 sets x² e^-2x(3-2x) = 0 and uses that e^-2x is never zero
(b) A1 x = 0 or x = 3/2, both values
(b) Answer: x = 0 or x = 3/2

Question 17

(a) M1 uses the quotient rule
(a) A1 dy/dx = (-5x² - 2x + 15)/(x²+3)² cao
(a) Answer: dy/dx = (-5x²-2x+15)/(x²+3)²
(b) M1 sets the numerator 5x²+2x-15 = 0 and applies the quadratic formula
(b) A1 x = (-1 ± 2sqrt(19))/5 cao
(b) Answer: x = (-1 ± 2sqrt(19))/5

Question 18

M1 differentiates implicitly: 2x + 2y(dy/dx) + 6 - 4(dy/dx) = 0
A1 rearranges to dy/dx(2y - 4) = -(2x + 6)
A1 cso: dy/dx = -(x+3)/(y-2)
Answer: dy/dx = -(x+3)/(y-2)

Question 19

M1 finds dy/dt = 2t + 3 and dx/dt = 2
A1 dy/dx = (2t+3)/2
A1 at t=2: dy/dx = 3.5 cao
Answer: dy/dx = (2t+3)/2; at t=2, dy/dx = 3.5

Question 20

(a) M1 writes V = π r² h = 16 π h and differentiates to get dV/dt = 16 π (dh/dt)
(a) A1 cso: dh/dt = 6/(16 π) = 3/(8 π)
(a) Answer: dh/dt = 3/(8pi)
(b) M1 recognises dh/dt is constant, so time = depth / (dh/dt)
(b) A1 awrt 20.9 minutes
(b) Answer: awrt 20.9 minutes

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Pure: Differentiation Depth: Fluency and Exam Drill - Worksheets, Questions and Revision

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P14D Pure: Differentiation Depth: Fluency and Exam Drill

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