Pure: Differentiation - Worksheets, Questions and Revision

13 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 5)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

A-Level · Pure Mathematics

P7 Pure: Differentiation

EDEXCEL 9MA0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question concerns differentiation from first principles.

(a)Using differentiation from first principles, show that if f(x) = x² then f'(x) = 2x.(4)

(b)Hence, or otherwise, find the gradient of the curve y = x² - 5x + 3 at the point where x = 4.(2)

(Total for Question 1 is 6 marks)

y = 3x⁴ - 2/x³ + 4sqrt(x) - 7, for x > 0.

(a)Find dy/dx, writing each term with a positive or negative index or a root as appropriate.(4)

(b)Hence find the exact value of dy/dx when x = 4, giving your answer as a single fraction in its simplest form.(3)

(Total for Question 2 is 7 marks)

The curve C has equation y = (3x² - 4x + 1)⁵.

(a)Using the chain rule, find dy/dx.(3)

(b)Find the exact value of the gradient of C at the point where x = 2.(2)

(c)Find the equation of the tangent to C at the point where x = 2, giving your answer in the form y = mx + c.(3)

(Total for Question 3 is 8 marks)

The curve C has equation y = (x² + 1)(x - 3)³.

(a)Using the product rule, show that dy/dx = (x - 3)² (5x² - 6x + 3).(5)

(b)Find the gradient of C at the point where x = 4.(2)

(Total for Question 4 is 7 marks)

The curve C has equation y = (2x - 1)/(x² + 3).

(a)Using the quotient rule, find dy/dx, giving your answer as a single fraction in simplest form.(5)

(b)Hence find the x-coordinates of the stationary points of C, giving your answers to 3 significant figures.(3)

(Total for Question 5 is 8 marks)

y = 2sin(3x) - cos²(x), where x is measured in radians.

(a)Find dy/dx.(4)

(b)Find the exact value of dy/dx when x = π/6.(3)

(c)Using the double angle identity sin(2x) = 2sin(x)cos(x), show that dy/dx = 6cos(3x) + sin(2x).(2)

(Total for Question 6 is 9 marks)

y = ln(3x² + 2) + e^{1 - 2x}.

(a)Find dy/dx.(4)

(b)Find the exact value of dy/dx when x = 0.(3)

(c)Find the equation of the tangent to the curve at the point where x = 0, giving your answer in the form y = mx + c, where c is given in terms of e and ln(2).(3)

(Total for Question 7 is 10 marks)

The curve C has equation y = 2x³ - 9x² + 12x + 5.

(a)Find dy/dx and d²y/dx².(3)

(b)Find the coordinates of the stationary points of C.(4)

(c)Using the second derivative, determine the nature of each stationary point.(2)

(Total for Question 8 is 9 marks)

The curve C has equation y = √2x + 5, for x > -2.5.

(a)Find dy/dx.(3)

(b)Find the equation of the tangent to C at the point where x = 2, giving your answer in the form ax + by + c = 0, where a, b and c are integers.(4)

(c)Find the equation of the normal to C at the point where x = 2, giving your answer in the form y = mx + c.(4)

(Total for Question 9 is 11 marks)

Water is poured into an inverted conical container at a constant rate of 30 cm³/s. The container has height 40 cm and radius 20 cm at the top. By similar triangles, when the depth of water is h cm the radius of the water surface is h/2 cm, so the volume of water is V = (π h³)/12 cm³.

(a)Show that dV/dh = (π h²)/4.(2)

(b)Given that dV/dt = 30, find an expression for dh/dt in terms of h.(3)

(c)Find the rate at which the depth of the water is increasing at the instant when h = 10 cm, giving your answer to 3 significant figures.(2)

(d)State one assumption made in modelling this situation that could affect the accuracy of your answer to part (c).(1)

(Total for Question 10 is 8 marks)

A curve has parametric equations x = t² + 1, y = t³ - 3t.

(a)Find dy/dx in terms of t, giving your answer as a single fraction in simplest factorised form.(4)

(b)Show that the point (5, 2) lies on the curve, and find the value of t at this point.(2)

(c)Find the gradient of the curve at the point (5, 2).(2)

(d)Find the equation of the tangent to the curve at the point (5, 2), giving your answer in the form y = mx + c.(3)

(Total for Question 11 is 11 marks)

The curve C is defined by the equation x² + 3xy - y² = 9. The point P(2, 1) lies on C.

(a)Show that P(2, 1) lies on C.(1)

(b)Find dy/dx in terms of x and y.(5)

(c)Find the gradient of C at the point P(2, 1).(2)

(d)Find the equation of the normal to C at P(2, 1), giving your answer in the form ax + by + c = 0, where a, b and c are integers.(3)

(Total for Question 12 is 11 marks)

An open-topped box is made from a rectangular sheet of card measuring 24 cm by 15 cm, by cutting a square of side x cm from each corner and folding up the sides, where 0 < x < 7.5.

Diagram NOT accurately drawn

(a)Show that the volume of the box, V cm³, is given by V = 4x³ - 78x² + 360x.(3)

(b)Find dV/dx.(2)

(c)Using dV/dx, find the value of x that maximises the volume of the box, justifying that it gives a maximum.(5)

(d)Hence find the maximum volume of the box.(2)

(Total for Question 13 is 12 marks)

Mark scheme · P7 Pure: Differentiation

Question 1

(a) M1 attempts f(x+h) = (x+h)² (oe, may be unsimplified)
(a) M1 forms [f(x+h) - f(x)]/h and expands numerator to 2xh + h² (oe)
(a) dM1 simplifies the quotient to 2x + h (dependent on previous M1)
(a) A1 cso: states as h -> 0, f'(x) = 2x (answer printed; limit statement required)
(a) Answer: f'(x) = 2x
(b) M1 dy/dx = 2x - 5
(b) A1 gradient = 3 (cao)
(b) Answer: 3

Question 2

(a) M1 attempts to differentiate at least one term of the form axⁿ using axⁿ -> anx^n-1
(a) A1 3x⁴ -> 12x³ correct
(a) A1 -2x^-3 -> 6x^-4 (oe 6/x⁴) correct
(a) A1 4x^1/2 -> 2x^-1/2 (oe 2/√x) correct, with -7 -> 0
(a) Answer: dy/dx = 12x³ + 6x^-4 + 2x^-1/2 (oe 12x³ + 6/x⁴ + 2/√x)
(b) M1 substitutes x = 4 into their dy/dx
(b) dM1 evaluates all three terms correctly: 768, 3/128, 1 (oe, dependent on M1)
(b) A1 cao: dy/dx = 98435/128
(b) Answer: 98435/128

Question 3

(a) M1 attempts chain rule of the form n(3x² - 4x + 1)^n-1 x (derivative of bracket)
(a) A1 5(3x² - 4x + 1)⁴ correct outer part
(a) A1 (6x - 4) correct derivative of the bracket, combined cao
(a) Answer: dy/dx = 5(3x² - 4x + 1)⁴ (6x - 4)
(b) M1 substitutes x = 2 into their dy/dx: 5(5)⁴(8)
(b) A1 cao: gradient = 25000
(b) Answer: 25000
(c) M1 finds y-coordinate at x = 2: y = 5⁵ = 3125
(c) M1 uses y - 3125 = 25000(x - 2) (ft their gradient)
(c) A1 cao: y = 25000x - 46875
(c) Answer: y = 25000x - 46875

Question 4

(a) M1 applies product rule u'v + uv' with u = x²+1, v = (x-3)³ correctly identified
(a) A1 u' = 2x correct
(a) A1 v' = 3(x-3)² correct
(a) dM1 factorises out the common factor (x-3)² from 2x(x-3)³ + 3(x²+1)(x-3)² (dependent on M1)
(a) A1 cso: simplifies fully to (x-3)²(5x² - 6x + 3)
(a) Answer: dy/dx = (x - 3)² (5x² - 6x + 3)
(b) M1 substitutes x = 4 into their dy/dx
(b) A1 cao: gradient = 59
(b) Answer: 59

Question 5

(a) M1 correct quotient rule statement with u = 2x-1, u' = 2, v = x²+3, v' = 2x identified
(a) A1 numerator (unsimplified): 2(x²+3) - (2x-1)(2x)
(a) dM1 expands and simplifies the numerator (dependent on M1)
(a) A1 numerator simplifies to -2x² + 2x + 6 (oe -2(x²-x-3))
(a) A1 cao: dy/dx = (-2x²+2x+6)/(x²+3)² (oe -2(x²-x-3)/(x²+3)²)
(a) Answer: dy/dx = -2(x² - x - 3)/(x² + 3)²
(b) M1 sets numerator of dy/dx equal to zero and simplifies to a 3-term quadratic, e.g. x² - x - 3 = 0
(b) A1 exact solutions x = (1+√13)/2 and x = (1-√13)/2 (oe, via quadratic formula)
(b) A1 awrt x = 2.30 and x = -1.30 (both required, ft)
(b) Answer: x = 2.30 or x = -1.30 (awrt, exact: x = (1 ± √13)/2)

Question 6

(a) B1 2sin(3x) -> 6cos(3x)
(a) M1 attempts chain rule on cos²(x): 2cos(x) x (-sin(x))
(a) A1 -cos²(x) -> 2sin(x)cos(x) (oe sin(2x)) correct
(a) A1 cao: dy/dx = 6cos(3x) + 2sin(x)cos(x) fully combined
(a) Answer: dy/dx = 6cos(3x) + 2sin(x)cos(x)
(b) M1 substitutes x = π/6 into their dy/dx
(b) M1 uses exact values sin(π/6) = 1/2, cos(π/6) = √3/2, cos(π/2) = 0
(b) A1 cao: dy/dx = √3/2
(b) Answer: √3/2
(c) M1 substitutes sin(2x) = 2sin(x)cos(x) into their expression from part (a)
(c) A1 cso: dy/dx = 6cos(3x) + sin(2x)
(c) Answer: dy/dx = 6cos(3x) + sin(2x) (shown)

Question 7

(a) M1 differentiates ln(3x²+2) using the chain rule, form (6x)/(3x²+2)
(a) A1 6x/(3x²+2) correct
(a) M1 differentiates e^1-2x using the chain rule, form e^1-2x x (-2)
(a) A1 -2e^1-2x correct, combined cao
(a) Answer: dy/dx = 6x/(3x²+2) - 2e^1-2x
(b) M1 substitutes x = 0 into their dy/dx
(b) M1 evaluates each term: 6(0)/2 = 0 and -2e¹ = -2e
(b) A1 cao: dy/dx = -2e
(b) Answer: -2e
(c) M1 finds y-coordinate at x=0: y = ln(2) + e
(c) M1 uses y - (ln2+e) = -2e(x - 0) (ft their gradient)
(c) A1 cao: y = -2ex + (e + ln2)
(c) Answer: y = -2ex + (e + ln2)

Question 8

(a) M1 attempts to differentiate term by term
(a) A1 dy/dx = 6x² - 18x + 12
(a) B1 ft: d²y/dx² = 12x - 18
(a) Answer: dy/dx = 6x² - 18x + 12, d²y/dx² = 12x - 18
(b) M1 sets dy/dx = 0 and simplifies to a 3-term quadratic, e.g. x² - 3x + 2 = 0
(b) M1 solves to x = 1 and x = 2 (factorising or formula)
(b) A1 y = 10 when x = 1
(b) A1 y = 9 when x = 2
(b) Answer: (1, 10) and (2, 9)
(c) M1 substitutes x = 1 and x = 2 into their d²y/dx² (ft)
(c) A1 cao: (1,10) is a maximum since d²y/dx² = -6 < 0; (2,9) is a minimum since d²y/dx² = 6 > 0
(c) Answer: (1,10) maximum; (2,9) minimum

Question 9

(a) M1 rewrites y = (2x+5)^1/2
(a) M1 applies chain rule: (1/2)(2x+5)^-1/2 x 2
(a) A1 cao: dy/dx = (2x+5)^-1/2 (oe 1/√2x+5)
(a) Answer: dy/dx = 1/√2x + 5
(b) M1 finds y-coordinate at x=2: y = √9 = 3
(b) M1 finds gradient at x=2: dy/dx = 1/3
(b) M1 uses y - 3 = (1/3)(x-2)
(b) A1 cao: x - 3y + 7 = 0 (oe with integer a,b,c)
(b) Answer: x - 3y + 7 = 0
(c) M1 normal gradient = -1/(their tangent gradient) = -3
(c) M1 uses y - 3 = -3(x - 2)
(c) A1 expands correctly to y = -3x + 6 + 3
(c) A1 cao: y = -3x + 9
(c) Answer: y = -3x + 9

Question 10

(a) M1 differentiates V = (π h³)/12 to get (π/12)(3h²)
(a) A1 cso: dV/dh = (π h²)/4
(a) Answer: dV/dh = (π h²)/4 (shown)
(b) M1 states the chain rule dh/dt = (dV/dt) / (dV/dh)
(b) M1 substitutes to get dh/dt = 30 / ((π h²)/4)
(b) A1 cao: dh/dt = 120/(π h²)
(b) Answer: dh/dt = 120/(π h²)
(c) M1 substitutes h = 10 into their dh/dt
(c) A1 awrt 0.382 cm/s
(c) Answer: 0.382 cm/s (awrt)
(d) B1 valid modelling assumption, e.g. the container is a perfect cone with no leaks/deformation, or water is poured in at a genuinely constant rate, or no water is lost to splashing/evaporation
(d) Answer: e.g. the cone is assumed to be a perfect rigid cone with no leaks and the inflow rate is exactly constant

Question 11

(a) M1 dx/dt = 2t
(a) A1 dy/dt = 3t² - 3
(a) M1 uses dy/dx = (dy/dt)/(dx/dt)
(a) A1 cao: dy/dx = 3(t-1)(t+1)/(2t) (oe 3(t²-1)/(2t))
(a) Answer: dy/dx = 3(t - 1)(t + 1)/(2t)
(b) M1 solves t² + 1 = 5 to get t = 2 or t = -2
(b) A1 cso: selects t = 2 using y = t³-3t = 2 (t=-2 gives y=-2, rejected)
(b) Answer: t = 2
(c) M1 substitutes t = 2 into their dy/dx
(c) A1 cao: gradient = 9/4
(c) Answer: 9/4
(d) M1 uses y - 2 = (9/4)(x - 5) (ft their gradient)
(d) M1 rearranges to isolate y
(d) A1 cao: y = (9/4)x - 37/4
(d) Answer: y = (9/4)x - 37/4

Question 12

(a) B1 cso: substitutes x=2, y=1 to get 4 + 6 - 1 = 9, confirming P lies on C
(a) Answer: 4 + 6 - 1 = 9 (confirmed)
(b) M1 differentiates x² implicitly to get 2x
(b) M1 differentiates 3xy using the product rule to get 3y + 3x(dy/dx)
(b) M1 differentiates -y² implicitly to get -2y(dy/dx)
(b) M1 collects dy/dx terms and factorises: dy/dx(3x - 2y) = -2x - 3y
(b) A1 cao: dy/dx = -(2x+3y)/(3x-2y) (oe (2x+3y)/(2y-3x))
(b) Answer: dy/dx = -(2x + 3y)/(3x - 2y)
(c) M1 substitutes x=2, y=1 into their dy/dx
(c) A1 cao: gradient = -7/4
(c) Answer: -7/4
(d) M1 normal gradient = -1/(their tangent gradient) = 4/7
(d) M1 uses y - 1 = (4/7)(x - 2) (ft)
(d) A1 cao: 4x - 7y - 1 = 0 (oe with integer a,b,c)
(d) Answer: 4x - 7y - 1 = 0