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GCSE · Calculus

C1 Differentiation of Polynomials

AQA 8365 · Calculator allowed · about 95 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Differentiate each of the following expressions with respect to x.

(a)y = x³(1)

(b)y = x⁵(1)

(c)y = 4x²(1)

(d)y = 7x(1)

(e)y = 9(1)

(f)y = 2x⁴(1)

(Total for Question 1 is 6 marks)

Find dy/dx for each of the following curves.

(a)y = x³ + 5x² - 3x + 7(2)

(b)y = 4x⁴ - 6x³ + x(2)

(c)y = 3x⁴ + 2x³ - 5x + 9(2)

(Total for Question 2 is 6 marks)

For each curve, expand the brackets first and then find dy/dx.

(a)y = (x + 4)(x - 2)(3)

(b)y = (2x - 1)(3x + 5)(3)

(Total for Question 3 is 6 marks)

For each curve, find the value of the gradient at the given point.

(a)Curve: y = x³ - 4x² + 5. Calculate the gradient of the curve at the point where x = 2.(3)

(b)Curve: y = 2x² - 7x + 3. Calculate the gradient of the curve at the point where x = -1.(3)

(Total for Question 4 is 6 marks)

Which of the following is the correct derivative of y = 5x⁴ - 3x² + 8?

(a)Select the correct option.(1)

A) 20x³ - 6x
B) 20x³ - 3x
C) 5x³ - 6x
D) 20x³ - 6x + 8

(Total for Question 5 is 1 mark)

For each curve, find the value(s) of x for which the gradient takes the stated value.

(a)Curve: y = x² - 6x + 10. Find the value of x for which the gradient of the curve is 8.(3)

(b)Curve: y = x³ - 3x. Find the value(s) of x for which the gradient of the curve is 9.(3)

(Total for Question 6 is 6 marks)

A curve has equation y = x³ - 3x² - 9x + 5. Find the coordinates of the stationary points of the curve.

(Total for Question 7 is 5 marks)

The curve C has equation y = x³ - 2x² - x + 4. Show that the gradient of C at the point where x = 3 is 14.

(Total for Question 8 is 3 marks)

A curve has equation y = x² - 5x + 6. Find the equation of the tangent to the curve at the point where x = 4.

(Total for Question 9 is 5 marks)

A curve has equation y = x³ - 2x + 1. The point P(2, 5) lies on the curve. Find the equation of the normal to the curve at P, giving your answer in the form x + ky + c = 0.

(Total for Question 10 is 5 marks)

The curve y = x² - 4x + 9 has a tangent that is parallel to the line y = 6x - 1. Find the coordinates of the point on the curve where this occurs.

(Total for Question 11 is 5 marks)

A curve has equation y = 2x³ + 3x² - 12x + 7. Find the x-coordinates of the stationary points of the curve and determine, using the second derivative, whether each is a maximum or a minimum point.

(Total for Question 12 is 6 marks)

A curve has equation y = x³ + 6x² + 15x - 2. Show that this curve is an increasing function for all values of x.

(Total for Question 13 is 4 marks)

A curve has equation y = x³ - 6x² + 2. Find the range of values of x for which the curve is a decreasing function.

(Total for Question 14 is 5 marks)

The volume of water, V cm³, in a container t seconds after a tap is turned on is modelled by V = 30t² - t³, for 0 ≤ t ≤ 30.

(a)Find dV/dt.(2)

(b)Find the rate of change of the volume of water when t = 10 seconds.(2)

(c)Find the value of t at which the volume of water is a maximum, for 0 < t ≤ 30, and find this maximum volume.(5)

(Total for Question 15 is 9 marks)

A curve has equation y = x⁴ - 8x² + 3. Find the coordinates of all the stationary points of the curve, and use the second derivative to determine the nature of each.

(Total for Question 16 is 7 marks)

A farmer has 40 metres of fencing to enclose a rectangular field. One side of the field runs along an existing straight wall, so fencing is only needed for the other three sides. Let x metres be the width of the field, measured perpendicular to the wall, and let A m² be the area enclosed.

(a)Show that A = 40x - 2x^2.(2)

(b)Find the value of x that gives the maximum area, and calculate this maximum area.(5)

(Total for Question 17 is 7 marks)

The cost, C pounds, of producing x items of a product is modelled by C = x³ - 30x² + 225x + 400, for 0 < x ≤ 20.

(a)Find dC/dx.(2)

(b)Find the value of x that minimises the cost, and find this minimum cost.(6)

(Total for Question 18 is 8 marks)

Mark scheme · C1 Differentiation of Polynomials

Question 1

(a) B1 dy/dx = 3x² cao
(a) Answer: dy/dx = 3x²
(b) B1 dy/dx = 5x⁴ cao
(b) Answer: dy/dx = 5x⁴
(c) B1 dy/dx = 8x cao
(c) Answer: dy/dx = 8x
(d) B1 dy/dx = 7 cao
(d) Answer: dy/dx = 7
(e) B1 dy/dx = 0 cao
(e) Answer: dy/dx = 0
(f) B1 dy/dx = 8x³ cao
(f) Answer: dy/dx = 8x³

Question 2

(a) M1 differentiates at least two terms correctly (e.g. 3x² and 10x seen)
(a) A1 dy/dx = 3x² + 10x - 3 cao
(a) Answer: dy/dx = 3x² + 10x - 3
(b) M1 differentiates at least two terms correctly (e.g. 16x³ and -18x² seen)
(b) A1 dy/dx = 16x³ - 18x² + 1 cao
(b) Answer: dy/dx = 16x³ - 18x² + 1
(c) M1 differentiates at least two terms correctly (e.g. 12x³ and 6x² seen)
(c) A1 dy/dx = 12x³ + 6x² - 5 cao
(c) Answer: dy/dx = 12x³ + 6x² - 5

Question 3

(a) M1 expands brackets to obtain a quadratic
(a) A1 correct expansion y = x² + 2x - 8 oe
(a) A1 dy/dx = 2x + 2 cao ft from expansion
(a) Answer: dy/dx = 2x + 2
(b) M1 expands brackets to obtain a quadratic
(b) A1 correct expansion y = 6x² + 7x - 5 oe
(b) A1 dy/dx = 12x + 7 cao ft from expansion
(b) Answer: dy/dx = 12x + 7

Question 4

(a) M1 differentiates to obtain dy/dx = 3x² - 8x
(a) M1 substitutes x = 2 into dy/dx
(a) A1 gradient = -4 cao
(a) Answer: gradient = -4
(b) M1 differentiates to obtain dy/dx = 4x - 7
(b) M1 substitutes x = -1 into dy/dx
(b) A1 gradient = -11 cao
(b) Answer: gradient = -11

Question 5

(a) B1 A cao
(a) Answer: A) 20x³ - 6x

Question 6

(a) M1 differentiates and sets 2x - 6 = 8
(a) M1 rearranges correctly to solve for x
(a) A1 x = 7 cao
(a) Answer: x = 7
(b) M1 differentiates and sets 3x² - 3 = 9
(b) M1 rearranges to x² = 4 and takes both square roots
(b) A1 x = 2 or x = -2 (both values required) cao
(b) Answer: x = 2 or x = -2

Question 7

M1 differentiates to obtain dy/dx = 3x² - 6x - 9
M1 sets dy/dx = 0 and factorises or solves the quadratic
A1 x = -1 and x = 3 (both values) cao
M1 substitutes both x-values back into y = x³ - 3x² - 9x + 5, dependent on previous M1 (dM1)
A1 stationary points (-1, 10) and (3, -22) (both required) cao
Answer: Stationary points at (-1, 10) and (3, -22)

Question 8

M1 differentiates correctly to obtain dy/dx = 3x² - 4x - 1
M1 substitutes x = 3 into dy/dx
A1 gradient = 14 fully shown cso (answer printed, all working correct)
Answer: 14 (shown)

Question 9

B1 y-coordinate at x = 4 is 2 (point (4, 2) found)
M1 differentiates to obtain dy/dx = 2x - 5
A1 gradient at x = 4 is 3
M1 forms tangent equation using y - 2 = 3(x - 4), dependent on gradient found (dM1)
A1 y = 3x - 10 oe (e.g. y - 3x + 10 = 0)
Answer: y = 3x - 10

Question 10

M1 differentiates to obtain dy/dx = 3x² - 2
A1 gradient of tangent at x = 2 is 10
M1 uses normal gradient = -1/(tangent gradient) = -1/10, dependent on tangent gradient found (dM1)
M1 forms normal equation using y - 5 = -1/10 (x - 2)
A1 x + 10y - 52 = 0 oe (e.g. y = -x/10 + 26/5)
Answer: x + 10y - 52 = 0

Question 11

M1 differentiates the curve to obtain dy/dx = 2x - 4
B1 identifies that the gradient of the line y = 6x - 1 is 6
M1 sets 2x - 4 = 6 and solves for x, dependent on both previous marks (dM1)
A1 x = 5
A1 y = 14, so the point is (5, 14)
Answer: (5, 14)

Question 12

M1 differentiates to obtain dy/dx = 6x² + 6x - 12
M1 sets dy/dx = 0 and factorises or solves the quadratic
A1 x = -2 and x = 1 (both values) cao
M1 finds the second derivative d2y/dx2 = 12x + 6
M1 evaluates d2y/dx2 at both x = -2 and x = 1, dependent on previous M1 (dM1)
A1 x = -2 gives a maximum (d2y/dx2 = -18 < 0) and x = 1 gives a minimum (d2y/dx2 = 18 > 0)
Answer: Maximum at x = -2; minimum at x = 1

Question 13

M1 differentiates to obtain dy/dx = 3x² + 12x + 15
M1 attempts to complete the square on 3x² + 12x + 15
A1 dy/dx = 3(x + 2)² + 3 oe, correctly completed
A1 concludes that since (x+2)² ≥ 0 for all x, dy/dx ≥ 3 > 0 for all x, so the curve is increasing for all x (cso)
Answer: dy/dx = 3(x+2)² + 3, which is always at least 3, so the curve is increasing for all values of x (shown)

Question 14

M1 differentiates to obtain dy/dx = 3x² - 12x
M1 sets dy/dx = 0 and factorises to find the critical values, dependent on previous M1 (dM1)
A1 x = 0 and x = 4
M1 tests the sign of dy/dx between and outside the critical values (or sketches the upward parabola dy/dx = 3x(x-4))
A1 0 < x < 4 oe
Answer: 0 < x < 4

Question 15

(a) M1 differentiates at least one term correctly
(a) A1 dV/dt = 60t - 3t² cao
(a) Answer: dV/dt = 60t - 3t²
(b) M1 substitutes t = 10 into dV/dt, ft from part (a)
(b) A1 rate of change = 300 cm³/s cao
(b) Answer: 300 cm³ per second
(c) M1 sets dV/dt = 0 and solves, e.g. 3t(20-t)=0, ft from part (a)
(c) A1 t = 20, rejecting t = 0 as it lies outside the range 0 < t ≤ 30
(c) M1 uses the second derivative d2V/dt2 = 60 - 6t (or a sign test) evaluated at t = 20
(c) A1 d2V/dt2 = -60 < 0, so t = 20 gives a maximum, dependent on previous M1 (dM1)
(c) A1 maximum volume = 4000 cm³ (substituting t = 20 into V)
(c) Answer: t = 20 seconds, maximum volume = 4000 cm³

Question 16

M1 differentiates to obtain dy/dx = 4x³ - 16x
M1 sets dy/dx = 0 and factorises fully, e.g. 4x(x-2)(x+2) = 0
A1 x = 0, x = 2 and x = -2 (all three values) cao
A1 y = 3, y = -13 and y = -13 respectively (ft from the x-values)
M1 finds the second derivative d2y/dx2 = 12x² - 16
M1 evaluates d2y/dx2 at all three x-values, dependent on previous M1 (dM1)
A1 (0, 3) is a maximum (d2y/dx2 = -16 < 0); (2, -13) and (-2, -13) are both minima (d2y/dx2 = 32 > 0)
Answer: Maximum at (0, 3); minima at (2, -13) and (-2, -13)

Question 17

(a) M1 expresses the length of the third fenced side as (40 - 2x) and multiplies by x
(a) A1 A = 40x - 2x² fully shown cso
(a) Answer: A = 40x - 2x² (shown)
(b) M1 differentiates to obtain dA/dx = 40 - 4x, ft from part (a)
(b) M1 sets dA/dx = 0 and solves for x, dependent on previous M1 (dM1)
(b) A1 x = 10
(b) M1 substitutes x = 10 back into A = 40x - 2x², dependent on x = 10 (dM1)
(b) A1 maximum area = 200 m²
(b) Answer: x = 10 m, maximum area = 200 m²

Question 18

(a) M1 differentiates at least two terms correctly
(a) A1 dC/dx = 3x² - 60x + 225 cao
(a) Answer: dC/dx = 3x² - 60x + 225
(b) M1 sets dC/dx = 0 and simplifies, e.g. dividing by 3 to give x² - 20x + 75 = 0, ft from part (a)
(b) M1 factorises or solves to find both stationary values, dependent on previous M1 (dM1)
(b) A1 x = 5 and x = 15 (both values)
(b) M1 finds the second derivative d2C/dx2 = 6x - 60 and evaluates it at x = 15, dependent on x = 15 (dM1)
(b) A1 d2C/dx2 = 30 > 0, so x = 15 gives the minimum cost (x = 5 gives a maximum, since d2C/dx2 = -30 < 0)
(b) A1 minimum cost = 400 pounds, ft from x = 15
(b) Answer: x = 15 items, minimum cost = 400 pounds

Differentiation of Polynomials - Worksheets, Questions and Revision