Find the gradient of the curve y = x3 - 4x + 1 at the point where x = 2.
(Total for Question 5 is 1 mark)
6
Differentiate y = sin(4x) with respect to x.
(Total for Question 6 is 1 mark)
7
Differentiate y = e3x with respect to x.
(Total for Question 7 is 1 mark)
8
Using the chain rule, find dy/dx for y = (2x + 3)4.
(Total for Question 8 is 2 marks)
9
Find the gradient of the curve y = (3x - 1)3 at the point where x = 1.
(Total for Question 9 is 2 marks)
10
Differentiate y = ln(5x) with respect to x, for x > 0.
(Total for Question 10 is 2 marks)
11
Find dy/dx for y = x2 * ex, using the product rule, giving your answer in a fully factorised form.
(Total for Question 11 is 2 marks)
12
Find dy/dx for y = (4x - 1)/(x + 2), using the quotient rule, giving your answer as a single fraction in simplest form.
(Total for Question 12 is 2 marks)
13
Find d2y/dx2 for y = x4 - 6x2 + 3x.
(Total for Question 13 is 2 marks)
14
Using differentiation from first principles, show that if f(x) = 3x2, then f'(x) = 6x. Hence find the gradient of the curve y = 3x2 - 5x + 2 at the point where x = -1.
(Total for Question 14 is 3 marks)
15
The curve C has equation y = √3x + 4, for x > -4/3. Find the equation of the normal to C at the point where x = 4, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(Total for Question 15 is 3 marks)
16
The curve C has equation y = x3 - 3x2 - 9x + 4. Find the x-coordinates of the stationary points of C, and use the second derivative to determine the nature of each.
(Total for Question 16 is 3 marks)
17
A spherical balloon is being inflated so that its volume increases at a constant rate of 200 cm3/s. Show that dr/dt = 50/(π*r2), where r cm is the radius of the balloon at time t seconds. Find the rate of increase of the radius at the instant when r = 10 cm, giving your answer to 3 significant figures.
(Total for Question 17 is 4 marks)
18
A curve has parametric equations x = t2 - 3, y = t3 + 2t. Show that the point (1, 12) lies on the curve, stating the value of t at this point. Find dy/dx in terms of t, and hence find the gradient of the curve at the point (1, 12).
(Total for Question 18 is 4 marks)
19
A closed cylindrical can has radius r cm and height h cm, and a fixed volume of 250 cm3. Show that the surface area S cm2 of the can is given by S = 2*π*r2 + 500/r. Find the value of r that minimises S, giving your answer to 3 significant figures, and find the minimum surface area, giving your answer to the nearest whole number.
(Total for Question 19 is 4 marks)
Mark scheme · P7D Pure: Differentiation: Fluency and Exam Drill
Question 1
B1 5x4 cao
Answer: dy/dx = 5x4
Question 2
B1 21x2 - 2 cao
Answer: dy/dx = 21x2 - 2
Question 3
B1 2x-1/2 cao, oe 2/√x
Answer: dy/dx = 2x-1/2
Question 4
B1 -12x-3 cao
Answer: dy/dx = -12x-3
Question 5
B1 8 cao
Answer: 8
Question 6
B1 4cos(4x) cao
Answer: dy/dx = 4cos(4x)
Question 7
B1 3e3x cao
Answer: dy/dx = 3e3x
Question 8
M1 differentiates using the chain rule, 4(2x+3)3 * 2
A1 8(2x+3)3 cao
Answer: dy/dx = 8(2x+3)3
Question 9
M1 uses the chain rule to find dy/dx = 9(3x-1)2
A1 36 cao
Answer: 36
Question 10
M1 uses the chain rule, dy/dx = 5/(5x)
A1 1/x cao
Answer: dy/dx = 1/x
Question 11
M1 applies the product rule, 2x*ex + x2*ex
A1 x(x+2)ex cao, oe
Answer: dy/dx = x(x+2)ex
Question 12
M1 applies the quotient rule, [4(x+2) - (4x-1)(1)] / (x+2)2
A1 9/(x+2)2 cao
Answer: dy/dx = 9/(x+2)2
Question 13
M1 finds dy/dx = 4x3 - 12x + 3
A1 12x2 - 12 cao
Answer: d2y/dx2 = 12x2 - 12
Question 14
M1 forms f(x+h) = 3(x+h)2 and expands, then finds f(x+h) - f(x) = 6xh + 3h2
M1 divides by h to get 6x + 3h and lets h -> 0 to show f'(x) = 6x
A1 -11 cao
Answer: f'(x) = 6x (shown); gradient at x = -1 is -11
Question 15
M1 differentiates using the chain rule, dy/dx = 3/(2*√3x+4)
M1 evaluates the gradient at x=4 as 3/8 and takes the negative reciprocal, -8/3, for the normal
A1 8x + 3y - 44 = 0 cao, oe
Answer: 8x + 3y - 44 = 0
Question 16
M1 sets dy/dx = 3x2 - 6x - 9 = 0 and factorises to 3(x-3)(x+1) = 0
A1 x = 3 and x = -1 cao
A1 d2y/dx2 = 6x-6 gives minimum at x=3 (d2y/dx2=12>0) and maximum at x=-1 (d2y/dx2=-12<0)
Answer: Minimum at x = 3, maximum at x = -1
Question 17
M1 differentiates V = (4/3)π*r3 to find dV/dr = 4*π*r2
M1 uses the chain rule dr/dt = (dV/dt)/(dV/dr) = 200/(4*π*r2), simplifying to 50/(π*r2)
M1 substitutes r = 10 into the expression for dr/dt