A Level Mathematics (MEI): Differentiation Topic Assessment – Calculus & Derivatives Practice Paper

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MEI A level Mathematics Differentiation

Topic assessment

1. Using the chain rule, differentiate ( x 2 − 1)6 . [3]

2. Show that the gradient of y = ( x 2 − 1)( x − 2)3 is given by
dy
= ( x − 2)2 (5 x 2 − 4 x − 3) . [4]
dx

x −1
3. Show that the gradient of the curve y = at the point where x = 2 is -3 [5]
x2 − 3

4. A potter is making an open topped vessel shaped as a right circular cylinder of radius r
and height 2r.
(a) Show that the volume V of the vessel is given by V = 2 r 3 [1]
(b) Find the rate at which the volume is increasing when the radius is 2 cm and
increasing at a rate of 0.25 cm/s. [4]
3
(c) Given that the volume is increasing at a rate of 5 cm /s when the radius is 5 cm,
find the rate at which the surface area is increasing at this point. [6]

5. In this question you must show detailed reasoning.
A curve has equation y = 3x 4 − 8 x3 + 6 x 2 + 1 .
(a) Find the coordinates of the stationary points and determine their nature. [6]
(b) Sketch the curve. [2]
(c) Find the values for x for which the curve is convex. [3]

6. In this question you must show detailed reasoning.
A log of wood is modelled as a cylinder with radius 10 cm and height 25 cm. It is to be
made into a cuboid with dimensions 2x cm by 2y cm by 25 cm by trimming the cylinder.
The cross-section is shown in the diagram.




(a) Find the value of x for which the area of the rectagle is maximum. [8]
(b) Calculate the volume of the largest cuboid that can be cut from the log. [2]


Total 44 marks




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MEI A level Maths Differentiation Assessment solutions


Topic assessment solutions

1. y = ( x 2 − 1)6
du
Let u = x 2 − 1  = 2x
dx
dy
y = u6  = 6u5
du
dy dy du
Using the chain rule: =  = 6u5  2 x
dx du dx
= 12 x ( x 2 − 1)5
[3]

2. y = ( x 2 − 1)( x − 2)3
du
Let u = x 2 − 1  = 2x
dx
dv
Let v = ( x − 2)3  = 3( x − 2)2
dx
dy dv du
Using the product rule: =u +v
dx dx dx
= ( x − 1)  3( x − 2)2 + ( x − 2)3  2 x
2


= ( x − 2)2 3( x 2 − 1) + 2 x ( x − 2)
= ( x − 2)2 (3x 2 − 3 + 2 x 2 − 4 x )
= ( x − 2)2 (5 x 2 − 4 x − 3)
[4]

x −1
3. y =
x2 − 3
du
Let u = x − 1  =1
dx
dv
Let v = x 2 − 3  = 2x
dx
du dv
v −u
dy dx dx
Using the quotient rule: =
dx v 2

( x − 3)  1 − ( x − 1)  2 x
2
=
( x 2 − 3)2
x 2 − 3 − 2x 2 + 2x
=
( x 2 − 3)2
− x 2 − 3 + 2x
=
( x 2 − 3)2


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