1. Differentiate the following functions with respect to x.
a) Use the quotient rule to differentiate
y = 2x^4 – 3x /4x – 1
y=
2x 4 − 3x
4x −1
dy d " 2x 4 − 3x %
= $
'
dx dx # 4x −1 &
dy (
=
dx
4x −1) *
d
d
2x 4 − 3x ) − ( 2x 4 − 3x ) * ( 4x −1)
(
dx
dx
2
( 4x −1)
3
4
dy ( 4x −1) (8x − 3) − ( 2x − 3x ) ( 4)
=
2
dx
( 4x −1)
4
3
4
dy (32x − 8x −12x + 3) − (8x −12x )
=
2
dx
( 4x −1)
dy 24x 4 − 8x 3 + 3
=
2
dx
( 4x −1)
b) Use the chain rule to differentiate y = 6 cos(x^3 + 3)
y = 6 cos ( x 3 + 3)
dy d
= 6 cos ( x 3 + 3)
dx dx
dy "
d
= #−6sin ( x 3 + 3)$% ( x 3 + 3)
dx
dx
dy "
= #−6sin ( x 3 + 3)$%(3x 2 )
dx
dy
= −18x 2 sin ( x 3 + 3)
dx
c) Select an appropriate rule to differentiate
y= (4x^2- e^2x) sin(3x)
y = ( 4x 2 − e−2 x ) sin (3x )
dy d
= ( 4x 2 − e−2 x ) sin (3x )
dx dx
dy
d
d
= sin (3x ) * ( 4x 2 − e−2 x ) + ( 4x 2 − e−2 x ) * sin (3x )
dx
dx
dx
dy
= sin (3x ) (8x 2 + 2e−2 x ) + ( 4x 2 − e−2 x ) (3cos3x )
dx
dy
= (8x 2 + 2e−2 x ) sin (3x ) + (12x 2 − 3e−2 x ) ( cos3x )
dx
2) The angular displacement, 0 radians, of the spoke of a wheel is given by the expression
0= 0.5t^4- t^3, where t is time in seconds. Find;
a) The angular velocity after 2 seconds.
θ = 0.5t 4 − t 3
ω=
d
d
θ = ( 0.5t 4 − t 3 )
dt
dt
ω = ( 0.5) ( 4) t 3 − 3 (t 2 )
ω = 2t 3 − 3t 2
3
ω (2) = 2 (2) − 3(2)
2
ω ( 2 ) = 16 −12
ω ( 2 ) = 4 rad/sec
b) The angular acceleration after 3 seconds.
α=
d
d
ω = ( 2t 3 − 3t 2 )
dt
dt
α = 2 (3) t 2 − 3 ( 2 ) t
α = 6t 2 − 6t
2
α (3) = 6 (3) − 6 (3)
α (3) = 54 −18
α (3) = 36 rad/sec 2
c) The time when the angular acceleration is zero.
α (t ) = 6t 2 − 6t = 0
6 (t 2 − t ) = 0
6 (t ) (t −1) = 0
The solutions are t = 0 and t = 1.
3
a) From a rectangular sheet of metal measuring 120 mm by 75 mm, equal squares of side
x are cut from each of the 4 corners. The remaining flaps are then folded upwards to form
an open box.
a) Draw a neat diagram of the rectangular sheet of metal and show the dimensions given
including the squares of side x.
b) Show that the volume of the box is given by :
V= 9000x – 390x^2 + 4x^3
The dimensions of the base of the box are (120 – 2x) by (75 – 2x).
The height of the box is x.
The volume is then:
Volume = (120 − 2x ) ( 75 − 2x ) x
= "#(120 ) ( 75) + (−2x ) ( 75) + (120 ) (−2x ) + (−2x ) (−2x )$%( x )
= ( 9000 −150x − 240x + 4x 2 ) ( x )
= ( 9000 − 390x + 4x 2 ) ( x )
= 9000x − 390x 2 + 4x 3
c) Find the value of x such that the volume is maximum:
V = 9000x - 390x^2 + 4x^3
dV d
= ( 9000x − 390x 2 + 4x 3 )
dx dx
dV
= 9000 − ( 2 ) (390x ) + 3 ( 4) ( x 2 )
dx
dV
= 9000 − 780x +12x 2
dx
The maximum will occur where dV/dx = 0:
dV
= 9000 − 780x +12x 2 = 0
dx
x=
x=
780 ±
2
(−780) − 4 (12) (9000)
2 (12 )
780 ± 176400
24
x = 50 and x = 15
The solution x = 50 must be discarded because a square of 50mm cannot be
removed from both corners along the edge measuring 75mm. After the first
square was removed, there would not be sufficient material remaining to remove
the other square.
The maximum value occurs when x = 15 mm.