Math 5A
Spring 2009
Answers and Solutions for Quiz #3
1
1 "2
1 !2 3 " 2
(x 2 " 3) # x 3 " x 3 # 2x
x #(x ! 3) ! 6x 2 $%
3
3
1. y! =
=
(x 2 " 3)2
(x 2 ! 3)2
!3 ! 5x 2
3 + 5x 2
=
OR
!
2
2
3x 3 (x 2 ! 3)2
3x 3 (x 2 ! 3)2
2.
d
(tan x) = sec 2 x
dx
d
(sec x) = sec x tan x
dx
3. Method I using the quotient rule:
d
d
sin x ! (1) "1! (sin x)
d
d ! 1 $
sin x ! (0) "1! cos x
dx
dx
(csc x) = #
=
&=
2
dx
dx " sin x %
(sin x)
sin 2 x
=
!cos x
!1 cos x
=
"
= ! csc x cot x
2
sin x
sin x sin x
. Method II using the chain rule:
d
d
d
!1
(csc x) = (sin x ) = !1" (sin x)!2 (sin x) = !1" (sin x)!2 (cos x) =
dx
dx
dx
=
!cos x
!1 cos x
=
"
= ! csc x cot x
2
sin x
sin x sin x
4. a) y! = cos(x 2 )"
b) y! = 2sin(x)"
d 2
(x ) = cos(x 2 )! (2x) = 2x cos(x 2 )
dx
d
(sin x) = 2sin(x)! cos(x) OR y! = 2sinx cosx
dx
OR using a trig identity y! = sin(2x)