MAT 131 Homework Solutions – Chapter 4
Selected Solutions WC 4th edition
Section 4.1
Problems
Page 304
1-37 every other odd, 53, 57, 59, 63, 69, 73, 77
Problem 1
f (x) = 3x ! f '(x) = 3
f (x) = 3" x ! f '(x) = 0 " x + 3"1 = 3
Problem 5
f (x) = x x + 3 = x 2 + 3x ! f '(x) = 2x + 3
( )
f (x) = x ( x + 3) ! f '(x) = 1( x + 3) + x (1) = 2x + 3
Problem 9
2
!2
s(x) = = 2x !1 " s'(x) = !2x !2 = 2
x
x
x 0 ! 2 1 !2
2
f (x) = " f '(x) =
= 2
x
x2
x
( ) ()
Problem 13
f (x) = 3x 4x 2 ! 1 " f '(x) = 3x 8x + 3 4x 2 ! 1 = 24x 2 + 12x 2 ! 3 = 36x 2 ! 3
(
)
( ) (
)
Problem 17
(
) (
)(
)
f '(x) = 2 ( 2x + 3) + ( 2x + 3) ( 2 ) = 4 ( 2x + 3) = 8x + 12
2
f (x) = 2x + 3 = 2x + 3 2x + 3
Problem 21
f (x) = x + 1 x 2 ! 1
(
(
)(
) (
)
)( )
f '(x) = 1 x 2 ! 1 + x + 1 2x = x 2 ! 1 + 2x 2 + 2x = 3x 2 + 2x ! 1
Problem 25
(
) (
2
)(
)
f (x) = 2x 2 ! 4x + 1 = 2x 2 ! 4x + 1 2x 2 ! 4x + 1
(
)(
) (
)(
) (
)(
)
f '(x) = 4x ! 4 2x 2 ! 4x + 1 + 2x 2 ! 4x + 1 4x ! 4 = 8 2x 2 ! 4x + 1 x ! 1
Problem 29
Use the formula on page 300
fgh ' = f ' gh + fg ' h + fgh'
( )
1
(
)(
)
f '(x) = 2x ( 2x + 3) ( 7x + 2 ) + x ( 2 ) ( 7x + 2 ) + x ( 2x + 3) ( 7 )
f (x) = x 2 2x + 3 7x + 2
2
Problem 33
f (x) =
f '(x) =
(
2
(
)
)(
)
)
1
1
!
1$
x + 1 # x + 2 & = x 2 + 1 x 2 + x '2
"
x %
( x )( x + x ) + ( x + 1)( x
'1
2
1
2
1
2
1
2
'2
! 1 $!
1$
=#
x
+
+
x 2 &%
" 2 x &% #"
(
1
2
'1
2
' 2x '3
)
! 1
2$
x +1 #
' 3&
"2 x x %
Problem 37
2x 2 + 4x + 1
y(x) =
=
3x ! 1
3x ! 1 4x + 4 ! 2x 2 + 4x + 1 3
y' =
2
3x ! 1
(
=
)(
(
) (
)( )
)
12x + 8x ! 4 ! 6x ! 12x ! 3
2
2
(3x ! 1)
2
=
6x 2 ! 4x ! 7
(3x ! 1)
2
Problem 53
! x $
y = x + 1+ 2#
" x + 1&%
(
)
! 1' x + 1 ( x '1$
2
& = 1+
y ' = 1+ 0 + 2#
2
2
#"
&%
x +1
x +1
(
)
(
)
Problem 57
d " 2
2
$ = d " x 4 ! x 2 $ = 4x 3 ! 2x
x
+
x
x
!
x
%
% dx #
dx #
(
)(
)
Problem 59
d " 3
x + 2x x 2 ! x $%
#
x=2
dx
" 3x 2 + 2 x 2 ! x + x 3 + 2x 2x ! 1 $
#
%x=2
(
(
)(
)(
)
) (
)(
)
(3(2) + 2)((2) ! (2)) + ((2) + 2(2))(2(2) ! 1) = 14 & 2 + 12 & 3 = 64
2
2
3
2
Problem 63
Equation of the line tangent to the graph at the point
f (x) = x 2 + 1 x 3 + x x = 1 f (1) = 4 ! x0 , y0 = 1,4
(
)(
)
f '(x) = ( 2x ) ( x + x ) + ( x + 1) ( 3x + 1)
3
2
(
) ( )
2
( )( ) ( )( )
= m ( x " x ) ! y = 12 ( x " 1) + 4 = 12x " 8
f '(1) = 2 2 + 2 4 = 12
y " y0
0
Problem 69
Rate of change of monthly sales = q'(t) =2000 - 200t.
When t = 5: q'(5) = 2000 - 200(5) = 1000 units/month
Therefore, sales are increasing at a rate of 1000 units per month.
Rate of change of price = p'(t) = -2t
When t = 5: p'(5) = -2(5) = -$10/month.
Therefore, The price of a sound system is dropping at a rate of $10 per month.
Revenue:
R(t) = p(t)q(t) = (1000 - t 2 )(2000t - 100t 2 )
Rate of change of revenue:
R'(t) = p '(t)q(t) + p(t)q '(t) = (-2t)(2000t - 100t 2 ) + (1000 - t 2 )(2000 - 200t)
R'(5) = [-2(5)][2000(5)-100(5)2]+ [1000-(5)2][2000-200(5)]= $900,000/month
Therefore, revenue is increasing at a rate of$900,000 per month.
Problem 73
Let S(t) be the number of T-shirts sold per day.
If t = 0 is now, we are told that S(0) = 20and S'(0) = -3.
Let p(t) be the price of T-shirts. We are told that p(0) = 7 and p'(0) = 1. The
revenue is then R(t) = S(t)p(t), so R'(0) =S'(0)p(0) + S(0)p'(0) = -3(7) + 20(1) = -1.
So, revenue is decreasing at a rate of $1 per day.
Problem 77
(
)
!3000 1 ! 3600x !2
3000
, so M'(10) is about0.7670
M (x) =
" M '(x) =
2
!1
x + 3600x !1
x + 3600x
(
)
mpg/mph.
This means that, at a speed of10 mph, the fuel economy is increasing at a rate
of 0.7670 miles per gallon per one mph increase in speed. M'(60) = 0 mpg/mph. This
means that, at a speed of 60 mph, the fuel economy is neither increasing nor decreasing
with increasing speed.
M'(70) is about -0.0540. This means that, at 70 mph, the fuel economy is decreasing at a
rate of 0.0540miles per gallon per one mph increase in speed.
60 mph is the most fuel-efficient speed for the car
3
Section 4.2 – The Chain Rule
Page 316
Problems 1-57 every other odd, 61, 63
Problem 1
2
f (x) = ( 2x + 1) ! u = 2x + 1 y = u 2
dy
du
dy du
= 2u
=2!
"
= 2u " 2
du
dx
du dx
f '(x) = 2 ( 2x + 1) 2 = 8x + 4
Problem 5
!2
f (x) = ( 2 ! x ) " u = 2 ! x
y = u !2
dy
= !2u !3
du
du
dy du
= !1 "
#
= !2u !3 # ( !1)
dx
du dx
2
!3
!3
f '(x) = !2 ( 2 ! x ) ( !1) = 2 ( 2 ! x ) =
( 2 ! x )3
Problem 9
!1
f (x) = ( 4x ! 1) " u = 4x ! 1 y = u !1
dy
= !1u !2
du
du
dy du
=4"
#
= !u !2 # ( 4 )
dx
du dx
f '(x) = ! ( 4x ! 1) # ( 4 ) = !4 ( 4x ! 1) =
!2
Problem 13
(
f (x) = x 2 + 2x
dy
= 4u 3
du
)
4
!2
! u = x 2 + 2x
!4
( 4x ! 1)2
y = u4
du
dy du
= 2x + 2 !
"
= 4u 3 " ( 2x + 2 )
dx
du dx
(
)
f '(x) = 4 x 2 + 2x " ( 2x + 2 ) = 8x 3 ( x + 2 ) ( x + 1)
Problem 17
(
3
)
f (x) = x 2 ! 3x ! 1
3
!5
(
)
" u = x 2 ! 3x ! 1
y = u !5
dy
= !5u !6
du
du
dy du
= 2x ! 3 "
#
= !5u !6 # ( 2x ! 3)
dx
du dx
!6
!5 ( 2x ! 3)
f '(x) = !5 x 2 ! 3x ! 1 # ( 2x ! 3) =
6
x 2 ! 3x ! 1
(
)
(
)
Problem 21
4
(
f (x) = 0.1x 2 ! 4.2x + 9.5
dy
= 1.5u 0.5
du
)
1.5
(
" u = 0.1x 2 ! 4.2x + 9.5
)
y = u1.5
du
dy du
= 0.2x ! 4.2 "
#
= 1.5u 0.5 # ( 0.2x ! 4.2 )
dx
du dx
(
f '(x) = 1.5 0.1x 2 ! 4.2x + 9.5
Problem 25
(
f (x) = 1 ! x 2 = 1 ! x 2
)
1
2
) # ( 0.2x ! 4.2 )
0.5
" u = 1 ! x2
y=u
1
2
dy 1 ! 1 2 du
dy du 1 ! 1 2
= u
= !2x "
#
= u # ( !2x )
du 2
dx
du dx 2
!1
!1
1
!x
f '(x) = 1 ! x 2 2 # ( !2x ) = !x 1 ! x 2 2 =
2
1 ! x2
(
)
(
)
Problem 29
h(x) = ( 3.1x ! 2 ) !
2
1
( 3.1x ! 2 )
= ( 3.1x ! 2 ) ! ( 3.1x ! 2 )
2
2
!2
d
d
( 3.1x ! 2 ) ! ( !2 ) ( 3.1x ! 2 )!3 ( 3.1x ! 2 )
dx
dx
!3
= 2 ( 3.1x ! 2 ) 3.1 + 2 ( 3.1x ! 2 ) 3.1
h '(x) = 2 ( 3.1x ! 2 )
= 6.2 ( 3.1x ! 2 ) + 6.2 ( 3.1x ! 2 ) = 6.2 ( 3.1x ! 2 ) +
!3
Problem 33
6.2
( 3.1x ! 2 )3
Product Rule and Chain Rule combined
(
) (1 ! x )
d "
d "
f '(x) =
x ! 2x ) $ & (1 ! x ) + ( x ! 2x )
(
(1 ! x ) $%
%
dx #
dx #
= !2 ( x ! 2x ) ( 2x ! 2 ) (1 ! x ) + ( x ! 2x ) 0.5 (1 ! x ) ( !2x )
= !2 ( x ! 2x ) ( 2x ! 2 ) (1 ! x ) ! x ( x ! 2x ) (1 ! x )
f (x) = x 2 ! 2x
2
!2
2 0.5
!2
2 0.5
!2
2
2
!3
2 0.5
2
!3
2 0.5
2 0.5
!2
2
2
2 !0.5
!2
2 !0.5
I could “go nuts” simplifying this one. I’ll do it “just for fun” but you really shouldn’t
5
(
) ( 2x ! 2 )(1 ! x ) ! x ( x ! 2x ) (1 ! x )
= ! # 2x ( x ! 2 ) 2 ( x ! 1) (1 ! x ) + x " x ( x ! 2 ) (1 ! x )
$
= ! # 4x ( x ! 2 ) ( x ! 1) (1 ! x ) + x ( x ! 2 ) (1 ! x ) %
$
&
= !x ( x ! 2 ) (1 ! x ) #$ 4 ( x ! 1) (1 ! x ) + x ( x ! 2 ) %&
4 ( x ! 1) (1 ! x ) + x ( x ! 2 )
=
!x ( x ! 2 ) (1 ! x )
!3
f '(x) = !2 x 2 ! 2x
2 0.5
!3
!3
!3
!3
2 0.5
2 0.5
2
%
&
2 !0.5
!2
2
2 !0.5
2
2
2 0.5
3
3
2 !0.5
!2
!2
!1
2 !0.5
!3
!3
!2
2
I could go even further, but I think you’ve got the point KISS (Keep it simple, student!)
Problem 37 chain rule and quotient rule
3
! z $
g(z) = #
" 1 + z 2 &%
d
d
2
[ z ] ' 1 + z 2 ( ( z ) dz )*1 + z 2 +,
! z $ dz
g '(z) = 3 #
2
" 1 + z 2 &%
1 + z2
(
(
)
)
(
)
(
)
2
2
2
2
2
! z $ 1' 1 + z ( ( z ) ( 2z )
! z $ 1 + z ( 2z
= 3#
= 3#
2
2
" 1 + z 2 &%
" 1 + z 2 &%
1 + z2
1 + z2
(
)
(
2
2
3z 1 ( z
! z $ 1( z
= 3#
=
2
4
2&
" 1 + z % 1 + z2
1 + z2
2
(
Problem 41
)
(
2
)
)
(
)
Chain Rule in a chain rule
t(x) = "# 2 + ( x + 1)
!0.1 4.3
$
%
d
" 2 + ( x + 1)!0.1 $
%
dx #
!0.1 3.3
!1.1 d
= 4.3 "# 2 + ( x + 1) $% ( !0.1) ( x + 1)
[ x + 1]
dx
!0.1
t '(x) = 4.3 "# 2 + ( x + 1) $%
= !0.43( x + 1)
Problem 45
!1.1
3.3
( 2 + ( x + 1) )
!0.1 3.3
This one is completely ridiculous! A chain of 3 powers!
6
(
))
d !
f '(x) = 3(1 + (1 + (1 + 2x ) ) )
1 + (1 + (1 + 2x ) ) $
%&
dx "#
d
!1 + (1 + 2x ) $
= 3(1 + (1 + (1 + 2x ) ) ) 3(1 + (1 + 2x ) )
%
dx "
d
= 3(1 + (1 + (1 + 2x ) ) ) 3(1 + (1 + 2x ) ) 3(1 + 2x )
!(1 + 2x ) $%
dx "
= 3(1 + (1 + (1 + 2x ) ) ) 3(1 + (1 + 2x ) ) 3(1 + 2x ) 2
= 54 (1 + 2x ) (1 + (1 + 2x ) ) (1 + (1 + (1 + 2x ) ) )
(
f (x) = 1 + 1 + (1 + 2x )
3
3 3
2
3 3
3 3
2
3 3
3 2
2
3 3
3 2
2
2
3 3
3 2
2
3 2
2
3
2
3 3
And I just sat here like an idiot typing all those powers…painful formatting.
Problem 49
ds ds dr
=
!
s = r "3 + r 0.5
dt dr dt
ds
= "3r "4 + 0.5r "0.5
dr
ds
dr
= "3r "4 + 0.5r "0.5
dt
dt
(
)
Problem 53
y = x3 +
1
x
(t, x ) = (1, 2 )
! dx $
#" dt &% = '1
t =1
! dy $
#" dt &%
t =1
dy dy dx
=
(
dt dx dt
dy
= 3x 2 ' x '2
dx
dy
dx
= 3x 2 ' x '2 (
dt
dt
'47
! dy $
2
'2
#" dt &% = 3 ( 2 ' 2 ( ( '1) = '11.75 = 4
t =1
(
(
)
)
Problem 57
7
x = 2 + 3t y = !5t
dy
dy
!5
= dx dt =
dx
3
dt
Problem 61
y = 35x !0.25
6.5 " x " 17.5
x = 7 + 0.2t
y = 35 ( 7 + 0.2t )
# y' = !8.75 ( 7 + 0.2t )
percentage points per month
November 1st is 10 months since January 1st
!1.25
y' = !1.75 ( 7 + 0.2 (10 ))
= !0.11226
!0.25
!1.25
( 0.2 ) = !1.75 ( 7 + 0.2t )!1.25
declining at 0.11 percentage points per month
Problem 63
P ( q ) = !100000 + 5000q ! 0.25q 2
q ( n ) = 30n + 0.01n 2
(
)
(
P ( n ) = !100000 + 5000 30n + 0.01n 2 ! 0.25 30n + 0.01n 2
(
)
2
)
dP
= 5000 ( 30 + .02n ) ! 0.50 30n + 0.01n 2 ( 30 + .02n )
dn
dP
n = 10 "
= 5000 ( 30 + .2 ) ! 0.50 ( 300 + 1) ( 30 + 0.2 ) = 146454.9
dn
8