Derivatives
–
The
4­Step
Process
(Algebraic
Method)
1.
The
 temperature
 during
 a
 typical
 day
 in
 May
 in
 a
 Midwestern
 city
 can
 by
 modeled
 with
 the
 function
( )
= −
0.804
x 2
+
21.291
x
−
60.69
°F
 x 
hours
after
midnight.

Use
the
algebraic
method
 to
estimate
the
rate
at
which
temperature
is
changing
at
10:30am.
€
2.
The
 table
 below
 shows
 the
 number
 of
 students
 per
 computer
 in
 US
 public
 schools
 in
 various
 years.

Find
 a
quadratic
model 
(round
coefficients
to
2
decimal
places)
and
use
it
to
algebraically
 estimate
the
rate
of
change
in
this
quantity
in
1987.
Year
 1986
 1988
 1990
 1992
 1994
 1996
#
Students
50
 32
 22
 18
 14
 10
3.
Find
the
derivative
formula
for
each
of
the
following
functions
using
the
4‐step
process.
(a)
( ) =
0.125
x 2
−
3.8
x
+
11.7
€
1
(b)
( ) = x
€

€
€
€
€
€
€
€
Solutions
1.
(
+ h )
= −
0.804
x 2
−
1.608
xh
−
0.804
h 2
( + h ) −
T x
= −
1.608
xh
−
0.804
h 2
+
21.291
x
+
21.291
h
−
60.69
+
21.291
h
( + h ) − ( ) h
= −
1.608
x
−
0.804
h
+
21.297
and
so
′ ( ) = −
1.608
x
+
21.291
So
T
 (
′
10.5
)
= −
1.608(10.5)
+
21.291
=
4.407
°F
per
hour
€
€
2.
( ) =
0.42
x 2
−
12.92
x
+
110.76
students
per
computer
 x 
years
after
1980
S ( 7
+ h ) =
0.42
h 2
−
7.04
h
+
40.9
°F/hour
€
S ( 7
+ h )
−
S 7
=
0.42
h 2
S ( 7
+ h ) −
S ( ) h
3.
 (a)
−
7.04
h
=
0.42
h
−
7.04
and
so
€
S
′ ( ) = −
7.04
students
per
computer
per
year
( + h ) =
0.125
x 2
+
0.25
xh
+
0.125
h 2
−
3.8
x
−
3.8
h
+
11.7
( + h ) − f x
=
0.25
xh
+
0.125
h 2
−
3.8
h
( + h ) − ( ) h
=
0.25
x
+
0.125
h
−
3.8
and
so
′
( )
=
0.25
x
−
3.8
€
€
(b)
€
1
( + h ) = x
+ h
€
( + h ) − ( ) =
1 x
+ h
−
1 x
= x
( + h )
− x
+ h
( + h )
=
− h
( + h )
€
( + h ) − ( ) h
=
−
1
( + h )
and
so
′ ( ) = −
1 x 2
€
€