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3.4 and 4.1. Defining and
Computing Derivatives
The derivative of f is the
instantaneous rate of change
0
()
(
)
'( )
lim
h
fx h fx
fx
h
→
+−
=
Derivative is the slope
of the tangent line
Slopes 23, 14, 7.52, 6
70
75
80
85
90
95
100
3.9
4
4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9
5
Constant functions: f(x)=5
0
55
lim
0
h
h
→
−
==
The curve is flat so the slope is 0
Linear functions: f(x) = 3x+2
3(
)
3
3
3
xh
x
h
hh
Constant slope =3, so derivative = 3
f(x) = x
2
22
2
2
2
0
2
2
2
2
lim
2
2
h
ab
a
a
bb
x
x
xh x
x
→
+=
+
+
+
+
+
+
f(x) = x
3
33
2
23
2
0
3
3
lim
3
3
3
h
x
x
h
x
h
h
x
xx
h
h
h
h
h
x
→
+
+
+
=++
++
=
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View Full DocumentPower functions
Function
Derivative
x1
x
2
2x
x
3
3x
2
x
p
px
p1
Constant Multiples
Rule 1: If f(x) = x
p
then f
′
(x) = px
p1
Rule 2: If c is a number, (cf)
′
= cf
′
.
f(x) = 4x
3
f
′
(x) = 4(3x
2
) = 12x
2
f(x) = 7x
5
f
′
(x) = (7)(5x
4
) = 35x
4
Derivative of Sums
Rule 1: (x
p
)
′
= px
p1
Rule 2: (cf)
′
= cf
′
Rule 3. (f+g)
′
= f
′
+ g
′
f(x) = 10 + 18x + 9x
2
–2x
3
f
′
(x) = 0 + 18(1) + 9(2x) – 2(3x
2
)
f
′
(4) = 18 + 72 – 96 = 6
Practice Problem
f(x) = x
4
–3x
3
+7x
2
+ 11
Find f
′
(x) and f
′
(1)
Weight of Bighorn sheep
on Ram Mountain in Alberta, Canada:
M(t) = 27.5 + 0.3 t – 0.001 t
2
M(t) is measured in kilograms
t = number of days since May 25.
Find the instantaneous rate of change
of weight for a sheep at t = 100 days.
M(t) = 27.5 + 0.3 t – 0.001 t
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 Spring '07
 DURRETT
 Calculus, Derivative, Rate Of Change, Slope

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