HWK #2. Due Wed 2/7, Thurs 2/8
4.1: 14, 18, 31, 71
4.2:
8, 12, 20, 28, 31
HWK #3. Due Tues 2/13, Wed 2/14
Solutions will be available 2/15
4.3: 22, 28, 36, 46
4.4: 2, 8, 18, 38b
Not HWK.
4.5: 1, 7, 9, 11
Prelim 1: Tues 2/20
covers 3.3-4.5
4.1. Computing Derivatives, part 2
Rule 1: (x
p
)
′
= px
p-1
p.214
Rule 2: (cf)
′
= cf
′
p.216
Rule 3. (f+g)
′
= f
′
+ g
′
p.217
Cost functions
Suppose the cost to produce x units of
a particular item is
C(x) = 9 + .3 x + .01 x
2
C(x+1)-C(x) = the actual cost of
producing one more item
≈
C
′
(x) = .3 + .02x,
the
marginal cost
Cost
C(x) = 9 + .3 x + .01 x
2
marginal cost
C
′
(x) = .3 + .02x
9 is a
fixed cost
that must be paid to
produce any items.
C
′
(0) = 0.3.
The first item costs
≈
30 cents to make.
C
′
(20) = 0.7
The 21
st
item costs
≈
70 cents to make
Tangent line
The equation of the tangent line to the
graph at the point x
0
is (p.189)
y = f(x
0
) + f
′
(x
0
) (x - x
0
)
Why? The value at x
= x
0
is
f(x
0
)
y = [f(x
0
) - f
′
(x
0
) x
0
] + f
′
(x
0
) x
So the slope is f
′
(x
0
)
If y = a + bx : a = intercept, b = slope
Two equations for a line
y = a + bx
Traditional form of line
b= slope,
a = intercept
y = f(x
0
) + f
′
(x
0
) (x - x
0
)
Easier to calculate
f
′
(x
0
) = slope, y = f(x
0
) at x
0