Learning Activity for MATH 115
Applied Math for Business Fall 2013
NAME:
Section:
Answer each of the questions below to the best of your ability. Be sure to answer all word problems
using full sentences that include units for the problem.
Theory
Composition of functions:
F (x) = (f ◦ g)(x) = f (g(x))
The chain rule can be used to differentiate a composition of functions. Thus for
F (x) = f (g(x))
then
d
F (x) = F 0 (x) = f 0 (g(x))g 0 (x)
dx
Questions
For each of the following functions find an expression for the instantaneous rate of change function
(the derivative).
• F (x) = (x3 + 2x)5
F 0 (x) = 5(x3 + 2x)4 (3x2 + 2)
• G(x) =
√
4x + 7
1
(4x + 7)−1/2 (4)
2
= 2 (4x + 7)−1/2
G0 (x) =
• f (x) =
3x−7
4x2 +1
4
0
f (x) = 4
3x − 7
4x2 + 1
3 (4x2 + 1)(3) − (3x − 7)(8x)
(4x2 + 1)2
• g(x) = (4x + 2)2 (3x + 1)5
g 0 (x) = (2(4x + 2)(4)) (3x + 1)5 + 5(3x + 1)4 (3)(4x + 2)2
• h(x) =
q
4
(x+7)
2x5
0
h (x) =
1
4
(x + 7)
2x5
−3/4 2x5 (1) − (x + 7)(10x4 )
(2x5 )2
7y 3
(5−9y)5
• Differentiate the function: b(y) =
b0 (y) =
(5 − 9y)5 (21y 2 ) + 7y 3 (5(5 − 9y)4 (−9)
(5 − 9y)10
• If $1000 is invested at interest rate i, compounded quarterly, in 5 years it will grow to an
amount of
i 20
A(t) = 1000 1 +
4
– Find the rate of change,
dA
di .
i 19 1
dA
= 20, 000 1 +
di
4
4
– What is the meaning of
dA
di ?
This gives the instantaneous rate of change in the investments value after five
years with respect to the interest rate used. Note the units of dA
di are in dollars
per percent.
• A total revenue function is given by:
p
R(x) = 1000 x2 − 0.1x
where R(x) is the total revenue, in thousands of dollars, from the sale of x items. Find the
rate at which total revenue is changing when 20 items have been sold.
Here we compute the derivative function first:
R0 (x) =
−1/2
1000 2
x − 0.1x
(2x − 0.1)
2
Next consider R0 (20).
−1/2
R0 (20) = 500 (20)2 − 0.1(20)
(2(20) − 0.1) = 1000
Thus, the revenue is increasing by $1,000,000 per item when twenty items have
been sold.