104
2.3. THE PRODUCT AND QUOTIENT RULES
which the value per share is changing. The big question we’ d like to answer is: how do
these respective rates of change affect the rate of change of the total value function?
To help better understand the relationship among changes in N, S, and V , let’s
consider some specific data. Suppose that on day 100, the investor owns 520 shares of
stock and the stock’s current value is $27.50 per share. This tells us that N(100) = 520
and S(100) = 27.50. In addition, say that on day 100, the investor purchases an additional
12 shares (so the number of shares held is rising at a rate of 12 shares per day), and that
on that same day the price of the stock is rising at a rate of 0.75 dollars per share per
day. Viewed in calculus notation, this tells us that N ′ (100) = 12 (shares per day) and
S ′ (100) = 0.75 (dollars per share per day). At what rate is the value of the investor’s total
holdings changing on day 100?
Observe that the increase in total value comes from two sources: the growing number
of shares, and the rising value of each share. If only the number of shares is rising (and
the value of each share is constant), the rate at which which total value would rise is found
by computing the product of the current value of the shares with the rate at which the
number of shares is changing. That is, the rate at which total value would change is given
by
S(100) · N
′ (100) = 27.50
dollars
share
· 12
shares
day
= 330
dollars
day
.
Note particularly how the units make sense and explain that we are finding the rate at
which the total value V is changing, measured in dollars per day. If instead the number of
shares is constant, but the value of each share is rising, then the rate at which the total
value would rise is found similarly by taking the product of the number of shares with the
rate of change of share value. In particular, the rate total value is rising is
N(100) · S
′ (100) = 520 shares · 0.75
dollars per share
day
= 390
dollars
day
.
Of course, when both the number of shares is changing and the value of each share is
changing, we have to include both of these sources, and hence the rate at which the total
value is rising is
V
′ (100) = S(100) · N
′ (100) + N(100) · S
′ (100) = 330 + 390 = 720
dollars
day
.
This tells us that we expect the total value of the investor’s holdings to rise by about
$720 on the 100th day. 5
5 While this example highlights why the product rule is true, there are some subtle issues to recognize. For
one, if the stock’s value really does rise exactly $0.75 on day 100, and the number of shares really rises by 12
on day 100, then we’ d expect that V (101) = N(101) · S(101) = 532 · 28.25 = 15029. If, as noted above, we
expect the total value to rise by $720, then with V (100) = N(100) · S(100) = 520 · 27.50 = 14300, then it
seems like we should find that V (101) = V (100) + 720 = 15020. Why do the two results differ by 9? One way
to understand why this difference occurs is to recognize that N ′ (100) = 12 represents an instantaneous rate of
2.3. THE PRODUCT AND QUOTIENT RULES
which the value per share is changing. The big question we’ d like to answer is: how do
these respective rates of change affect the rate of change of the total value function?
To help better understand the relationship among changes in N, S, and V , let’s
consider some specific data. Suppose that on day 100, the investor owns 520 shares of
stock and the stock’s current value is $27.50 per share. This tells us that N(100) = 520
and S(100) = 27.50. In addition, say that on day 100, the investor purchases an additional
12 shares (so the number of shares held is rising at a rate of 12 shares per day), and that
on that same day the price of the stock is rising at a rate of 0.75 dollars per share per
day. Viewed in calculus notation, this tells us that N ′ (100) = 12 (shares per day) and
S ′ (100) = 0.75 (dollars per share per day). At what rate is the value of the investor’s total
holdings changing on day 100?
Observe that the increase in total value comes from two sources: the growing number
of shares, and the rising value of each share. If only the number of shares is rising (and
the value of each share is constant), the rate at which which total value would rise is found
by computing the product of the current value of the shares with the rate at which the
number of shares is changing. That is, the rate at which total value would change is given
by
S(100) · N
′ (100) = 27.50
dollars
share
· 12
shares
day
= 330
dollars
day
.
Note particularly how the units make sense and explain that we are finding the rate at
which the total value V is changing, measured in dollars per day. If instead the number of
shares is constant, but the value of each share is rising, then the rate at which the total
value would rise is found similarly by taking the product of the number of shares with the
rate of change of share value. In particular, the rate total value is rising is
N(100) · S
′ (100) = 520 shares · 0.75
dollars per share
day
= 390
dollars
day
.
Of course, when both the number of shares is changing and the value of each share is
changing, we have to include both of these sources, and hence the rate at which the total
value is rising is
V
′ (100) = S(100) · N
′ (100) + N(100) · S
′ (100) = 330 + 390 = 720
dollars
day
.
This tells us that we expect the total value of the investor’s holdings to rise by about
$720 on the 100th day. 5
5 While this example highlights why the product rule is true, there are some subtle issues to recognize. For
one, if the stock’s value really does rise exactly $0.75 on day 100, and the number of shares really rises by 12
on day 100, then we’ d expect that V (101) = N(101) · S(101) = 532 · 28.25 = 15029. If, as noted above, we
expect the total value to rise by $720, then with V (100) = N(100) · S(100) = 520 · 27.50 = 14300, then it
seems like we should find that V (101) = V (100) + 720 = 15020. Why do the two results differ by 9? One way
to understand why this difference occurs is to recognize that N ′ (100) = 12 represents an instantaneous rate of