1.5. INTERPRETING, ESTIMATING, AND USING THE DERIVATIVE
45
take the limit to get f ′ (x), we get these same units on the derivative f ′ (x): units of f per
unit of x. Regardless of the function f under consideration (and regardless of the variables
being used), it is helpful to remember that the units on the derivative function are “units
of output per unit of input,” in terms of the input and output of the original function.
For example, say that we have a function y = P(t), where P measures the population
of a city (in thousands) at the start of year t (where t = 0 corresponds to 2010 AD), and we
are told that P ′ (2) = 21.37. What is the meaning of this value? Well, since P is measured
in thousands and t is measured in years, we can say that the instantaneous rate of change
of the city’s population with respect to time at the start of 2012 is 21.37 thousand people
per year. We therefore expect that in the coming year, about 21,370 people will be added
to the city’s population.
Toward more accurate derivative estimates
It is also helpful to recall, as we first experienced in Section 1.3, that when we want to
estimate the value of f ′ (x) at a given x, we can use the difference quotient
f (x+h)− f (x)
h
with
a relatively small value of h. In doing so, we should use both positive and negative values
of h in order to make sure we account for the behavior of the function on both sides of
the point of interest. To that end, we consider the following brief example to demonstrate
the notion of a central difference and its role in estimating derivatives.
Example 1.4. Suppose that y = f (x) is a function for which three values are known:
f (1) = 2.5, f (2) = 3.25, and f (3) = 3.625. Estimate f ′ (2).
Solution. We know that f ′ (2) = lim h→0
f (2+h)− f (2)
h
. But since we don’t have a graph for
y = f (x) nor a formula for the function, we can neither sketch a tangent line nor evaluate
the limit exactly. We can’t even use smaller and smaller values of h to estimate the limit.
Instead, we have just two choices: using h = −1 or h = 1, depending on which point we
pair with (2, 3.25).
So, one estimate is
f
′ (2) ≈
f (1) − f (2)
1 − 2
=
2.5 − 3.25
−1
= 0.75.
The other is
f
′ (2) ≈
f (3) − f (2)
3 − 2
=
3.625 − 3.25
1
= 0.375.
Since the first approximation looks only backward from the point (2, 3.25) and the second
approximation looks only forward from (2, 3.25), it makes sense to average these two
values in order to account for behavior on both sides of the point of interest. Doing so, we
45
take the limit to get f ′ (x), we get these same units on the derivative f ′ (x): units of f per
unit of x. Regardless of the function f under consideration (and regardless of the variables
being used), it is helpful to remember that the units on the derivative function are “units
of output per unit of input,” in terms of the input and output of the original function.
For example, say that we have a function y = P(t), where P measures the population
of a city (in thousands) at the start of year t (where t = 0 corresponds to 2010 AD), and we
are told that P ′ (2) = 21.37. What is the meaning of this value? Well, since P is measured
in thousands and t is measured in years, we can say that the instantaneous rate of change
of the city’s population with respect to time at the start of 2012 is 21.37 thousand people
per year. We therefore expect that in the coming year, about 21,370 people will be added
to the city’s population.
Toward more accurate derivative estimates
It is also helpful to recall, as we first experienced in Section 1.3, that when we want to
estimate the value of f ′ (x) at a given x, we can use the difference quotient
f (x+h)− f (x)
h
with
a relatively small value of h. In doing so, we should use both positive and negative values
of h in order to make sure we account for the behavior of the function on both sides of
the point of interest. To that end, we consider the following brief example to demonstrate
the notion of a central difference and its role in estimating derivatives.
Example 1.4. Suppose that y = f (x) is a function for which three values are known:
f (1) = 2.5, f (2) = 3.25, and f (3) = 3.625. Estimate f ′ (2).
Solution. We know that f ′ (2) = lim h→0
f (2+h)− f (2)
h
. But since we don’t have a graph for
y = f (x) nor a formula for the function, we can neither sketch a tangent line nor evaluate
the limit exactly. We can’t even use smaller and smaller values of h to estimate the limit.
Instead, we have just two choices: using h = −1 or h = 1, depending on which point we
pair with (2, 3.25).
So, one estimate is
f
′ (2) ≈
f (1) − f (2)
1 − 2
=
2.5 − 3.25
−1
= 0.75.
The other is
f
′ (2) ≈
f (3) − f (2)
3 − 2
=
3.625 − 3.25
1
= 0.375.
Since the first approximation looks only backward from the point (2, 3.25) and the second
approximation looks only forward from (2, 3.25), it makes sense to average these two
values in order to account for behavior on both sides of the point of interest. Doing so, we