As we saw in Chapters I and II the derivative can be a powerful analytic tool for any quantifiable scientific discipline. This concept is used in many interpretations to consolidate information about changing relations between variables. The symbolic and numeric methods for finding or estimating the derivative as a number or as a function make it efficient, eliminating repetitive arguments and computations, and allow the focus of its use to remain on the applications, keeping most mathematical manipulations routine and secondary. The visual tools of graphs and transformation figures provide additional force to the concept by supplying alternative meanings for more abstract contexts. Applications of the derivative are not limited to those previewed in chapter I or to those that we will discuss in this chapter. In fact one could easily describe everything in this text as an application of the derivative in some fashion! We will analyze three types of applications in this section. Each has a wide scope of impact extending well beyond the specific context in which we will see them initially.
A. Estimations: Understanding estimations is a critical part of any science that uses measurements. In using the derivative to estimate values the key lies in the assumption that a function with a derivative behaves like a function with a constant rate of change for some small interval. In this chapter we will use the derivative in two common estimation procedures. First, we will estimate the value of a function thinking of it as a position function in a motion interpretation. Then we will discuss an algorithmic process called Newton's method to estimate a zero of a function or root of an equation thinking of this as a time when a position function is at the initial distance or graphically as the Xintercept of a curve.
B. Graphs: The important role played by graphs in current science is undeniable. It is the chief tool of visualization of data and provides an efficient vehicle for suggesting relations between variables. With the power that technology adds to the creation and display of graphs, the need for better understanding of graphical features and their analysis has risen while the need to analyze functions represented symbolically for quick and accurate drawings has diminished only slightly. As we saw in Chapter I, an understanding of the derivative can bring with it a richer appreciation for graphic function properties.
C. Modelling: It is in the use of the derivative in the sciences that this mathematical concept pays for the time and effort it takes to master its notation and rules. Without the ability to bring this abstraction into more then just the problems generated by mathematics alone, it would be hard to see why mathematics would be as prominent as the "language of science." Even with the rise of computational power through technology, the use of the derivative continues as before to describe and investigate the world through the measurement of variable. In the next chapter we will look more extensively at the ramifications of describing a context by relating the rates at which variables change. In this chapter we will look at models where the relation between the variables can be expressed in some equation or with a function with one controlling variable. Though this does seem limited in scope the questions we will examine for these models are of sufficient generality to make them good prototypes for more general models. The questions are simply how to use the information available to predict the behavior of variables under specific constraints. Just as with the graphical applications, the derivative can inform us of extremes, intervals where variables will increase and decrease, the rates at which the rates of change change.
III.A.1. THE DIFFERENTIAL
MOTIVATION:
Consider a jogger running on a straight track so that after 2 seconds
the jogger is 10 meters from the starting point P and at that moment is
moving away from P with a velocity of 3 meters/sec. I would like
to approximate the position of the jogger .4 seconds later. It seems
reasonable to assume for this estimation that the velocity wouldn't change
much in .4 of a second. So we'll treat the velocity of the jogger as a
constant. Now it should be apparent that in .4 of a second the jogger would
move approximately (.4)(3) meters further away from P so that the jogger
would be approximately 10 + 1.2 = 11.2 meters from P. We can express this
analysis more technically using some function notation. Let t denote the
time in seconds and s(t) denote the jogger's distance from P at time t.
Then the initial facts were that s(2) = 10 and that s'(2)=3. To estimate
the change in the value of s for .4 seconds we multiplied s'(2), the rate
at which the jogger was running, by .4, the time the jogger would be running.
In symbolic form we had s(2.4)s(2) »
s'(2) .4 = 3 (.4) = 1.2. Now we complete the analysis by adding the
estimate of the change to the runners position at 2 seconds giving
s(2.4) = s(2) + {s(2.4)  s(2)} » 10 +
1.2 = 11.2.
The simple technique
we used here generalizes to a method for estimating values
of any differentiable function based on information about the value of
the function and its derivative at a single point. The key is using the
product of the value of the derivative with a small change in the controlling
variable to estimate the change in thecorresponding change in the function's
value. The word that has been used since Leibniz to describe the estimate
for the change is the "differential."

Definition and Notation: Suppose f is a function that is differentiable at a and h is any real number. We'll write
df is
called "the differential of f at a." As the more complete notation
df (a,h) suggests, the quantity df depends
on both the numbers a and h. When the notation y = f (x)
is used to describe the function, then we also denote the differential
with dy as well as df.
Interpretation:
We can interpret the number a as a time at which we know information about
the value of f. The number h can be thought of as measuring a time
interval we add to (or subtract from) a to determine a later (or earlier)
time at which we would like to know the function's value. The value of
df is then an estimate of the change in the function's value at
this later (or earlier) time, a+h. Thus df(a,h) »
f(a+h)  f(a).


More on notation: The notation of the differential was introduced by Leibniz with a different view of what it represented from its current use. In particular Leibniz used the symbols dy and dt to represent the measures of very small I mean extremely small segments measuring the rise and run of very short sections of a curve. Thus it appears that Leibniz was interested in finding the slope of a curve by inspecting the curve very closely. This is analogous to what we might do today with the ability to zoom in with graphical technology so that the graph of a curve would appear to be indistinguishable from the graph of a line. We will continue in this section to develop the notation for differentials to allow us to make some sense out of this older view. The Leibniz notation has proved particularly successful in connecting many concepts to scientific applications.
What does dx
denote? Suppose x(t) = t for all t. Then dx=
dx(a,h) = x'(a)^{ .}h. But x'(t)
= 1 for all t, so dx= h. This bizzare consequence of the
notation justifies the abuse of notation in saying that when y =
f (x),
It is possible now
to make sense in many situations of Leibniz notation even though the original
use of this notation most likely had a very different though consistent
meaning to Leibniz and others historically. For example, with
this notation df /dx and dy/dx can be interpreted
as quotients so that
More Notation:
In Chapter I we used h for some of our initial derivative estimations.
We suppose again that y = f(x) and let
By our previous comments
then
dx = Dx = h and Dy
= f(a+dx) f(a).


EXAMPLE: (Let's try it.) Find dy when y = f(x) = x^{ 3}  5x + 7. Evaluate dy when x = 2 and dx = .3. Use dy to estimate y when x = 2.3. Find f(2.3) and Dy exactly.
SOLUTION: Using y = f(x), we have that f '(x) = 3x^{ 2}  5 so
dy = f '(a)dx = ( 3a^{ 2}  5 ) dx.
Noticing that f(2) = (8  10 + 7) = 5, we estimate
It is not hard to find f(2.3) =(2.3)^{ 3}  5(2.3) + 7 = 7.667, so that Dy = 7.667 5 = 2.667.
It is worth noting here that the size of the error in the differential estimate of f(2.3) is the difference between



0.1

9.9833416647e02

0.1

0.01

9.9998333342e03

0.01

0.001

9.9999983333e04

0.001

0.0001

9.9999999833e05

0.0001

0.00001

9.9999999998e06

0.00001

0.000001

1.0000000000e06

0.000001






0.5

1.5

0.6666667

0.5

33.33%

0.1

1.1

0.9090909

0.9

9.09%

0.01

1.01

0.990099

0.99

0.99%

0.001

1.001

0.999001

0.999

0.10%

0.5

0.5

2

1.5

100.00%

0.1

0.9

1.1111111

1.1

11.11%

0.01

0.99

1.010101

1.01

1.01%

0.001

0.999

1.001001

1.001

0.10%

Interpretation
(The graph and the tangent line): The differential of f
at a can also be visualized using the interpretation of the derivative
as the slope of the line tangent to the graph of y=f(x) at the point
(a, f(a)). Figure 2 shows the lengths of the key elements
used in determining the differential, namely the points on the graph of
f, (a,f(a)) and (a+h,f(a+h)), and on the tangent line
(a+h, f(a)+f '(a)^{ .} h).
Note that the second coordinate of the point on the tangent line was
determined from the fact that the slope is f '(a). When h is small
we have that the slope of the secant line determined by (a,f(a))
and (a+h,f(a+h)) is a good estimate for the slope of the tangent
line, f '(a), i.e,

Interpretation: (Economics): Consider a function model for the cost C of producing x kilograms of a perfectly divisible commodity. As we saw in chapter I, when we produce a kilograms, the marginal cost is C'(a). If we decide to produce an additional h kilograms of our product, we can estimate the change in our costs and our new costs with the differential at a. Thus ?C . dC= dC(a,h) = C'(a)A h and C(a+h).C(a) + dC=C(a) + C'(a)A h. [The marginal cost at a, which we denoted MC(a) in chapter I, originally meant the change in the cost for a change of one unit of production. Now the MC(a) can be connected more directly to the derivative by using the differential estimation, giving MC(a) . dC(a, 1)=C'(a).] See exercises.....
Application: Use the differential to estimate 9 ^{1/3} from the fact that 8 ^{1/3} = 2.
Solution: Consider f(x) = x^{ 1/3}. f(9) = f(8+1) so in the notation we've established we let a=8 and h=1. Thus f(a+h) .f(a) + f '(a)^{ .} h = f(8) + f '(8)^{ .} 1 = 2 + (1/3)(8)^{ 2/3}
so f(9) . 2 + 1/12 = 25/12.
Notice that what made this solution possible was the ability to evaluate both f(8) and f '(8). This ease of computation was what actually led to the choice of a = 8.
Application: Use the differential to estimate 98^{ 1/2} .
Solution: Consider f(x) = x^{ 1/2}. f(98) = f(100  2) so in the notation we've established we let a=100 and h=2. Thus
f(a+h) .f(a) + f '(a) ^{.} h = f(100) + f '(100) ^{.}(2) = 10  (100)^{ ?1/2}
and f(98) . 10  1/10 = 9.9 .
The calculus of differentials.
Since the differential of a function is directly related to the derivative of the function, we can write formulas for a calculus of differentials each of which can be justified by reference to the appropriate derivative rule. For example, if u and v are both functions of x, then d(u^{.}v)=udv+vdu. This is justified by considering the derivative product rule D_{x}(u^{.}v)=uD_{x}(v)+vD_{x}(u).
Hence d(u^{ .}v) = D_{x}(u^{.}v)dx = [uD_{x}(v) + vD_{x}(u)]dx = uD_{x}(v)dx + vD_{x}(u)dx
= udv + vdu.
In the exercises for this section you will find many similar results for the "differential calculus" which you are asked to justify.
For each of the functions in problems 712 use the differential to estimate the value of (a) f(1.1) and (b) f(.95) .
7. f(x) = x^{ 2} + 3x 8. f(x) = 5x^{ 2} + 3x
9. f(x) = x^{ 3} + 3x 10. f(x) = 5x^{ 3} + 3x
11. f(x) = x^{ 3} + x 12. f(x) = 5x^{ 3} + x
In problems 1320, use the differential to estimate the indicated value.
13. (82)^{ 1/2} 14. (63)^{ 1/2} 15. (127)^{ 1/3} 16. (25)^{ 1/3} 17. 1/103 18. 1/998 19. (33)^{ 1/5} 20. (29)^{ 1/5}
21. Use the differential to give a formula for estimating x^{ 1/2}
when x is close to (a) 100 (b) 25 (c) 81 and (d) t.
22. Use the differential to give a formula for estimating x^{ 1/3}
when x is close to (a) 1000 (b) 125 (c) 27 and (d) t.
23. A circle of radius 3 meters is painted red with a edge of 10 centimeters painted blue. Use the differential to estimate the area of the region that is painted blue.
24. A spherical ball of radius 20 centimeters is coated with a shell of plastic .5 cm in thickness. Estimate the volume of plastic of the plastic shell.
25. A closed cylindrical tin can has radius 4 cm. and height 6 cm. Estimate the volume of the tin if the tin is 3 mm in thickness.
26. A rectangular poster that is 2 feet by 3 feet has a border of red that is 1/2 inch wide. Estimate the area of the border using the differential. Find the exact area of the border.
Justify the differential calculus formulae in problems 2733.
Assume that u and v are differentiable functions of x.
27. d(au) = adu where a is any real number.
28. d(u + v) = du + dv.
29. d(1/v) = dv/v2.
30. d(u/v) =[vdu  udv]/v2 .
31. d(sin u) = cos u du
32. d(sec u) = sec u tan u du.
33. Suppose that w=f(u) and u=g(x) and y = f(g(x)). Prove dy = dw/du ^{ .} du when interpreted appropriately.
b) Based on your estimate for L(1/4), estimate L(1/2).
c) Continue. Use the estimate of L(1/2) and then L(3/4) to estimate L(1).
d) Based on this work, suggest a method to estimate L(1) more accurately. Explain with an example using your method.
37. Project: Suppose P(x) is a differentiable function with P(0)=1 and for every x, P'(x)=3P(x).
a) Estimate P(1/4).
b) Based on your estimate for P(1/4), estimate P(1/2).
c) Continue. Use the estimate of P(1/2) and then P(3/4) to estimate P(1).
d) Based on this work, suggest a method to estimate P(1) more accurately. Explain with an example using your method.