Introduction:
As we saw in Chapters I and II, the derivative can be a powerful
analytic tool for any quantifiable scientific discipline. In many
interpretations this concept consolidates 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 mapping diagrams provide additional
power 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. The key in using the derivative to estimate values 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 number as a time when a position function is at the initial distance or graphically as the Xintercept coordinate 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. Modeling: It is in the sciences that the use of the
derivative as a 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 variables.
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 may 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, even 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 0.4 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 0.4 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 $0.4$ seconds we multiplied $s'(2)$, the rate at which the jogger was running, by $0.4,$ the time the jogger would be running. In symbolic form we had $s(2.4)s(2) \approx s'(2) .4 = 3 (.4) = 1.2$. Now we complete the analysis by adding the estimate of the change to the runners position at $2$ seconds giving $s(2.4) = s(2) + {s(2.4)  s(2)}\approx 10 + 1.2 = 11.2$. 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 the corresponding 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
$df = df(a,h) = f '(a ) \cdot h$
.
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 $dx=
dx(a,h) = x'(a) \cdot h$. 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 $ \frac {df}{dx}$ and $\frac {dy}{dx}$ 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 $dx
= \Delta x = h$ and $\Delta y
= f(a+dx) f(a)$.


EXAMPLE: (Let's try it.) Find $dy$ when $y = f(x) = x^3  5x + 7$. Evaluate $dy$ when $x = 2$ and $dx = 0.3$. Use $dy$ to estimate $y$ when $x = 2.3$. Find $f(2.3)$ and $\Delta y$ 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 $\Delta y = 7.667 5 = 2.667$.
It is worth noting here that the size of the error in the differential estimate of $f(2.3)$ is the difference between
EXAMPLES:To see the relative quality of the estimate for values of a function using the differential let's look at the sine function values using the differential at 0. We compare these in Table 1 with the estimation values that arise from the differential. Since $\sin(0)=0$ and $\sin'(0)= \cos(0)=1$, the estimate for $\sin(0+h)$ is $\sin(0)+\sin'(0)\cdot h = h$. So Table 1 demonstrates that when $h \approx 0, sin(h) \approx h$. [This should remind you of the fact we demonstrated in chapter I, namely that $\lim_{h \to 0} \frac {\sin(h)} h =1$.] The errors in the estimates of this table are clearly smaller when $h$ is closer to $0$. [Can you see why we haven't considered relative errors here?] 


A similar comparison for $f(x)=1/x$ using the differential at $1$ shows a less symmetric situation. Here $f(1)= 1$ and $f '(1)= 1$ so $f(1+h)$ is approximated by $f(1) + f '(1) \cdot h = 1  h$. Table 2 shows how these estimates compare with some function values close to $1$. Again the errors in the estimates are smaller when $h$ is closer to $0$, as are the relative errors which are shown as percentages in the table. 

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
$(a+h, f(a)+f '(a) \cdot h)$.
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, $\frac {f(a+h)f(a)} h =\frac {\Delta y}{\Delta x} \approx f'(a)$. 

Notice how the points on the graph are paired with the arrows on the mapping diagram. Martin Flashman, Oct 2014, Created with GeoGebra 
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 $\Delta C \cdot dC= dC(a,h) = C'(a) \cdot h$ and $C(a+h) \cdot C(a) + dC = C(a) + C'(a)\cdot h$. [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 $MC(a) \cdot dC(a, 1)=C'(a)$.] See exercises.....
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 $a = 8$.
$f(a+h) \approx f(a) + f '(a) \cdot h = f(100) + f '(100) \cdot (2) = 10 + 1/2(100) ^{1/2}\cdot (2)$
and so $f(98) \approx 10  1/10 = 9.9 $.
The relative error of a value, $V$ in a computation is determined by
measuring the error, $\Delta V$, in the computation and comparing
that as a ratio with the computed value, i.e., $\frac {\Delta V}
V$.
To make an estimate of the relative error, it makes sense to use $ dV
\approx \Delta V$ when $V$ is a differentiable function of a controlling
variable.
Example: Suppose the side of a cube is measured to be 5 meters
with a possible error in the measurement of at most 1 centimeter = 0.01
meters. Estimate the relative error in using the measurement of the
length of the side to compute the volume of the cube.
Solution: Let $s=5$ be the measurement of the side of the
cube and $V = s^3$ be the volume of the cube, which gives us $V =
5^3 = 125$ cubic meters. The largest possible error is $\Delta s = ds =
.01$.
We use $dV$ to estimate the error, $\Delta V \approx dV = 3s^2 \cdot ds$, so we
find that $dV = 3 \cdot 5^2 \cdot 0.01 = 0.75$ so the relative error is
approximately $ \frac {dV} V = \frac {0.75}/{125} = 0.006 = 0.6\%$
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 $d(u \cdot v) = D_x(u\cdot v)dx = [uD_x(v) + vD_x(u)]dx = uD_x(v)dx + vD_x(u)dx$ $= udv + vdu$.
In
the exercises for this section you will find 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) $f(.95)$ .Check you work with the GeoGebra Applet.
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.Check you work with the GeoGebra Applet.
13. $(82) ^{1/2}$ 14. $(63)^{1/2}$ 15. $(127) ^{1/3} $ 16.$(25) ^{1/3} $ 17. $\frac 1{103}$ 18. $\frac 1{998} $ 19. $(33)^{1/5}$ 20. $(29^{1/5}$
21. Use the differential to give a formula for estimating $\sqrt{x} = x^{1/2}$
when $x$ is close to (a) 100 (b) 25 (c) 81 and (d) $t$.
22. Use the differential to give a formula for estimating $\sqrt[3]{x} = x^{1/3}$
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 0.5 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. $d(1/v) = dv/v^2$.
30. $d(u/v) =[vdu  udv]/v^2$ .
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 $dy
= dw/du \cdot du$ 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 $x$, $P'(x)=3\cdot P(x)$.
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.