Chapter 3
Three faces of the
derivative: geometric,
analytic, and
computational
In Chapter 2 we bridged two concepts: the average rate of change (slope of secant line)
and the instantaneous rate of change (the derivative). We arrived at a “recipe” for calculating a derivative algebraically. We thereby introduced the idea of limits, a concept that
merits further discussion. One goal of this chapter is to consider the technical aspects of
limits - a requirement if we are to use the definition of the derivative to determine derivatives of common functions.
Before doing so, we consider a distinct approach, which is geometric in flavour.
Namely, we show that the local behaviour of a continuous function is described by a tangent line at a point on its graph, We can visualize the tangent line by zooming into the
graph of the function. This duality - the geometric (graphical) and analytic (algebraic calculation) views - will form important themes throughout the discussions to follow, and are
two complementary, but closely related approaches to calculus.
3.1
The geometric view: Zooming into the graph of a
function
Section 3.1 Learning goals
1. Understand the connection between the local behaviour of a function (seen by zooming into the graph) at a point and the tangent line to the graph of the function at that
point.
2. Given the graph of a function, be able to sketch its derivative.
3.1.1
Locally, the graph of a function looks like a straight line
In this section we consider well-behaved functions whose graphs are “smooth”, in contrast
to the discrete data points of Chapter 2. We connect the derivative to the local shape of
41
42Chapter 3. Three faces of the derivative: geometric, analytic, and computational
the graph of a given function. By local behaviour we meet the behaviour seen when we
magnify the graph by zooming into a point. Imagine using a high-powered magnifying
glass or a microscope. The center of the field of vision is the point of interest. As we zoom
in, the graph looks flatter, till we observe a straight line, as shown in Fig. 3.1.
Definition 3.1 (Tangent line). The straight line that we see when we zoom into the graph
of a smooth function at some point x0 is called the tangent line at x0 .
Definition 3.2 (Geometric definition of the derivative). The slope of the tangent line at
the point x will be denoted as derivative of the function at the given point.
6.0
-2.0
y
4.0
2.0
-6.0
y
2.2
x
0.0
0.5
2.0
x
1.6
1.4
1.6
x
Figure 3.1. Zooming in on the graph of y = f (x) = x3 − x at the point x = 1.5
makes the graph “look like” a straight line, the tangent line. The slope of that tangent line
is the derivative of the function at x = 1.5.
Example 3.3 (Zoom 1) Consider the function y = f (x) = x3 − x and the point x = 1.5
(Fig. 3.1). Find the tangent line to the graph of this function by zooming in at the given
point.
Solution: The graph of the function is shown in Figure 3.1(a), where we have indicated
the point of interest with a red dot. Now zoom in, and magnify the graph, centered on the
given point. Locally, the graph resembles a straight line.
Example 3.4 (Zooming into the sine graph at the origin:) Determine the derivative of the
function y = sin(x) at x = 0 by zooming into the origin on the graph of this function. Then
write down the equation of the tangent line at that point.
Solution: In Figure 3.2 we show a zoom into the graph of the function
y = sin(x)
at the point x = 0. The sequence of zooms leads to a straight line (far right panel) that
3.1. The geometric view: Zooming into the graph of a function
y
1.0
1.0
-3.14
3.14
-1.0
y
0.3
x
43
y
x
-1.0
x
-0.3
-1.0
1.0
-0.3
Figure 3.2. Zooming into the graph of the function y = f (x) = sin(x) at the
point x = 0 . Eventually, the graph resembles a line of slope 1. This is the tangent line at
x = 0 and its slope, the derivative of y = sin(x) at x = 0 is 1.
we identify once more as the tangent line to the function at x = 0. From the graph it is
apparent that the slope of this tangent line is 1. We say that the derivative of the function
y = f (x) = sin(x) at x = 0 is 1, and write f ! (0) = 1 to denote this fact. As this line goes
through (0, 0) and has slope 1, its equation is simply y = x. We can also say that close to
x = 0 the graph of y = sin(x) looks a lot like the line y = x.
3.1.2
At a cusp or a discontinuity, the derivative is not defined
y
Cusp
x
Figure 3.3. If we zoom into a function at a cusp, there is no one straight line
that describes local behaviour. No matter how far we zoom in, we see two distinct lines
meeting at a sharp “corner”. We say that the function has no tangent line at a a cusp and
the derivative is not defined at that point.
3.1.3
From a function to a sketch of its derivative
The tangent line to the graph of a function varies from point to point along the graph of the
function - what we see when zooming in depends on the point at which we zoom in. This
0.3
44Chapter 3. Three faces of the derivative: geometric, analytic, and computational
is equivalent to saying that the derivative f ! (x) is, itself, also a function. Here we consider
the connection between these two functions by using the graph of one to sketch the other.
The sketch is meant to be approximate, but will contain some important elements.
f(x)
x
Figure 3.4. The graph of a function. We will sketch its derivative.
Example 3.5 Consider the function graphed in Fig. 3.4. Reason about the tangent lines at
various points along this curve to arrive at a sketch of the derivative f ! (x).
f(x)
x
Tangents
2
1
0
-1 -0.5 0
1
2
3
Slopes
f ' (x)
x
Figure 3.5. Sketching the derivative of a function
Solution: In Figure 3.5 we first sketch a few tangent lines along the graph of f (x). Pay
special attention to the slopes (rather than height, length, or any other property) of these
dashes. Copying these lines in a row along the direction of the x axis, we estimate their
slopes with rather crude numerical values.
From left to right, slopes are first positive, decrease to zero, become negative, and
then increase again through zero back to positive values. (We see precisely two dashes that
are horizontal, and so have slope 0.) Next, we plot the numerical values (for slopes) that
3.1. The geometric view: Zooming into the graph of a function
45
we have recorded on a new graph. This is the beginning of the graph of the derivative,
f ! (x). Only a few points have been plotted for f ! (x), but the trend is clear: The derivative
function has two zeros It dips below the axis between these places. In Figure 3.5 we show
the original function f (x) and its derivative f ! (x). We have aligned these graphs so that
the slope of f (x) matches the value of f ! (x) shown directly below.
Example 3.6 Sketch the derivative of the function shown in Fig. 3.6.
y=f (x)
x
Figure 3.6. Sketching the derivative of a function.
Solution: See Fig. 3.7 for the entire process, and note that this time, we represent only the
sign of the derivative (positive or negative) and places where it is zero. The thin vertical
lines demonstrate that f ! (x) = 0 coincides with tops of hills or bottom of valleys on the
graph of the function f (x).
3.1.4
Constant and linear functions and their derivatives
Example 3.7 (Derivative of y = C) Use a geometric argument to determine the derivative
of the function y = f (x) = C at any point x0 on its graph.
Solution: This function is a horizontal straight line, whose slope is zero everywhere. Thus
“zooming in” at any point x, leads to the same result, so the derivative is 0 everywhere.
Example 3.8 (Derivative of y = Bx)
Solution: The function y = Bx is a straight line of slope B. At any point on its graph, it
has the same slope, B. Thus the derivative is equal to B at any point on the graph of this
function.
The reader will notice that in the above two examples, we have thereby found the
derivative for the two power functions, y = x0 and y = x1 . We summarize:
The derivative of any constant function is zero. The derivative of the
function y = x is 1.
46Chapter 3. Three faces of the derivative: geometric, analytic, and computational
Function
y=f (x)
x
Tangent lines
x
Slopes
+ 0 -
0
Derivative
+
0
-
0
+
+
y=f (x)
x
Figure 3.7. Sketching the derivative of a function for Example 3.6
3.1.5
Molecular motors
Cells such as neurons can be up to 1 meter in length. This makes for a tough engineering
problem of getting material from the cell body, where it is made, to the cell ends where
it is needed for communication, repair, and metabolism. It is essential to have a transport
3.1. The geometric view: Zooming into the graph of a function
47
mechanism! This is accomplished by proteins that act like “molecular motors”, that carry
cargo (vesicles) along a system of “molecular highways” called microtubules (MT). Each
microtubule “road” is a one-way street with two distinct ends (called the “plus” and “minus” ends). Some motors specialize in walking one-way towards + end, while others walk
towards the - end. In Fig. 3.8(a), we show a schematic diagram of kinesin (represented
by the letter k), a motor that walks towards the + MT end. We represent vesicles by the
little spheres in the same figure. Motors can hop off one microtubule and onto another, so
a motor moving one way can suddenly be seen to switch directions. In Example 3.9, we
study a sample vesicle track (displacement, y over time t) and decipher the sequence of
motor events that caused that motion.
(a)
vesicle
kinesin k
k
(c)
velocity, v
(b)
displacement, y
+
0
0
+
microtubules
t
t
Figure 3.8. The molecular motor kinesin walks towards a microtubule “plus” end.
It can detach and reattach to another microtubule. (b) The displacement of the vesicle. (c)
The velocity of the vesicle.
Example 3.9 (Motion of molecular motors) The displacement y(t) of a vesicle is shown
in Fig. 3.8(b). Sketch the corresponding instantaneous velocity v(t) for the vesicle and use
your sketch to explain what was happening to the kinesin carrying that vesicle.
Solution: The plot in Fig. 3.8(b) consists of straight line segments with sharp corners
(cusps). Over each of these line segments, the slope dy/dt, which corresponds to the instantaneous velocity, v(t), is constant. Segments with positive slope correspond to motion
towards the right (as in the top MT track in Fig. 3.8a). Over times where the slope is negative, the motion is to the left. Where the slope is zero (flat graph), the vesicle was stationary.
In Fig. 3.8(c), we sketch the graph of the instantaneous velocity, v(t). Observe that v(t),
which is the derivative of y(t), is not defined at the points where y(t) has “corners”.
48Chapter 3. Three faces of the derivative: geometric, analytic, and computational
Based on Fig. 3.8(c), the kinesin motor was moving on a right-facing microtubule,
then hopped onto a left-facing MT, and then hopped back to a right -facing MT. For a brief
time it was either stuck or detached from the MT tracks (stationary part). Finally, it hopped
onto a left-moving MT. We summarize the velocities in Fig. 3.8(c).
3.2 Analytic view: calculating the derivative
Section 3.2 Learning goals
1. Be able to explain the definition of a continuous function, and to identify functions
that have various types of discontinuities.
2. Understand how to evaluate simple limits of rational functions.
3. Be able to calculate the derivative of a simple function using the definition of the
derivative, 2.18.
3.2.1
Technical matters: continuous functions and limits
So far, we encountered two distinct types of functions. In Chapter 2, we considered functions made up of discrete data points (e.g. temperature defined at a finite set of time points).
In Chapter 1 we introduced continuous functions, such as the height of a falling ball (2.1),
or power functions more generally. Intuitively, for a continuous function, every point on
the graph is connected to neighbouring points. We also have some notion of where a function has a “break” or is undefined. For example, the function (2.1) which
√ is just a power
function, is continuous for all values of t, whereas a function such as y = x is continuous
and defined as a real number only for x ≥ 0. The function y = 1/(x + 1) is defined and
continuous for x $= −1 (since division by zero is undefined).
Here we extend the intuitive discussion with a more formal definition, based on the
concept of a limit. We first define what it means for a function to be continuous, and then
show how limits are computed to test that definition.
Definition 3.10 (Continuous function). We say that y = f (x) is continuous at a point
x = a in its domain if
lim f (x) = f (a).
x→a
By this we mean that the function is defined at x = a, that the above limit exists, and that
it matches with the value that the function takes at the given point.
This definition has two important parts. First, the function should be defined at the
point of interest, and second that the value assigned by the function has to “fit the local
behaviour” in the sense of the limit. This rules out a “jump” or break in the graph. When
the above is not true at some point xs , we say that the function is discontinuous at xs .
To demonstrate what types of discontinuities exist, we provide a few examples, and, at the
same time, illustrate how limits are calculated.
3.2. Analytic view: calculating the derivative
49
Function with a hole in its graph
Consider a function of the form
f (x) =
(x − a)2
.
(x − a)
Then if x $= a, we can cancel a common factor, and obtain (x − a). At x = a, the function
is not defined (0/0). In short, we have
!
(x − a)2
x−a
x $= a
f (x) =
=
undefined x = a.
(x − a)
Even though the function is not defined at x = a, we can evaluate the limit of f as x
approaches a. We write
(x − a)2
= lim x − a = 0,
x→a
x→a (x − a)
lim f (x) = lim
x→a
and say that “the limit as x approaches a” exists and is equal to 0. We also say that the
function has a removable discontinuity. If we add the point (a, 0) to the set of points at
which the function is defined then we obtain a continuous function identical to the function
x − a. See also Appendix D.
Function with jump discontinuity
Consider the function
!
−1 x ≤ a,
1
x > a.
We say that the function has a jump discontinuity at x = a. As we approach the point
of discontinuity we observe that the function has two distinct values, depending on the
direction of approach. We formally capture this observation using right and left hand
limits,
lim f (x) = −1,
lim f (x) = 1.
f (x) =
x→a−
x→a+
Since the left and right limits are unequal, we say that “ the limit does not exist” (abbreviated DNE).
Function with blow up discontinuity
Consider the function
1
.
x−a
Then as x approaches a, the denominator approaches 0, and the value of the function goes
to ±∞. We say that the function “blows up” at x = a and that the limit limx→a f (x) does
not exist. We also write
lim f (x) = ∞
f (x) =
x→a
to denote the same thing.
Figure 3.9 illustrates the differences between functions that are continuous everywhere, those that have a hole in their graph, and those that have a jump discontinuity or a
blow up at some point a.
50Chapter 3. Three faces of the derivative: geometric, analytic, and computational
y
y
continuous
x
y
hole
a
x
y
jump
a
x
blow-up
a
x
Figure 3.9. Left to right: a continuous function, a function with a removable
discontinuity, a jump discontinuity, and a function with blow up discontinuity.
Examples of Limits
We now examine several examples of computations of limits. More details about properties
of limits are provided in Appendix D.
By Definition 3.10, to calculate the limit of any function at a point of continuity, we
simply evaluate the function at the given point.
Example 3.11 (Simple limit of a continuous function) Find the following limits:
x
1
(c) lim
.
(a) lim x2 + 2 (b) lim
x→10 1 + x
x→3
x→1 x + 1
Solution: In each of the above, the function is continuous at the point of interest (at x =
3, 1, 10, respectively). Thus, we simply “plug in” the values of x in each case to obtain
1
1
=
x→1 x + 1
2
(a) lim x2 + 2 = 32 + 2 = 11 (b) lim
x→3
x
10
=
.
x→10 1 + x
11
(c) lim
Example 3.12 (Hole in graph limits) Calculate the limits of the following functions. Note
that each has a removable discontinuity (“a hole in its graph”).
x2 − 6x + 9
x2 + 3x + 2
(a) lim
(b) lim
.
x→3
x→−1
x−3
x+1
Solution: We first simplify algebraically by factoring the numerator, and then evaluate the
limit. Note that the simplification is possible so long as we evaluate the limit, rather than
the actual function, at the point of discontinuity.
(x − 3)2
x2 − 6x + 9
= lim
= lim (x − 3) = 0
x→3 (x − 3)
x→3
x→3
x−3
(a) lim
x2 + 3x + 2
(x + 1)(x + 2)
= lim
= lim (x + 2) = 1.
x→−1
x→−1
x→−1
x+1
(x + 1)
(b) lim
Example 3.13 (Limit involving sin(x)) Use the observation made in Example 3.4 to arsin(x)
rive at the value of lim
x→0
x
3.2. Analytic view: calculating the derivative
51
Solution: Example 3.4 illustrated the fact that close to x = 0 the function sin(x) has the
following behaviour:
sin(x)
≈ 1.
sin(x) ≈ x, or
x
This is equivalent to the result
lim
x→0
sin(x)
= 1.
x
(3.1)
We read this “as x approaches zero, the limit of sin(x)/x is 1.” We will find this limit useful
in later calculations involving derivatives of trigonometric functions.
3.2.2
Computing the derivative
As discussed in Section 2.5, calculating a derivative requires the use of limits. The two
statements below emphasize this fact.
1. A secant line connects two points on the graph of a function (e.g. x and x + h). In
the limit that those points get closer together (h → 0), we obtain a tangent line.
2. The slope of a secant line is an average rate of change, but in the limit (h → 0), we
obtain an instantaneous rate of change, which is the slope of the tangent line, namely
the derivative.
We now apply the techniques so far to calculating a few derivatives based on the
definition of the derivative. We start with several simple examples, and work our way up to
more interesting cases.
Example 3.14 (Derivative of a linear function) Using Definition 2.18 of the derivative,
compute the derivative of the function y = f (x) = Bx + C.
Solution: We have already used a geometric approach to find the derivative of related
functions in Examples 3.7-3.8. Here we do the formal calculation as follows:
f (x + h) − f (x)
(start with the definition)
h→0
h
[B(x + h) + C] − [Bx + C]
= lim
(apply it to the function)
h→0
h
Bx + Bh + C − Bx − C
(expand the numerator)
= lim
h→0
h
Bh
= lim
(simplify)
h→0 h
= lim B (cancel a factor of h)
f ! (x) = lim
h→0
=B
(evaluate the limit).
(3.2)
52Chapter 3. Three faces of the derivative: geometric, analytic, and computational
Hence, we confirmed that the derivative of f (x) = Bx + C is f ! (x) = B. This agrees with
the sum of the derivatives of the two parts, Bx and C found in Examples 3.7-3.8. Indeed,
as we will establish shortly, the derivative of the sum of two functions is the same as the
sum of their derivatives.
Example 3.15 (Derivative of the cubic power function) Compute the derivative of the function y = f (x) = Kx3 .
Solution: For y = f (x) = Kx3 we have
dy
f (x + h) − f (x)
= lim
dx h→0
h
K(x + h)3 − Kx3
= lim
h→0
h
3
(x + 3x2 h + 3xh2 + h3 ) − x3
= lim K
h→0
h
2
2
(3x h + 3xh + h3 )
= lim K
h→0
h
2
= lim K(3x + 3xh + h2 )
h→0
= K(3x2 ) = 3Kx2 .
(3.3)
Thus the derivative of f (x) = Kx3 is f ! (x) = 3Kx2 .
Example 3.16 Use the definition of the derivative to compute f ! (x) for the function y =
f (x) = 1/x at the point x = 1.
Solution: We write down the formula for this calculation at any point x and then simplify
algebraically, using common denominators to combine fractions, and then, in the final step,
calculate the limit formally. Then we substitute the value x = 1.
f (x + h) − f (x)
(the definition)
h→0
h
1
1
(x+h) − x
(applied to the function)
= lim
h→0
h
f ! (x) = lim
= lim
[x−(x+h)]
x(x+h)
(common denominator)
h
−h
(algebraic simplification)
= lim
h→0 hx(x + h)
−1
= lim
(cancel factor of h)
h→0 x(x + h)
1
= − 2 (limit evaluated)
x
h→0
(3.4)
(3.5)
3.3. Computational face of the derivative: software to the rescue!
53
Thus, the derivative of f (x) = 1/x is f ! (x) = −1/x2 and at the point x = 1 it takes the
value f ! (1) = −1.
In Problem 9 we apply similar techniques to the derivative of the square-root function
to show that
√
1
(3.6)
y = f (x) = x ⇒ f ! (x) = √ .
2 x
In the next chapter, we formalize some observations about derivatives of power funcitons
and rules of differentiation. This will allow us to avoid such tedious calculations in finding
simple derivatives.
3.3
Computational face of the derivative: software to
the rescue!
Mathematical analysis benefits greatly from computational methods that complement algebraic and geometric approaches. Using various software, we can gain insight, experiment,
as well as design methods for accurate computations that would be too tedious to carry out
by hand. Here we show a third face of the derivative: its numerical implementation using a
simple spreadsheet. The ideas introduced here will reappear in a variety of problems where
repetitive calculations are needed to arrive at the solution to one or another problem.
Section 3.3 Learning goals
1. Appreciate the fact that software can numerically compute an approximation to the
derivative.
2. Understand that the approximation replaces a (true) tangent line with an (approximating) secant line.
3. Be able to use a spreadsheet (or your favorite software aid) to graph a function and
its derivative.
4. Be able to explain verbally how that the derivative shape is connected with the shape
of the original function.
5. Interpret the differences between two types of biochemical kinetics, MichaelisMenten and Hill function.
Definition 3.17 (Numerical derivative). By a numerical derivative calculation we mean a
numerical approximation to the value of the derivative, obtained by using a finite value,
f ! (x)numerical ≈
∆f
∆x
rather than the actual value f ! (x)actual = lim
∆x→0
∆f
.
∆x
The numerical derivative approximates the true derivative provided ∆x is small relative to the range over which the function f changes dramatically. Since ∆f is the difference
54Chapter 3. Three faces of the derivative: geometric, analytic, and computational
of two values of f , (∆f = f (x + ∆x) − f (x)) it follows that the numerical derivative is the
same as the slope of a secant line. This important realization, associated with LG 2, means
that a secant line is often used to approximate a tangent line, and the slope of a secant line
is used to approximate a derivative in numerical computations. We will see this idea again
in several contexts.
3.3.1
Derivative of Michaelis-Menten and Hill functions
A spreadsheet can be used to numerically approximate derivatives. We illustrate this using
as examples the reaction speeds for Michaelis-Menten (1.8) and Hill function kinetics, (1.7)
(see Section 1.5), repeated below:
vMM = f1 (c) =
Kc
,
a+c
(3.7)
Kcn
.
(3.8)
+ cn
(Both f1 (c), f2 (c) are shown as functions of c in Fig. 3.10a.) Our purpose is to use a
spreadsheet to compute a numerical approximation of the derivative of these two reaction
speeds with respect to the chemical concentration c.
vHill = f2 (c) =
an
Example 3.18 (The derivative on a spreadsheet) Use a spreadsheet (or your favorite software) to plot the derivatives of the functions vMM = f1 (c), vHill = f2 (c).
concentration, c
a
0.0000
0.1000
0.2000
0.3000
0.4000
0.5000
0.6000
0.7000
0.8000
0.9000
1.0000
vMM = f1 (c)
b
0.0000
0.4545
0.8333
1.1538
1.4286
1.6667
1.8750
2.0588
2.2222
2.3684
2.5000
vHill = f2 (c)
c
0.0000
0.0005
0.0080
0.0402
0.1248
0.2941
0.5737
0.9681
1.4529
1.9809
2.5000
∆f1 /∆c
d
4.5455
3.7879
3.2051
2.7473
2.3810
2.0833
1.8382
1.6340
1.4620
1.3158
1.1905
∆f2 /∆c
e
0.0050
0.0749
0.3219
0.8463
1.6931
2.7954
3.9441
4.8483
5.2796
5.1914
4.7086
Table 3.1. On a spreadsheet, we pick discrete points (values of c at increments
of ∆c along the a column) at which to compute the functions (f1 , f2 ) in columns b, c. We
then calculate the quantities ∆fi /∆c to approximate the derivatives. Results are plotted
in Fig. 3.10.
Solution: Figure 3.11 demonstrates typical spreadsheet manipulation that we use to numerically plot the two functions and their derivatives. Steps are as follows
3.3. Computational face of the derivative: software to the rescue!
5.5
5.5
Hill Function
Hill Function
Michaelis-Menten
dv/dc
Reaction Rate, v
55
Michaelis-Menten
0.0
0.0
0.0
chemical concentration, c
(a)
5.0
0.0
chemical concentration, c
5.0
(b)
Figure 3.10. (a) A plot of f1 (c), f2 (c) produced by graphing the points in the
columns b, c of the spreadsheet against the values in column a (see Table 3.1). (b) A plot
of both the functions and their (approximate) derivatives. The latter are plots of points in
columns d, e against the values in column a.
(A) Input the two functions.
(1) Because numerical calculations are based on arithmetical calculations rather
than qualitative analysis, we have to specify values of any constants or parameters in the functions of interest. Let us set K = 5, a = 1 for both functions and
n = 4 for the latter.
(2) Spreadsheet cells are labeled by row and column (see Table 3.1). In column a,
indicate the values of independent variable at which functions and derivatives
will be computed. This represents a subdivision of the “x axis” (which is the
chemical concentration axis, c in this example). We use steps of size ∆c = 0.1.
To do so, type 0 in cell a0 and a0 + 0.1 in cell a1.
(3) In cell b0 type the formula for the desired function: 5 ∗ a0/(1 + a0).
(4) Generate an entire set of values (e.g. for f1 at increments ∆c) by highlighting the cells b0, clicking on the dot in the lower right corner, and dragging
it down the column (Fig. 3.11). This will populate all cells with the desired
computations, saving you much tedious work(!)
(5) Repeat the process for the second function 5 ∗ a0∧ 4/(1 + a0∧ 4), in column c.
The symbol ∧ denotes a power, ∗ denotes multiplication, and braces are used
as needed in quotients or products of terms. The columns a-c now contain the
coordinates of points along the functions. To get a reasonable approximation
for the derivative, the points along the x axis should be close together, so that
∆c is small.
56Chapter 3. Three faces of the derivative: geometric, analytic, and computational
(6) Plotting the above data produces the graph shown in Fig. 3.10a.
(B) Next, we compute the desired numerical approximations of the derivatives.
(7) Use column d for the numerical derivative of f1 (c). To do so, approximate the
actual derivative with a finite difference14 ,
∆f1
df1
≈
.
∆c
dc
We picked ∆c = 0.1, which is stored in cell a1. Pointing to cell d0, we type
(b1 − b0)/$a$1), which computes successive values of the function in the b
column and divides by the same value of ∆c each time15 . Dragging the black
dot down the d column generates the desired list of values of the numerical
derivative.
(8) The process is repeated to generate the derivative of the function f2 in the e
column. Observe that we use the same absolute reference $a$1 for ∆c and
successive differences of the function in the c column (by typing (c1−c0)/$a$1
in cell e1.
Results of the above process lead to the graphs shown on the right panel of Fig. 3.10.
Example 3.19 Interpret the graphs of the derivatives in Fig. 3.10b in terms of the way that
reaction speed increases as the chemical concentration is increased in each of MichaelisMenten and Hill function kinetics
Solution: Both derivatives are positive everywhere, since both f1 (c) and f2 (c) are increasing functions. For Michaelis-Menten, the derivative is always decreasing. This agrees with
the observation that f1 (c) (thin red curve) gradually levels off and flattens as c increases.
While the reaction rate vMM increases with c, the rate of increase, dv/dc, slows due to
saturation at higher c values.
In contrast, the Hill function derivative starts at zero, increases sharply, and only then
decreases to zero. Correspondingly, the Hill function (thin blue curve) is flat at first, then
becomes steeply increasing, and finally flattens to an asymptote. We can summarize this
biochemically by saying that the initial reaction rate vHill is small and hardly changes near
c ∼ 0. For intermediate range of c, the reaction rate depends sensitively on c (evidenced by
large dv/dc). As c increases to higher values, saturation slows down the rate of reaction,
leading to the drop in dv/dc.
14 Importantly, the two expressions are not equal (!) However, for sufficiently small ∆c, they approximate one
another well.
15 To ensure we point only to cell a1, we use an absolute reference (typical syntax $a$1.
3.3. Computational face of the derivative: software to the rescue!
57
(A) Define the two functions
(1)
(2)
(4)
(3)
(B) Compute the two “derivatives”
(5)
Figure 3.11. Spreadsheet used to calculate a derivative.
(6)
58Chapter 3. Three faces of the derivative: geometric, analytic, and computational
Exercises
3.1. Sketching the derivative (geometric view): Sketch the graph of the derivative of
the function shown in Figure 3.12.
y
x
Figure 3.12. Figure for Problem 1
3.2. Sketching the derivative (Geometric view): Shown in Figure 3.13 below are three
functions. Sketch the derivatives of these functions.
y
y
y
x
x
x
Figure 3.13. Figure for problem 2
3.3. Sketching the function given its derivative: You are given the following information about the the values of the derivative of a function, g(x). Use this information
to sketch (very rough) graph the function for −3 ≤ x ≤ 3.
x
g (x)
!
-3
-1
-2
0
-1
2
0
1
1
0
2
-1
3
-2
3.4. What the sign of the derivative tells us: You are given the following information
about the signs of the derivative of a function, f (x). Use this information to sketch
a (very rough) graph of the function for −3 ≤ x ≤ 3.
x
f ! (x)
-3
0
-2
+
-1
0
0
-
1
0
2
+
3
+
3.5. Shallower or steeper rise: Shown in Fig. 3.14 are two similar functions, both increasing from 0 to 1 but at distinct rates. Sketch the derivatives of each one. Then
Exercises
59
comment on what your sketch would look like for a discontinuous “step function”,
defined as follows:
!
0 x<0
f (x) =
1 x ≥ 0.
(a)
(b)
Figure 3.14. Figure for Problem 5.
3.6. Introduction to velocity and acceleration: The acceleration of a particle is the
derivative of the velocity. Shown in Figure 3.15 is the graph of the velocity of a
particle moving in one dimension. Indicate directly on the graph any time(s) at
which the particle’s acceleration is zero.
v
t
Figure 3.15. Figure for Problem 6
3.7. Velocity, continued: The vertical height of a ball, d (in meters) at time t (seconds)
after it was thrown upwards was found to satisfy d(t) = 14.7t − 4.9t2 for the first 3
seconds of its motion.
(a) What is the initial velocity of the ball (i.e. the instantaneous velocity at t = 0)?
(b) What is the instantaneous velocity of the ball at t = 2 seconds?
3.8. Geometric view, continued:
(a) Given the function in Figure 3.16(a), graph its derivative.
(b) Given the function in Figure 3.16(b), graph its derivative
60Chapter 3. Three faces of the derivative: geometric, analytic, and computational
(c) Given the derivative f ! (x) shown in Figure 3.16(c) graph the function f (x).
(d) Given the derivative f ! (x) shown in Figure 3.16(d) graph the function f (x).
1.0
10.0
y=f(x)
y=f(x)
-0.5
-10.0
0.0
2.3
0.0
10.0
(a)
10.0
(b)
3.0
f '(x)
f '(x)
-10.0
-2.0
0.0
10.0
-1.3
(c)
1.3
(d)
Figure 3.16. Figures for Problem 8.
3.9. Computing the derivative of square-root (from the definition): Consider the
function
√
y = f (x) = x.
(a) Use the definition of the derivative to calculate f ! (x). You will need to use the
following algebraic simplification:
√ √
√
√
( a − b)( a + b)
a−b
√
√ .
= √
√
( a + b)
( a + b)
(b) Find the slope of the function at the point x = 4.
(c) Find the equation of the tangent line to the graph at this point.
3.10. Computing the derivative: Use the definition of the derivative to compute the
derivative of the function y = f (x) = C/(x + a) where C and a are arbitrary
constants. Show that your result is f ! (x) = −C/(x + a)2 .
x
.
3.11. Computing the derivative: Consider the function y = f (x) =
(x + a)
Exercises
61
a
.
(x + a)
(b) Use the results of Problem 10 to determine the derivative of this function.
(Note: you do not need to use the definition of the derivative to do this coma
putation.) Show that you get f ! (x) = (x+a)
2.
(a) Show that this same function can be written as f (x) = 1 −
3.12. Molecular motors: Fig. 3.17 (a) shows the displacement of a vesicle carried by a
molecular motor. The motor can either walk right (R), left (L) along one of the microtubules or it can unbind (U) and be stationary, then rebind again to a microtubule.
Sketch a rough graph of the velocity of the vesicle v(t) and explain the sequence of
events (using the letters R, L, U) that resulted in this motion. Fig. 3.17 (b) shows the
velocity v(t) of another vesicle. Sketch a rough graph of its displacement starting
from y(0) = 0.
y
v
(a)
(b)
0
0
t
t
Figure 3.17. Figure for problem 12
3.13. Concentration gradient: Certain types of tissues, called epithelia are made up
of thin sheets of cells. Substances are taken up on one side of the sheet by some
active transport mechanism, and then diffuse down a concentration gradient by a
mechanism called facilitated diffusion on the opposite side. Shown in Figure 3.18
is the concentration profile c(x) of some substance across the width of the sheet (x
represents distance). Sketch the corresponding concentration gradient, i.e. sketch
c! (x), the derivative of the concentration with respect to x.
c(x)
facilitated
diffusion
active
transport
distance across the sheet
x
Figure 3.18. Figure for Problem 13
3.14. Tangent line to a simple function: What is the slope of the tangent line to the
function y = f (x) = 5x + 2 when x = 2? when x = 4 ? How would this slope
62Chapter 3. Three faces of the derivative: geometric, analytic, and computational
change if a negative value of x was used? Why?
3.15. Slope of the tangent line: Use the definition of the derivative to compute the slope
of the tangent line to the graph of the function y = 3t2 − t + 2 at the point t = 1.
3.16. Tangent line: Find the equation of the tangent line to the graph of y = f (x) =
x3 − x at the point x = 1.5 shown in Fig. 3.1. You may use the fact that the tangent
line goes through (1.7, 1.47) as well as the point of tangency.
3.17. Numerically computed derivative: Consider the two Hill functions
H1 (x) =
x2
,
0.01 + x2
H2 (x) =
x4
0.01 + x4
(a) Sketch a rough graph of these two functions on the same plot and/or describe
in words what the two graphs would look like.
(b) On a second plot, sketch a rough graph of both derivatives of these functions
and/or describe in words what the two derivatives would look like.
(c) Using a spreadsheet or your favorite software, plot the two functions over the
range 0 ≤ x ≤ 1.
(d) Use the spreadsheet to calculate an approximation for the derivatives H1 , (x), H2! (x)
and plot these two functions together. (NOTE: In order to have a reasonably
accurate set of graphs, you will need to select a small step size of ∆x ≈ 0.01.)
3.18. More numerically computed derivatives: As we will later find out, trigonometric
functions such as sin(t) and cos(t) can be used to describe biorhythms of various
types. Here we numerically compute the first and second derivative of y = sin(t)
and show the relationships between the trigonometric functions and their derivatives.
We will use only numerical methods (e.g. a spreadsheet), but later, in Chapter 14,
we will also study the analytical calculation of the same derivatives.
(a) Use a spreadsheet (or your favorite software) to plot, on the same graph the
two functions
y1 = sin(t), y2 = cos(t),
0 ≤ t ≤ 2π ≈ 6.28.
Note that you should use a fairly small step size, e.g. ∆t = 0.01 to get a
reasonably accurate approximation of the derivatives.
(b) Use the same spreadsheet to (numerically) calculate (an approximate) derivative y1! (t) and add it to your graph.
(c) Now calculate y1!! (t), that is (an approximation to) the derivative of the derivative of the sine function and add this to your graph.