Calculus: From A to TheFull Frontal Calculus Chapter 1: The Fundamental Ideas 2015, Seth Braver 1...
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Calculus: From A to The
A calculus is a set of symbolic rules for manipulating objects of some specified type. If youβve studied
statistics, youβve probably used the calculus of probabilities.* If youβve studied formal logic, it follows
that youβve met the propositional calculus.β After youβve mentally summed up all the little bits of
knowledge in this book, youβll have learned the calculus of infinitesimals.
An infinitesimal is an infinitely small number β smaller than any positive real number, yet greater
than zero. Like square roots of negatives (the regrettably-named βimaginary numbersβ), infinitesimals
seem paradoxical: They manifestly do not belong to the familiar system of real numbers.
Mathematicians and scientists have, nonetheless, used them for centuries, because infinitesimals help
us understand functions of real numbers β those faithful tools with which we model, among other
things, the so-called real world of experience. To gain perspective on our home world of real numbers, it
helps to survey this world from without. Such a justification for working with numbers beyond the reals
should suffice for the hard-nosed pragmatist; for another, equally valid, justification, we need only
observe our fellow mammals the dolphins and whales at play. There is pleasure to be had in breaching
the surface of a world that normally confines us.
Biologists know that for some purposes, the naked eye is perfectly serviceable. For others,
microscopes are necessary. Similarly, users of mathematics know that for all of our pragmatic
measurement needs, the real numbers suffice. However, for certain theoretical developments,
observing the mathematical world through infinitesimal-sensitive lenses sharpens the pixilated image
presented by the naked eye of the reals. Of course, this latter microscope is purely mental; cultivating a
sense of what it reveals requires practice and imagination. The essence of the idea, however, is simple:
Magnitudes which appear equal to the naked eye of real numbers may turn out, when viewed through
our infinitesimal-sensitive microscope, to differ by an infinitesimal amount. Conversely, if you observe
two magnitudes under the microscope that differ by an infinitesimal, these will always correspond,
when viewed with the naked eye, to one and the same real number.
βAn interesting idea,β you may reflect, βBut why bother? The reals are familiar, simple, and useful. I
wish we just stick to the reals.β Be careful what you wish for! The real numbers are, despite what you
might think, exceedingly, disturbingly strange, though their weirdness is rarely brought out into the light
in elementary mathematics classes. Adding infinitesimals into the mix actually clarifies some of this
uncomfortable weirdness. Here are two examples of real weirdness, and how thinking about
infinitesimals can help us make better intuitive sense of βthe real world.β
Example 1. Everyone is dubious when first told the infinite decimal 0.99999β¦ equals 1. Even after
learning why this is so (just multiply both sides of 0.33333333 = 1/3 by 3), some skeptics still
feel in their bones that nonetheless, 0.99999β¦ must somehow be less than 1.
Infinitesimals can alleviate the skepticβs discontent. Yes, 0.99999β¦ and 1 do correspond to
the same real number (i.e. there is no real difference between them), but this need not preclude
the possibility of an infinitesimal difference between them.
* The calculus of probabilities contains rules such as π(π΄ βͺ π΅) = π(π΄) + π(π΅) β π(π΄ β© π΅). β Such a calculus would contain rules such as π β§ π βΉ π.
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Example 2. Suppose we have 8 points, one of which is painted red. If we choose one at random,
then the probability of picking the red one is obviously 1/8. Now suppose that of a line segmentβs
infinitely many points, one is red; in a random selection, how likely are we to pick the red one?
In the case of the line segment, there are over 106 points, so the probability of choosing the
red one must be under 10β6. Indeed, a little thought reveals that the probability of choosing the
red point must be under ππβπ for every whole number π, since there are, for any π, over 10π
points on the segment. This condition in boldface, however, is satisfied by only one nonnegative
real number: zero. Hence, if we confine ourselves to the real numbers, we have no choice but to
conclude that the probability of selecting the red point is precisely zero.
Naturally, the argument against the red pointβs being chosen applies to every other point on
the segment as well. Some point on the segment, however, must be selected, so if we accept only
the real numbers we are faced with a deeply counterintuitive dilemma β an event whose
probability of occurrence is zero can, nevertheless, occur! If, however, we accept infinitesimals,
this paradox vanishes. For then we can then maintain that the probability of selecting the red
point from the segmentβs points is infinitesimal β unfathomably minuscule, impossible to
represent as a decimal, indistinguishable from zero in the real world, and yet⦠not quite zero.
Everyone and his mother βknowsβ that a circleβs area is ππ2, where π is the circleβs radius. But why is this
so? Infinitesimals will help you understand this familiar fact.
Example 3. The area of a circle with radius π is ππ2.
Proof. Inscribe a regular π-sided polygon in the circle. Clearly, the greater π
is, the closer the polygon will cleave to the circle. Even when π is relatively
small (say, π = 50), distinguishing the polygon from the circle becomes
difficult for the naked eye; when π is a million, the task is hopeless, even
with a microscope. Our proof of the circleβs area formula hinges on a radical
reconceptualization: We shall think of the circle as a regular polygon with
infinitely many infinitesimally small sides. This way of thinking will enable us
to use facts about polygons (straight, simple objects) to learn about circles
(curved, hence difficult objects).
Since our regular π-gonβs area is π times that of the triangle in the figure above, its area must
be π(πβ/2). In fact, since ππ = π (the polygonβs perimeter), this expression simplifies to πβ/2.
Thus the circle, being a polygon (of infinitely many infinitesimally small sides), has area πβ/2.
But here, π is the circleβs circumference (which is 2ππ), and β is the circleβs radius (which is π.)
Substituting these values into πβ/2, we conclude that the circleβs area is (2ππ)π 2β , which
simplifies to ππ2, as claimed. β
Please dwell on this surprising, beautiful, disconcerting argument. When mathematicians began to use
infinitesimals, even philosophers and theologians took note. Is a proof that uses infinitesimals a genuine
proof? Is a circle really a polygon of infinitely many sides? Is it wise for finite man to reason about the
infinite? Even as such philosophical debates raged (from the 17th century on), mathematicians paid only
halfhearted attention, busy as they were developing a potent calculus of infinitesimals. That the calculus
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worked no one questioned; that it lacked a comprehensible foundation no one denied. Its triumphs
were astonishing. The infinitesimal calculus helped physics break free of its static Greek origins to
become a dynamic modern science. And yet... all attempts to establish fully comprehensible logical
foundations for the calculus failed. Since at least the time of Euclid (c. 300 BC), mathematics had been
viewed as the archetype of logical reasoning. Small wonder then that, despite the undeniable utility of
the infinitesimal calculus, its murky basis received stinging criticism. Most famously, philosopher George
Berkeley suggested in 1734 that anyone who could accept the shaky foundations of the infinitesimal
calculus βneed not, methinks, be squeamish about any point in Divinity.β To compare mathematics β the
traditional rock of logical certainty β to theology, to suggest that mathematicians, far from proceeding
by perfectly rigorous thought βsubmit to authority, take things on trust, and believe points
inconceivableβ (as Berkeley would have it) is to shake one of the very pillars of Western civilization.*
Not until the late 19th century did mathematicians discover a perfectly rigorous method (the theory
of limits, which youβll learn about in Chapter π) to set the results of the infinitesimal calculus on solid
foundations. That it took so long to develop these foundations is understandable, given the surprising
sacrifice involved: To transfer the massive body of results onto the long-desired secure logical
foundations, mathematicians had to sacrifice nothing less than the infinitesimals themselves!
On these new foundations, the results developed over the previous centuries were finally secure
(there was nothing to fear from the philosophers anymore), but the infinitesimals that had given birth to
these results were ruthlessly expunged in the victory celebration. The very notion of an infinitesimal
came to be viewed as an embarrassment to the brave new calculus, as though βinfinitesimalβ were a
discredited religious idea from a more primitive time whose abandonment was necessary for the further
progress of humanity. Even the subjectβs name was changed. What had once proudly been βThe
Infinitesimal Calculusβ thenceforth became known simply as The Calculus, a name whose very
emptiness spoke β to those with ears to hear β of the ghosts of departed quantities.β
Infinitesimal ghosts continued to haunt the calculus, for while the theory of limits had brilliantly
disposed of a logical problem, it had introduced a psychological problem. In the minds of many who
used calculus purely as a scientific tool (but who had no particular concern for the subjectβs esoteric
logical foundations), infinitesimals remained more intuitive than limits. For this reason, much notation
that originally referred to infinitesimals was, surprisingly, retained in the calculus even after the great
infinitesimal purge. Naturally, the notation was all reinterpreted in terms of limits, but this led to a sort
of mathematical schizophrenia. One would use the notation of infinitesimals, one would think in terms
of infinitesimals, but good mathematical hygiene demanded that one refrain from actually mentioning
infinitesimals in public. To be sure, textbook authors would sometimes timidly advise their readers that
it might be helpful to think of such and such an expression in terms of infinitesimals, but such advice was
invariably followed (as if an authority figure had just returned to the room) by a stern warning that
actually, infinitesimals donβt exist, and one shouldnβt speak about such things in polite society.
* Berkeley was a masterful shaker of pillars. His denial that matter exists outside of minds paved the way for David Humeβs
philosophical demolition of causality and personal identity, and hence to Immanuel Kantβs subsequent reconstruction of these ideas on a radically new philosophical basis (transcendental idealism) that he developed to refute Hume, and which then became a cornerstone of modern philosophy.
β Since βThe Calculusβ is only slightly more expressive than βThe Thing,β there was no real loss when, in time, even the definite article was shed. Hence todayβs unadorned Calculus (on tap at a college near you).
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Resurrection βI think in coming centuries it will be considered a great
oddity in the history of mathematics that the first exact
theory of infinitesimals was developed 300 years after the
invention of the differential calculus.β
- Abraham Robinson
In his landmark book Nonstandard Analysis (1966), from which the preceding epigraph was taken,
Abraham Robinson astonished the mathematical world by using tools from 20th-century logic to
construct the Holy Grail of Calculus: a perfectly rigorous way to make infinitesimals logically respectable!
Robinson used his newly vindicated infinitesimals (which joined the familiar real numbers to
produce an extended system of hyperreals) to construct an alternate foundation for calculus. His
infinitesimal-based, 20th-century foundation was every bit as solid as the limit-based, 19th-century
foundation, but the mathematical world, in the intervening years, had grown secular; its desire for the
Infinitesimal Grail had been nearly extinguished. Though duly impressed by Robinsonβs intellectual
achievement, mathematicians were largely unmoved by it, for the theory of limits (itself a century old by
that time) was not only fully rigorous, but also fully entrenched. The logical foundations of calculus had
long since ceased to be an active field of inquiry, and few mathematicians (who were busy, naturally,
exploring other problems) cared to revisit it. Their position was quite understandable: The theory of
limits, which they, their teachers, and their teachersβ teachers had all mastered long ago as
undergraduates, worked. From a strictly logical standpoint, it did not need to be replaced, and
mathematicians are no less inclined than laymen to hearken unto the old saw: If it ainβt broke, donβt fix
it. Thus they tended β and tend β to view Robinsonβs work as a remarkable curiosity. All mathematicians
know of Robinsonβs achievement. Few have studied it in detail.
O intended reader of this book, you are not a professional mathematician. You are a student in a
freshman-level class. You need not, at this point in your academic career, concern yourself with the full
details of calculusβs logical foundations, except to be reassured that these exist and are secure; you can
study them (in either the limit-based version or the infinitesimal-based version) in the appropriate books
or classes should you feel so inclined in the future. One does not take a course in Driverβs Education to
learn the principles of the internal combustion engine, and one does not take freshman calculus to learn
the subjectβs logical foundations. Rather, one takes a course such as this to learn the calculus itself β to
learn what it is, how to use it, and how to think in terms of it, for calculus is as much a way of thinking as
it is a collection of computational tricks. For you (and your teacher), the importance of Robinsonβs work
lies not in its formidable logical details, but rather in the retrospective blessing it bestows upon the
centuries-old tradition of infinitesimal thinking, a tradition that will help you understand how to think
about calculus β how to recognize when calculus is an appropriate tool for a problem, how to formulate
such problems in the language of calculus, how to understand why calculusβs computational tricks work
as they do. All of this becomes considerably easier when we allow ourselves the luxury of working with
infinitesimals. We need no longer, as in the 1950βs, blush to say βthe i-word.β In accordance with this
bookβs title, infinitesimals shall parade proudly through its pages, naked and unashamed.
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The World of Calculus: An Overview βIβm very good at integral and differential calculus,
I know the scientific names of beings animalculous.
In short in matters vegetable, animal, and mineral,
I am the very model of a modern major general.β
- Major General Stanley, in Gilbert and Sullivanβs Pirates of Penzance.
Calculus is traditionally divided into two branches, integral and differential calculus.
Integral calculus is about mentally decomposing something into infinitely
many infinitesimal pieces, and then re-integrating the pieces (i.e. summing
them up to reconstitute the whole). The spirit of integral calculus hovered
over our proof of the circleβs area formula, when we reimagined the circleβs
area as the sum of the areas of infinitely many infinitesimally thin triangles.
Integral calculus is thus a way of thinking β or perhaps a way of seeing. One
might, with βintegral calculus eyes,β view a sphere as a stack of infinitely
many infinitesimally thin discs, as suggested by the figure at right. Alternately,
one might imagine the sphere as a nested series of hollow, infinitesimally thin
spheres, resembling the nested layers of an onion.
Differential calculus is about rates of change. An objectβs speed, for example, is a rate of change
(the rate at which its distance from a fixed point changes in time). A bankβs interest rate is a rate of
change (the rate at which dollars in a bank account change in time). Chemical reactions have rates of
change (the rate at which iron rusts when left in water, for instance). The slope of a line is a rate of
change (the rate at which the line rises as one runs along it). Where there is life, there is change; where
there is change, there is calculus. Differential calculus is specifically concerned with rates of change on
an infinitesimal scale. Thus it is not concerned with how the temperature is changing over a period of
weeks, years, or centuries, but rather with how the temperature is changing at a given instant.
Naturally, the two branches of calculus work together: To understand large-scale global change, we
mentally disintegrate it into an infinite sequence of local instantaneous changes; we scrutinize these
infinitesimal changes with differential calculus, and then we re-integrate them with integral calculus, so
as to see the whole again with new eyes and new insights.
In this book, weβll begin with differential calculus, and then move on to integral calculus.
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Differential Calculus: The Key Geometric Idea
Differential calculus grows from a single idea: On an infinitesimal scale, curves are straight.
To see this, imagine zooming in on a point π lying on
a curve. As you do so, the part of the curve you can see
(an ever-shrinking βneighborhoodβ of π) becomes less
and less curvy. In an infinitesimally small neighborhood
of π, the curve coincides with (part of) a straight line.
We call this straight line the curveβs tangent at π·.
Thus, in the infinitesimal neighborhood of a point, a
curve and its tangent are indistinguishable. Outside of
this neighborhood, of course, the
curve and its tangent usually go
their separate ways, as happens
in the figure at left. Nonetheless,
simply by recognizing the βlocal
linearityβ of curves, we can answer seemingly tricky questions in geometry and
physics. Consider the following two examples, in which tangent lines make unexpected appearances. No
calculations are involved, just thinking in terms of infinitesimals.
Example 1. (An Illuminating Tangent on Optics)
Light reflects off of flat surfaces in a simple manner: Each
ray βbounces offβ the surface at the same angle at which it
initially struck it. (Or as a physicist would say, the rayβs angle
of reflection equals its angle of incidence.) What happens,
however, when a light ray hits a curved surface?
To answer this question, you must first grasp that when a ray reflects off of a flat surface, the
angles are determined by a mere infinitesimal bit of surface.
If, for instance, we were to erase most of the βsurfaceβ in the
previous figure, and then aim another light ray at the same
point (and from the same angle), the geometry of the
reflection obviously would not change one bit.
What happens when a light ray strikes a curve? Since the reflectionβs geometry is determined
locally, let us mentally visit an infinitesimal neighborhood of
the point of impact. There, the curve coincides with its
(straight) tangent, which means that our curvy conundrum
has simply vanished! The curve is straight, so the old rule
(angle of reflection = angle of incidence) still holds. Newly
enlightened, we zoom back out to our usual perspective,
extend the tangent line as in the figure at right, and measure
our reflection angles against it, knowing that when we do so the usual reflection rule will hold.
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Example 2. (So Long and Thanks for All the Fish)
If the Sun were to vanish, where would the Earth go? Newton taught
us that Earth is kept in its orbit by the Sunβs gravitational force. He also
taught us (following Galileo) that when no force acts on an object moving
in a straight line, it will continue moving in a straight line. But what about
an object suddenly freed from all forces that had been moving on a
curved path? Well, if we think infinitesimally, we recognize that during
any given instant β any infinitesimal interval of time β an object moving on a curved path is actually
moving on a straight path. Thus when the Sun vanishes, Earth will continue to move along the
straight path on which it was traveling in that same instant. That is, the Earth will move along the
tangent to its elliptical orbit.
Tangents, in short, are important. Let us pause for a few exercises.
Exercises.
1. Let π be a point on a straight line. Describe the tangent to the line at π.
2. Let π be a point on a circle. Describe the tangent to the circle at π. [Hint: Consider the diameter with π as an
endpoint. Because the circle is symmetric about this diameter, the tangent line at π must also be symmetric
about it. (Any βlopsidednessβ of the tangent would indicate an asymmetry in the circle, but, of course, there is
no such thing.)]
3. a) We normally measure angles where two straight lines cross. How would we define the measure of the angle
at which two curves cross? Explain why this definition is reasonable.
b) How large is the curved angle between a circle and a tangent to the circle? (cf. Euclid, Elements 3.16.)
4. If you stand in the open country in eastern Washington, the earth looks like a flat plane (hence βthe plains.β) Of
course, it isnβt flat; you are actually standing on a sphere. Explain why the Earth looks flat from that perspective,
and what this has to do with the key idea of differential calculus.
5. Some graphs lack tangents at certain points. Explain why the graph of π¦ = |π₯| lacks a tangent at its vertex. [Hint:
Reread the first few paragraphs of this section.] The gleaming calculus machine does what it was designed for
phenomenally well, but it was not built to handle corners. At corner points (which, fortunately, are rare on the
graphs of the most commonly encountered functions), differential calculus breaks down.
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Rates of Change
Once we recognize the local straightness of curves, it affects even the way we think about functions. On
an infinitesimal scale, any functionβs graph is a straight line, which, in turn, is the graph of a linear
function. Hence, on an infinitesimal scale, all functions are linear!*
This is excellent news, for linear functions are baby simple. They are
simple because the output of any linear function changes at a fixed rate,
which we call its slope. (βSlopeβ and βrate of changeβ are more or less
synonymous.) If, say, a car travels 60π‘ + 10 miles in π‘ hours, this
distance changes at the fixed rate of 60 miles per hour. Since English has
a word for the rate at which distance changes in time, we might as well
use it: The carβs speed is fixed at 60 miles per hour.
Few phenomena we wish to study are so obliging as to conform to a strict linear relationship. No
one actually drives in accordance with the linear function in the previous paragraph β at least not for
any appreciable length of time. Rather, a driverβs speed varies from moment to moment, even if, on
average, he drives 60 miles per hour. But if physical phenomena are rarely linear in a global sense, all
functions, even nasty ones, are locally linear, which is precisely why linear functions are so important:
Linear functions underlie all functions.
Accordingly, it makes sense to speak of a nonlinear functionβs local rate of change or slope. This
notion of a local (also called instantaneous) rate of change is familiar to every driver; when you glance
down at your speedometer, its reading of β63 mphβ indicates that your car is moving down the highway
at that particular rate at that particular instant. Even a few seconds later, your speed may differ. When a
policeman aims his speed gun at your car, he is measuring its instantaneous rate of change.
To sum up: Locally, the curved graph of a nonlinear function coincides with a straight tangent line.
The tangentβs slope is the functionβs local slope, its rate of change near
the point of tangency. With this insight, we may visually estimate a
nonlinear functionβs local rate of change. For example, consider the
figure at right, which is a more realistic nonlinear graph of a carβs distance
travelled (in miles) as a function of time (in hours). The slopes of the two
tangents shown in the figure measure the distance functionβs local rate of
change (i.e. the carβs speed) at two different times. The graph thus shows
us quite plainly that the car was moving faster after one hour than it was
after half an hour.
* Exceptions to this rule occur in infinitesimal neighborhoods of βcorner pointsβ of the sort described in exercise 5. For the sake
of readable (if slightly inexact) exposition, I shall continue to make universal statements about βall functions,β trusting the intelligent reader, who has been forewarned, always to bear in mind that calculus breaks down where no tangent line exists.
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The Derivative of a Function
We are finally ready to define the central object of differential calculus, the derivative of a function.
This idea is best understood through a few examples.
Example 1. If a function π is given by the graph at
right, then according to the derivativeβs definition,
πβ²(0) measures the graphβs slope (i.e. the slope of
its tangent) when π₯ = 0. Clearly, in an infinitesimal
neighborhood of (0, β1), this graph resembles a
line whose slope is approximately 1, so
πβ²(0) β 1.
Similarly, we see that πβ²(5) β 0, and πβ²(4) < 0.
Example 2. After π‘ hours of driving on a straight highway, Jehuβs car has travelled π(π‘) miles. By
definition of the derivative, πβ²(π‘) represents the rate at which this distance is changing at time π‘;
that is, it represents Jehuβs speed (in miles per hour) at time π‘.
If, for example, πβ²(1/2) = 112, then we know that Jehu was driving furiously (112 mph) at
the instant occurring exactly 30 minutes after he first drove onto the highway. If, three hours
later, traffic brought his car to a standstill, then we know that πβ²(7 2β ) = 0.
A graph of Jehuβs distance function π would have a slope of 112 when π‘ = 1/2, and a flat
tangent line when π‘ = 7/2.
Most people use βspeedβ and βvelocityβ synonymously, but these words have distinct meanings in
physics. To a physicist, βspeedβ is a magnitude (i.e. a positive number); βvelocityβ is a magnitude and a
direction. To distinguish two opposed directions (e.g. up/down, left/right, North/South) mathematically,
we use positive and negative numbers. For example, if we throw a ball upwards, and take up to be the
positive direction, then saying βthe ballβs velocity is βππ feet/secβ is equivalent to saying βthe ball is
descending at a speed of 50 feet/sec.β
In such contexts, we also use plus and minus to distinguish distance from a point (e.g. β3 feet awayβ)
from position relative to that point (e.g. β3 feet to the West.β) For example, if a ladybug paces back and
forth on a line, one point of which we call βthe origin,β one direction of which we deem βpositive,β her
position might be +8 inches at one moment, and β8 at another, depending on which side of the origin
she happens to be. (In both cases, her distance from the origin is 8 inches.)
The rate of change of position with respect to time is velocity. Consequently, the derivative of an
objectβs position function describes the objectβs velocity.
Definition. The derivative of a function is a new function whose output is the original functionβs
local slope (i.e. local rate of change) at the given input value.
If π is a function, its derivative is denoted πβ² (which we read βπ-primeβ.)
(There are other notations for a functionβs derivative, which youβll meet in due time.)
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Example 3. When Eris tosses a golden apple in the air, its height after π‘ seconds is β(π‘) meters.
Since β is the appleβs position function (in the up/down dimension), its derivative, ββ²(π‘), gives the
appleβs velocity (in m/s) after π‘ seconds.
Thus, if β(1) = 12 and ββ²(1) = 5, we know that after one second, the apple is 12 meters
high and is rising at 5 m/s. If β(2) = 12 and ββ²(2) = β5, then two seconds after leaving Erisβs
hand, the apple is again 12 meters high, but is descending at 5 m/s.
At the instant the apple reached its zenith (which clearly occurred sometime during its second
second of flight), its velocity was zero: It was neither ascending nor descending, but hanging.
Thus, if the appleβs maximum height occurred when π‘ = 3/2, weβd have ββ²(3 2β ) = 0.
Velocity is the rate at which position changes. Acceleration is the rate at which velocity changes. Hence,
we measure acceleration in units of velocity per unit time, such (π/π )/π (abbreviated, alas, as π/π 2.*)
Note the chain of three functions linked by derivatives: The derivative of an objectβs position function is
its velocity; the derivative of its velocity function is its acceleration.
Acceleration is thus the second derivative of position, where βsecond derivativeβ simply means βthe
derivative of the derivative.β The second derivative of a function π(π₯) is denoted, unsurprisingly, πβ²β²(π₯).
Example 4. If we neglect air resistance, any body in free fall (i.e. with no force but gravity acting
on it) near Earthβs surface accelerates downwards at a constant rate of 32 ft/s2. Consequently, if
π(π‘) describes the height (in feet) of such an object, and π‘ is measured in seconds, we
immediately know that the second derivative of π is a constant function: πβ²β²(π‘) = β32.
Position and velocity are rare β but vitally important β examples of functions whose derivatives have
special names. Even when such names do not exist, you can interpret any given derivative as the rate at
which its output changes with respect to its input.
Example 5. Suppose π(π‘) represents the volume (in π3) of beer in a large vat, where π‘ is the
number of hours past noon on a certain day. Throughout the day, beer leaves the vat (as people
drink it), but new beer is also poured in by Ninkasi, the Sumerian beer goddess.
Here, πβ²(π‘) represents the rate at which the volume of beer in the vat is changing at time π‘.
If, say, πβ²(5.5) = β0.5, then at 5: 30 pm, the vatβs volume is decreasing at half a cubic meter per
minute. (Not even Ninkasi can keep up with thirsty Sumerians after the 5 oβclock whistle.) If
π(6) = 0.0001 and πβ²(6) = 2, then at 6: 00, even though the vessel is almost empty, it is β at
that very instant β filling at a rate of 2 cubic meters per minute. All praise to Ninkasi!
Whatever the graph of π(π‘) looks like overall, it must pass through (6, 0.0001), where it
comes dangerously close to touching the π‘-axis (which would signify an empty vat), but at that
very point, it will exhibit a strong sign of recovery: an upward-thrusting slope of 2.
* Squared seconds are not physically meaningful units, hence the parenthetical sigh. One must, like Leopold Bloom, remember
what π/π 2 actually means: βThirtytwo feet per second, per second. Law of falling bodies: per second, per second. They all fall to the groundβ¦ Per second, per second. Per second for every second it means.β (James Joyce, Ulysses)
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Exercises. 6. Judging by the graph at rightβ¦
a) What is the approximate numerical value of πβ²(3)?
b) Arrange in numerical order: πβ²(1), πβ²(2), πβ²(3), πβ²(4), πβ²(6).
c) How many solutions does π(π₯) = 0 have in the interval [0,10]?
d) How many solutions does πβ²(π₯) = 0 have in the interval [0,10]?
e) Which integers in [0,10] satisfy the inequality π(π₯)πβ²(π₯) < 0?
f) True or false: πβ²(π) > 0. g) True or false: [π(5)]/3 > πβ²(7).
7. True or false: If π(π₯) = π₯2, π(π₯) = ln π₯, and β(π₯) = 1/π₯, thenβ¦
a) πβ²(0) = π(0). b) π(π₯) > 0 for all π₯ in πβs domain. c) πβ²(π₯) > 0 for all π₯ in πβs domain.
d) π(1) = πβ²(1). e) ββ²(π₯) < 0 for all π₯ in ββs domain. f) ββ²(β1) = ββ²(1)
g) πβ²(π₯) > πβ²(π₯) for all positive values of π₯.
8. Given the functions in the previous problem, arrange the following in numerical order: πβ²(5), πβ²(5), ββ²(5).
9. Consider the constant function π(π₯) = 5. What is πβ²(0)? What is πβ²(5)? What is πβ²(π₯)?
What can be said about the derivative of a constant function in general?
10. Consider the general linear function π(π₯) = ππ₯ + π. What is πβ²(π₯)?
11. The function β(π₯) = |π₯| is defined for all real values of π₯, but ββ²(π₯) has a slightly smaller domain. What is it?
[Hint: See exercise 5.] Also, thinking geometrically, write down a formula for ββ²(π₯).
12. If π(π₯) = β4 β π₯2, find the exact values of the following.
a) πβ²(0) b) πβ²(1) c) πβ²(β2) d) πβ²(β3)
[Hint: Recall exercise 2, and recall how the slopes of perpendicular lines are related.]
13. If π(π₯) = ββ16 β π₯2, find the values of the following.
a) πβ²(0) b) πβ²(2)
c) If we take values of π closer and closer to 4, what happens to the corresponding values of π(π)?
d) If we take values of π closer and closer to 4, what happens to the corresponding values of πβ²(π)?
e) Sketch graphs of π and πβ² on the same set of axes.
14. Rube Waddell throws a baseball at the full moon. Let π(π‘) be the ballβs height (in feet) π‘ seconds after it leaves
his hand. In terms of physicsβ¦
a) What does the quantity πβ²(2) represent? b) What does the solution to the equation πβ²(π‘) = 0 represent?
c) What is the formula for πβ²β²(π‘), and why is this so? d) If πβ²(π) < 0, then what is happening at time π?
15. Buffalo Bill goes ice skating. More particularly, he skates along a narrow frozen river running East/West, often
reversing his direction by executing beautifully precise 180Β° turns. If we take a cigar that he dropped on the ice
to be the origin, and we let East be the positive direction, then his position after π‘ minutes can be described by
the function π(π‘), where distances are measured in meters. In physical termsβ¦
a) What does πβ²(π‘) represent? b) What does |πβ²(π‘)| represent? c) What does |π(π‘)| represent?
d) Could there be a time π at which π(π) > 0, but πβ²(π) < 0? Explain.
e) If πβ²β²(π‘) = 0 for all π‘ in some interval (π, π), what is happening between π‘ = π and π‘ = π?
f) If πβ²(π‘) = 0 for all π‘ in some interval (π, π), what is happening between π‘ = π and π‘ = π?
g) Suppose πβ²(π‘) < 0 for all π‘ in some interval (π, π), that πβ²(π) = 0, and that πβ²(π‘) > 0 for all π‘ in some
interval (π, π). What happened when π‘ = π ?
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16. Physicists call the rate of change of acceleration the jerk. (Thus, the jerk is positionβs third derivative.)
a) If distance is measured in meters, and time in seconds, what are the units of the jerk?
b) What, if anything, can be said about a freely-falling objectβs jerk?
17. Suppose that the function π(β) gives the temperature on 11/1/2015 at 10:46 am (the time at which Iβm typing
these words) β meters above my house in Olympia, WA. Suppose further that the function πβ²(β) is strictly
negative (meaning that its value is negative for all heights β). What does this strictly negative derivative signify
physically?
18. Buzz Aldrin is walking clockwise around the rim of a perfectly circular crater on the moon, whose radius is 2
miles. Let π (π‘) be the distance (in miles, as measured along the craterβs rim) he has walked after π‘ minutes.
a) If π β²(π‘) = π/40 for all π‘ during his first lap, then how long will it take him to complete one lap?
b) If π β²β²(π‘) < 0 during his second lap, will Buzz be moving faster when he begins this lap or when he ends it?
19. A mysterious blob from outer space has volume π(π‘) after π‘ hours on Earth, where π is measured in cubic feet.
a) What does πβ²(5) represent physically, and in what units is it measured?
b) If the graph of π(π‘) shows that π is decreasing between π‘ = 24 and π‘ = 48, what do we know about πβ²(π‘)?
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A Gift From Leibniz: π -Notation
Now that you know what a derivative is, you are ready for the special notation introduced by Gottfried
Wilhelm Leibniz, one of the intellectual giants in the story of calculus.
You are no doubt familiar with the βdelta notationβ commonly used to describe the change in a
variable.* Recall in particular that we find a lineβs slope as follows: from the coordinates of any two of its
points, compute the βriseβ (βπ¦) and βrunβ (βπ₯); the slope is βthe rise over the run,β βπ¦/βπ₯.
In calculus, we use Leibnizβs analogous βπ-notationβ to
describe infinitesimal change. If π represents a magnitude
(length, temperature, or what have you), then the symbol
π π represents an infinitesimal change in π.
Since a curve is straight on an infinitesimal scale, the
slope of its tangent (i.e. its derivative) at any given point is
given by the ratio of infinitesimals π π/π π.
In fact, if π¦ = π(π₯), then ππ¦/ππ₯ itself is an alternate,
and excellent, notation for the derivative πβ²(π₯).
The two derivative notations peacefully coexist. Each has its advantages and disadvantages, and
those who use calculus employ whichever notation happens to be more convenient in a given situation.
Prime notationβs prime advantage is that it provides a natural place for the derivativeβs input. In the
figure above, when the input is π, the derivative of π is approximately 3/4. To express this in prime
notation we simply write πβ²(π) β 3/4. In contrast, to express this same fact in Leibniz notation (as his d-
notation is now generally called), we must supply a brief explanatory phrase:
ππ¦
ππ₯β 3/4 when π₯ = π.
The concluding phrase is essential, since ππ¦/ππ₯βs values vary.β
Leibniz notationβs many advantages will become ever clearer as you learn more calculus. Above all,
it emphasizes the derivativeβs meaning. An infinitesimal rise over an infinitesimal run β such as Leibniz
notation holds constantly before our faces β obviously represents a slope, a rate of change. Leibniz
notation is expressive. Prime notation is not. Why a prime and not, say, a dot?β‘ The prime is arbitrary.
Leibniz put a great deal of thought into his notation; as we proceed through the course, youβll see
instances where it does half your thinking for you. Be thankful for Leibnizβs gift. Not all mathematicians
are so thoughtful about the notation they introduce.
* Examples: If β, a sunflowerβs height in feet, changes from 3 to 7 over a period of time, then ββ = 4 feet.
If the temperature π drops from 9Β° to β11Β°, then βπ = β20Β°.
β The notation ππ¦
ππ₯|π is occasionally used, but is very awkward. I wonβt use it in this book.
β‘ Newton, in fact, did use a dot for derivatives. Newton and Leibniz are usually credited as the independent co-creators of
calculus, which oversimplifies history a great deal, but is still a good first approximation to the truth. A bitter priority dispute arose between these two men and their followers.
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Exercises. 20. If π¦ = π(π₯), rewrite the following statements in Leibniz notation: πβ²(2) = 3, πβ²(π) = π.
21. If π¦ = π(π₯), rewrite these statements in prime notation: ππ¦
ππ₯= 1 when π₯ = β2,
ππ¦
ππ₯= ln 5 when π₯ = β5.
22. If π(2) = 6 and πβ²(2) = 3 for a function π, then in an infinitesimal neighborhood around π₯ = 2, the functionβs
output increases by 3 units for each 1 unit of increase in the input. (Be sure you see why: Think geometrically.)
For most functions, the preceding statement about infinitesimal neighborhoods still holds, at least
approximately, for small real (i.e. not just infinitesimal) neighborhoods of π₯ = 2. Thus, increasing the input
variable from 2 to 2.001 should cause the output variable to increase from 6 to approximately 6.003.
a) Explain why the approximation described in the preceding paragraph should be reasonable (even though the
neighborhood isnβt infinitesimal), provided the graph of π isnβt too intensely curved near the point (2,6).
b) If π(5) = 8, and πβ²(5) = 2.2, approximate the value of π(5.002).
c) If π(0) = 2, and πβ²(0) = β1.5, approximate the value of π(0.003).
d) If π (π₯) = sin π₯ (where π₯ is measured in radians), weβll soon be able to prove that π β²(π) = β1. Assuming this
fact for now, approximate the value of sin(3.19). Then check your answer with a scientific calculator.
23. Expressed in Leibniz notation, the previous problemβs big idea becomes algebraically obvious. For instance, if,
when π₯ = 5, we have ππ¦/ππ₯ = 3.7, basic algebra yields π π = (π. π)π π when π = π. (Just clear the fraction!)
That is, the infinitesimal change in output, ππ¦, will be 3.7 times the infinitesimal change in input, ππ₯, in an
infinitesimal neighborhood of π₯ = 5. As discussed in the previous exercise, this result also gives us a very good
approximation in small real neighborhoods of π₯ = 5.
a) Suppose that ππ¦ ππ₯β = π when π₯ = 10. Rewrite this statement so that ππ¦ is a function of ππ₯, taking care
not to omit the qualifying phrase βwhen π₯ = 10β, and interpret this as in the preceding paragraph.
b) Suppose that ππ¦ ππ₯β = β2 when π₯ = 3. Rewrite this statement so that ππ¦ is a function of ππ₯, taking care
not to omit the appropriate qualifying phrase, and interpret the resulting statement.
c) Suppose that ππ¦ ππ₯β = β2 when π₯ = 3, and suppose further that π¦ = 7 when π₯ = 3.
Find approximate values for π¦ when π₯ = 3.05 and when π₯ = 2.96.
24. Unlike prime notation, Leibniz notation explicitly names the derivativeβs input variable. We often read ππ¦/ππ₯
aloud as βthe derivative of π¦ with respect to π₯.β Similarly, ππ§/ππ is the derivative of π§ with respect to π. This
helps us keep track of our thoughts in applications. If, for example, πΎ represents an objectβs kinetic energy (in
joules) at time π‘ (in seconds), then ππΎ/ππ‘ is the derivative of kinetic energy with respect to time, and is
therefore measured in joules per second.
a) If, in the preceding kinetic energy example, we know that when π‘ = 60, we have πΎ = 80 and ππΎ/ππ‘ = 12,
then roughly how much kinetic energy do we expect the object to have when π‘ = 60.5?
b) Suppose that π΄ represents an area that grows and shrinks over time. Use Leibniz notation to express the
following: After 5 minutes, the area is shrinking at a rate of 2 square meters per minute.
c) If π§ = π(π‘), rewrite πβ²(6) = 4 in Leibniz notation.
d) Let πΆ be the total cost in dollars of producing π widgets per year at a factory. To boost π (which stands for
βquantityβ) past certain values may require, say, purchasing more (or better) machinery, hiring more
employees, or making other sorts of changes that lead to sudden lurches in πΆ, the total cost. The derivative
ππΆ/ππ is known as the βmarginal cost function.β In what units would values of ππΆ/ππ be measured?
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25. Leibniz notationβs explicit reference to the derivativeβs input variable (see exercise 24) is especially useful when
βtheβ input variable can be viewed in multiple ways.
Consider, for instance, a conical vat. Suppose water pours into the (initially empty) vat
at a constant rate of 3 ft3/min. Let π be the volume of water in the cone, and let β be the
waterβs βheightβ, as indicated in the figure. Naturally, we can view π as a function of the
time, π‘, elapsed since the water began pouring in. However, we can also view π as a
function of β; if the waterβs height is known, then the volume of the water is, in principle,
determined β whether or not we can determine its numerical value ourselves.
Since π can be considered a function of π‘ or β, we can distinguish between two different derivatives of π:
ππ/ππ‘ and ππ/πβ. The former measures the rate at which the volume changes with respect to time. This, we
are told, is constant: ππ ππ‘β = 3 ft3/min. The latter, ππ/πβ, measures the rate at which the volume changes
with respect to the waterβs height. A little thought will convince you that this is not a constant rate of change.
a) Have the little thought mentioned in the previous sentence. Namely, to convince yourself that ππ/πβ is not
a constant function of β, imagine two different situations corresponding to different values of β. First, let β
be very small, so that there is hardly any water in the vat. If, in this case, we increase β by a tiny amount πβ,
consider the resulting change in volume, ππ. Draw a picture, and indicate what ππ represents geometrically.
(You need not calculate anything.) Second, let β be relatively large, so that the vat is, say, 3/4 full. Again,
imagine increasing β by the same tiny amount πβ. Draw another picture and think about what ππ represents
geometrically. Since applying the same little nudge to the input variable πβ yields different changes to the
output variable, ππ, the ratio ππ/πβ has a different value in each case. Hence, ππ/πβ is not constant.
b) Which is greater: ππ/πβ when β is small, or ππ/πβ when β is large?
c) If the cone were upside down, so that water poured into its vertex at a constant rate, which would be
greater: ππ/πβ when β is small, or ππ/πβ when β is large? Draw a picture.
d) If the water were pouring into a spherical tank of radius 10 feet, rank the following in numerical order:
ππ/πβ when β = 1, ππ/πβ when β = 5, ππ/πβ when β = 10, ππ/πβ when β = 17.
e) Give an example of a shape for a tank that would ensure that ππ/πβ is constant if ππ/ππ‘ is constant.
26. When π¦ = π(π₯), we can write ππ¦/ππ₯ in the form π(π(π₯))/ππ₯. Thus, for example, we may rewrite
If π¦ = tan π₯, then ππ¦
ππ₯= 4 when π₯ =
π
3,
in the following more concise form:
π(tanπ₯)
ππ₯= 4 when π₯ =
π
3.
Naturally, all the usual interpretations hold. Here, for example, the notation is telling us that in the infinitesimal
neighborhood of π₯ = π/3, the output value of tan π₯ increases by four units for each unit by which its input π₯ is
increased. Use this notation to rewrite the following statements.
a) If π¦ = ln π₯, then ππ¦
ππ₯= 5 when π₯ =
1
5. b) If π¦ = 3π₯3 + 1, then
ππ¦
ππ₯= 36 when π₯ = 2.
c) If π¦ = 2π₯, then ππ¦
ππ₯= ln 16 when π₯ = 2. d) If π¦ = β4π₯ + 2, then
ππ¦
ππ₯= β4.
e) In part d, why wasnβt it necessary to include a qualifying statement about an π₯-value? Think geometrically.
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An Infinitesimal Bit of an Infinitesimal Bit
βSo naturalists observe, a flea
Hath smaller fleas that on him prey;
And these have smaller fleas to bite 'em.
And so proceed ad infinitum.β
- Jonathan Swift, βOn Poetry: A Rhapsodyβ
We are ultimately interested in phenomena on a real scale. Infinitesimals (such as ππ, ππ€, or ππ§) can
help us understand real phenomena, so we use them, but we draw the line at infinitesimal bits of real
magnitudes. Infinitesimal bits of infinitesimals (βsecond-order infinitesimalsβ such as (ππ₯)2, (ππ¦)2, or
ππ’ β ππ£) are so insanely small that we simply disregard them as though they were zeros. For instance, if
we expand the binomial (π₯ + ππ₯)2 to obtain
(π₯ + ππ₯)2 = π₯2 + 2π₯(ππ₯) + (ππ₯)2,
we treat the second-order infinitesimal (ππ₯)2 as a zero, and thus write (π₯ + ππ₯)2 = π₯2 + 2π₯(ππ₯).
It may help to imagine, as we did on this chapterβs first page, that you have a βcalculus microscope.β
Looking through it, first-order infinitesimals are magnified to visibility, but second-order infinitesimals
remain undetectable. Could a more powerful microscope make sense of them? Perhaps, but we need
not concern ourselves with such questions here; we are interested in infinitesimals not for their own
sake, but for what they tell us about ordinary, real-scale phenomena. For that purpose, first-order
infinitesimals suffice.
Exercises. 27. Expand the following.
a) (π₯ β ππ₯)2 b) (π₯ + ππ₯)2 β π₯2 c) (π₯ + ππ₯)3 d) (π’ + ππ’)(π£ + ππ£)
28. If we increase a function πβs input from π₯ to π₯ + ππ₯, its output changes from π(π₯) to π(π₯ + ππ₯). Consequently,
the expression π (π(π)), which denotes the infinitesimal change in πβs value, is π(π + π π) β π(π).
[Thus, for example, π(π₯2) = (π₯ + ππ₯)2 β π₯2. And so, by exercise 27π, π(π₯2) = 2π₯ππ₯.]
Your problem: Show thatβ¦
a) π(π₯3) = 3π₯2ππ₯ b) π(3π₯2) = 6π₯ππ₯ c) π(ππ₯2 + ππ₯ + π) = (2ππ₯ + π)ππ₯.
29. Expand the following infinitesimal changes by writing them as differences.
a) π(π(π₯)) b) π(5π(π₯)) c) π(π(π₯) + π(π₯)) d) π(π(π₯)π(π₯)) e) π (π(π(π₯)))
30. On the graph of π¦ = π₯2, Let π be the point (3,9). Let π be a point, infinitesimally close to π, whose coordinates
are (3 + ππ₯, (3 + ππ₯)2). As we move from π to π, the infinitesimal change in π₯ is ππ₯.
a) Express ππ¦, the corresponding infinitesimal change in π¦, in terms of ππ₯.
b) Since π¦ = π₯2, we have ππ¦ = π(π₯2). Use this, and your result from part (a) to compute π(π₯2)
ππ₯ when π₯ = 3.
c) Use the ideas in this problem to find π(π₯2)
ππ₯ when π₯ = β1/2.
31. a) Show that
π’+ππ’
π£+ππ£ β π’
π£
ππ₯=π£ππ’ β π’ππ£
π£2ππ₯ b) Show that the right-hand side can be rewritten as
π£(ππ’
ππ₯) β π’(
ππ£
ππ₯)
π£2.
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The Derivative of π = ππ
Differential calculus teaches us nothing about linear functions; putting the graph of a linear function
under the calculus microscope reveals the same old line visible to the naked eye. In exercise 10, you
learned the entire story of linear functionsβ derivatives: Linear functions have constant slopes, so their
derivatives are constant functions.
Calculus exists to analyze nonlinear functions. Perhaps the simplest nonlinear function is π¦ = π₯2,
whose derivative β our first nontrivial example of a derivative β we will now compute. Note well: We are
not merely looking for this functionβs derivative at a particular point (say, ππ¦ ππ₯β when π₯ = 3), which is
a number (you found it in exercise 30); we seek the derivative ππ¦ ππ₯β itself, which is a function. Once we
have an explicit formula for ππ¦/ππ₯, weβll be able to evaluate it at any point we wish.
Problem. Find the derivative of the function π¦ = π₯2.
Solution. At any point π₯ in the functionβs
domain, the derivative measures ππ¦/ππ₯. To
compute this ratio, we observe that when π₯
is increased by an infinitesimal amount ππ₯
(which must obviously be exaggerated in the
figure!), the corresponding change in π¦ is
ππ¦ = (π₯ + ππ₯)2 β π₯2
= 2π₯(ππ₯).*
Thus, for any π₯ in the domain,
ππ¦
ππ₯=2π₯(ππ₯)
ππ₯= ππ.
Weβve just proved that the derivative of π₯2 is 2π₯. That is,
π(π₯2)
ππ₯= 2π₯.
We often state results of this type (βthe derivative of π΄ is π΅β) in the following alternate form:
π
ππ₯(π₯2) = 2π₯.
Here, we think of the symbol π /π π as an operator that turns a function into its derivative.β
Our first substantial project in differential calculus, which weβll begin after the next set of exercises,
will be to discover a method for rapidly finding the derivative of any polynomial whatsoever.
* See exercise 27π above. β Just as a function turns numbers into numbers, an operator turns functions into functions.
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Exercises.
32. Knowing π(π₯2)
ππ₯= 2π₯ in general, you should need only seconds to find
π(π₯2)
ππ₯ when π₯ = 4 in particular. Find it.
33. Find the equation of the line tangent to the graph of π¦ = π₯2 at point (β3,9).
34. a) Find the coordinates of the one point on the graph of π¦ = π₯2 at which the functionβs slope is exactly β5.
b) Find the equation of the tangent to the graph at the point you found in part (a).
35. Is there a non-horizontal tangent to π¦ = π₯2 that passes through (1 3β , 0)? If so, find the point of tangency.
36. Use the ideas in this section to find π(ππ₯2+ππ₯+π)
ππ₯.
37. a) Use the ideas in this section to find π(π₯3)
ππ₯. b) Use part (a) to find
π(π₯3)
ππ₯ when π₯ = β2, and when π₯ = π.
c) Find the equation of the line tangent to the graph of π¦ = π₯3 at point (1,1).
d) Does the tangent line from part (c) cross the graph again? If so, where? If not, how do you know?
[Hint: At some point, youβll need to solve a cubic equation. Even though you (presumably) do not know how
to solve cubics in general, you can solve this particular one because you already know one of its solutions.]
38. The notations π(π(π₯))
ππ₯ and
π
ππ₯(π(π₯)) are equivalent and are used interchangeably. To accustom yourself to
these notational dialects, rewrite the following expressions in the other form. (Yes, this exercise is trivial.)
a) π(π₯2)
ππ₯ b)
π(sinπ₯)
ππ₯ c)
π
ππ₯(ln π₯) d)
π
ππ₯(1
π₯)
Derivatives of Polynomials βHe is like a mere x. I do not mean x the kiss symbol but, as we allude in algebra terminology, to denote an unknown quantity.β
βWhat the hell this algebra's got to do with me, old feller?β
- All About H. Hatterr, G.V. Desani.
Let us revisit some basic algebra you learned on your motherβs knee.
How does one multiply algebraic expressions such as (π + π + π)(π + π + π)? Well, each term in
the first set of parentheses must βshake handsβ with each term in the second set. The sum of all such
βhandshakesβ (i.e. multiplications) is the product we seek.
For example, to multiply (π + π + π)(π + π + π), first π shakes hands with each term in the second
set (yielding ππ, ππ, and ππ), then π does (ππ, ππ, and ππ), and finally, π does (ππ, ππ, and ππ.) Thus,
(π + π + π)(π + π + π) = ππ + ππ + ππ + ππ + ππ + ππ + ππ + ππ + ππ.
Naturally, this works regardless of how many terms are in each parenthetical expression. Applied, for
instance, to the very simple product (π + π)(π + π), it produces the familiar βFOILβ expansion you
learned in your first algebra course.
If one wishes to multiply not two, but three expressions, then each βhandshakeβ must be a three-
way handshake, with one handshaker drawn from each of the three expressions. For example,
(π + π)(π + π)(π + π) = πππ + πππ + πππ + πππ + πππ + πππ + πππ + πππ.
Similarly, multiplying four expressions requires four-way handshakes; multiplying π expressions requires
π-way handshakes. Let us rest our hands and return to calculus.
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Problem. Find the derivative of the function π¦ = π₯π, where π is a whole number.
Solution. We begin by noting that
π(π₯π)
ππ₯=(π₯ + ππ₯)π β π₯π
ππ₯.
To proceed, observe that the binomial in the numerator is
(π₯ + ππ₯)π = (π₯ + ππ₯)(π₯ + ππ₯)β― (π₯ + ππ₯)β π times
.
To multiply this out, we must sum all possible π-way βhandshakesβ of the sort described above.
What will the final product be? Clearly, one of the π-way handshakes involves all π of the π₯βs.
Since their product is π₯π, the expansion of (π₯ + ππ₯)π will certainly contain the term ππ.
Among the other handshakes, some will involve (π β 1) of the π₯βs and a single ππ₯. In fact,
there will be exactly π handshakes of this sort: The first of these π handshakes involves the ππ₯
from the first parenthetical expression and the π₯βs from all the other groups, the second involves
the ππ₯ from the second parenthetical expression and the π₯βs from all the others, and so forth.
Each of these π handshakes yields an π₯πβ1ππ₯ term. Collecting these like terms together, we
conclude that the expansion of (π₯ + ππ₯)π contains the term πππβππ π.
Every other π-way handshake involves at least two ππ₯βs, and so will yield results containing
factors of ππ₯2, which means that these terms β these handshakes β can be disregarded! This
being so, the overall expansion we seek is simply (π₯ + ππ₯)π = π₯π + ππ₯πβ1ππ₯.
Consequently,
π(π₯π)
ππ₯=(π₯ + ππ₯)π β π₯π
ππ₯
=(π₯π + ππ₯πβ1ππ₯) β π₯π
ππ₯
= ππ₯πβ1 .
Functions of the form π¦ = π₯π are called βpower functions,β because they raise their inputs to some
fixed power. The result weβve just established is important enough to merit its own box.
This is a preliminary version inasmuch as we will eventually prove that the power rule holds for all real
values of π: not just whole number values.
Example. By the power rule, π
ππ₯(π₯7) = 7π₯6.
The Power Rule (Preliminary Version).
For all whole numbers π,
π
π π(ππ) = πππβπ.
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A quick note on language: We do not βderiveβ π₯7 to get its derivative 7π₯6. Many students make this
grammatical mistake. Donβt be one of them. To pass from π to πβ², we βtake the derivative of π.β We
donβt βderiveβ π.*
Over the next few chapters, youβll learn how to take many functionsβ derivatives. Along with
derivatives of specific functions (π₯π, sin π₯, ln π₯, etc.), youβll learn structural rules that let you take
derivatives of nasty functions built up from simple ones (such as π₯2 ln π₯ + 4 sin(5π₯).) The first structural
rules weβll need are the derivativeβs linearity properties.
Proof. To prove the first linearity property, we just follow our noses:
π
ππ₯(ππ(π₯)) =
π(ππ(π₯))
ππ₯=ππ(π₯+ππ₯)βππ(π₯)
ππ₯
= π (π(π₯+ππ₯)βπ(π₯)
ππ₯) = π (
π(π(π₯))
ππ₯) = π
π
ππ₯(π(π₯)).
Proving the second property is just as simple.
π
ππ₯(π(π₯) + π(π₯)) =
π(π(π₯)+π(π₯))
ππ₯=(π(π₯+ππ₯)+π(π₯+ππ₯)) β (π(π₯)+π(π₯))
ππ₯
=(π(π₯+ππ₯)βπ(π₯)) + (π(π₯+ππ₯)βπ(π₯))
ππ₯=π(π₯+ππ₯)βπ(π₯)
ππ₯+π(π₯+ππ₯)βπ(π₯)
ππ₯
=π(π(π₯))
ππ₯+π(π(π₯))
ππ₯=
π
ππ₯(π(π₯)) +
π
ππ₯(π(π₯)). β
If that proof gave you any trouble, please work through it again after revisiting exercises 28, 29, and 38.
The justification for each equals sign in the proof should be crystal clear to you.
Using the power rule and the linearity properties, we can find any polynomialβs derivative in a
matter of seconds, as the following example and exercises should convince you.
* One may also use the awkward verb differentiate here (e.g. βIf we differentiate π₯2, we obtain 2π₯.β) Since this sense of
βdifferentiateβ has nothing in common with the verbβs ordinary meaning, those who use it must take care to differentiate differentiate from differentiate. History explains this quirk: βDifferentiateβ derives from βdifferential,β a term Leibniz introduced for any expression of the form π(π(π₯)). Thus, to βdifferentiateβ π₯2 originally meant to turn π₯2 into its differential, π(π₯2) = 2π₯ππ₯. (Hence βdifferential calculus.β)
I generally prefer βtake the derivative ofβ to βdifferentiate.β The longer phrase is a bit clunky, but it conforms to a common mathematical pattern (βtake the reciprocal ofβ βtake the logarithm ofβ βtake the square root,β etc.)
Linearity Properties of the Derivative.
π. The derivative of a constant times a function is the constant times the functionβs derivative:
π
ππ₯(ππ(π₯)) = π
π
ππ₯(π(π₯)).
ππ. The derivative of a sum of functions is the sum of their derivatives:
π
ππ₯(π(π₯) + π(π₯)) =
π
ππ₯(π(π₯)) +
π
ππ₯(π(π₯)).
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Example. Find the derivative of π¦ = 5π₯6 + 3π₯4.
Solution. π
ππ₯(5π₯6 + 3π₯4) =
π
ππ₯(5π₯6) +
π
ππ₯(3π₯4) (by one of the linearity properties)
= 5π
ππ₯(π₯6) + 3
π
ππ₯(π₯4) (by the other linearity property)
= 5(6π₯5) + 3(4π₯3) (by the power rule.)
= 30π₯5 + 12π₯3 .
Exercises. 39. The power rule is geometrically obvious in the special cases when π = 1 or π = 0. Explain why.
40. If we combine the power rule and the first linearity property, we find that π
ππ₯(ππ₯π) is equal to what?
41. Use the result of exercise 40 to find the following derivatives in one step:
a) π
ππ₯(5π₯3) b)
π
ππ₯(β3π₯7) c)
π
ππ₯(π
2π₯2)
42. By combining the result of exercise 40 and the second linearity property, find the following derivatives.
a) π
ππ₯(3π₯3 + 2π₯2) b)
π
ππ₯(β10π₯5 +
1
4π₯3) c)
π
ππ₯(β2π₯ + β3)
43. Prove that the derivative of a difference of functions is the difference of their derivatives.
44. Use the result of the previous exercise to find the following derivatives.
a) π
ππ₯(5π₯3 β 2π₯4) b)
π
ππ₯(3π₯2 β 5) c)
π
ππ₯(β
2
5π₯10 β π)
45. Convince yourself that the derivative of a sum of three (or more) functions is the sum of their derivatives. Then
use this fact to compute the following, ideally writing down each derivative in a single step:
a) π
ππ₯(2π₯3 + 4π₯2 + 5π₯ + 1) b)
π
ππ₯(3
4π₯3 β 9π₯2 β β5π₯ + 2) c)
π
ππ₯(βπ₯6 + π₯3 β 4π₯2 + 3)
46. The derivative of a product of functions is not the product of their derivatives! Show, for example, that
π
ππ₯(π₯5π₯3) β (
π
ππ₯(π₯5)) (
π
ππ₯(π₯3)) .
47. Compute the derivative of π¦ = (π₯ + 1)4.
48. The derivative of a quotient of functions is not the quotient of their derivatives! Demonstrate this by providing
a counterexample, as you did in exercise 46.
49. If π¦ = 2π3, what is ππ¦/ππ₯?
50. What does the power rule tell us about π
ππ₯(3π₯)?
51. Using symbols other than π₯ and π¦ for a functionβs independent and dependent variables does not change the
formal rules for finding derivatives. With this in mind, find the derivatives of these functions:
a) π(π‘) = 2π‘3 β 3π‘2 + 5 b) π(π§) =1
4π§4 +
1
3π§3 +
1
2π§2 + π§ c) π΄(π) = ππ2
52. Recall that the graph of any quadratic function (of the form π¦ = ππ₯2 + ππ₯ + π) is a parabola whose axis of
symmetry is parallel to the π¦-axis. Clearly, such a graph has a horizontal tangent only at its vertex. This
observation yields a quick way to find the vertexβs π₯-coordinate: set the quadraticβs derivative equal to zero. Be
sure you understand this idea, then use it β together with the fact that a parabola of this sort opens up or
down according to whether its leading coefficient is positive or negative β to sketch graphs of the following
quadratics. Include the coordinates of each parabolaβs vertex and of any intersections with the axes.
a) π¦ = π₯2 + 3π₯ + 4 b) π(π₯) = β2π₯2 + 3π₯ β 4 c) π(π₯) = ππ₯2 + ππ₯ + β2
53. Find the equation of the line tangent to π¦ = π₯3 β 2π₯2 + 3π₯ + 1 at (1,3).
54. There is exactly one tangent to π¦ = π₯3 that passes through (0,2). Find the point of tangency.
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One Last Example
Your new ability to take polynomialsβ derivatives lets you solve otherwise intractable applied problems.
Example. Cliff Clifford stands at the edge of a cliff, 100 feet above the ocean. He hurls a rock
upwards. It climbs for a while, comes down, narrowly misses the cliffβs edge, and plunges into the
ocean below. If, π‘ seconds after it is thrown, the rockβs height (relative to the ocean) is given by
π (π‘) = β16π‘2 + 64π‘ + 100 ,
then answer the following questions.
a) Find the rockβs velocity when π‘ = 0.5 and when π‘ = 2.5.
b) What is the rockβs maximum height?
c) How fast is the rock moving at the instant when it hits the water?
Solution. Since π (π‘) gives the rockβs position (its height in feet) after π‘ seconds, its derivative,
π β²(π‘) = β32π‘ + 64, gives the rockβs velocity (in ft/sec) at time π‘.
In particular, half a second after it leaves Cliffβs hand, the rockβs velocity is π β²(0.5) = 48. That
is, it is moving upwards at 48 ft/sec at that instant. Similarly, after 2.5 seconds, the rockβs velocity
is π β²(2.5) = β16. Hence, the rock is moving downwards at 16 ft/sec at that instant.
Clearly, the rock will move upwards (i.e. have a positive velocity) for a little while and then
move downwards (i.e. have a negative velocity) thereafter. Its maximum height will therefore
occur at the instant when its velocity is zero β the moment when its velocity passes from positive
to negative. Solving the equation π β²(π‘) = 0, we find that this occurs when π‘ = 2. Thus, the
maximum height attained by the rock will be π (2) = 164 feet above the ocean.
The rock will hit the water when π (π‘) = 0. This is a quadratic equation; substituting its sole
positive solution, π‘ = 2 + β41/2 into our velocity function, we find that the rockβs velocity upon
impact is π β²(2 + β41/2) β β102.4 ft/sec.
As discussed earlier, velocityβs rate of change is acceleration. In the preceding example, the rockβs
position was given by π = β16π‘2 + 64π‘ + 100, from which we deduced its velocity function:
π£ =ππ
ππ‘= β32π‘ + 64 .
By taking the derivative of the velocity function, we can now determine the rockβs acceleration function:
π =ππ£
ππ‘= β32 ,
which confirms Galileoβs famous discovery: any object in free fall (i.e. with no force other than gravity
acting upon it) accelerates downwards at a constant rate of 32 feet per second per second.
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Exercises. 55. Suppose that a point is moving along a horizontal line, on which we will call right the positive direction. If the
pointβs position (relative to some fixed βoriginβ) after π‘ seconds is given by π = β5 + 4π‘ β 3π‘2, find the time(s)
at which the point is momentarily at rest (i.e. when its velocity is zero), the times when the point is moving to
the right, and the times when it is moving to the left.
56. Suppose two points are moving on the line from the previous problem, and their positions are given by
π 1(π‘) = π‘2 β 3π‘ and π 2(π‘) = β2π‘
2 + 5π‘.
a) Which point is initially moving faster?
b) When, if ever, will the two points have the same velocity?
c) When the clock starts, the two points occupy the same position, but they separate immediately thereafter.
Where and when will they next coincide? What will their velocities be then? Will they meet a third time?
d) True or false: one of the two points is perpetually accelerating and the other is perpetually decelerating.
57. Molly Bloom throws an object down from the rock of Gibraltar in such a manner that the distance (in meters) it
has fallen after π‘ seconds is given by the function π = 30π‘ + 4.9π‘2. How fast is the object moving after 5
seconds? What is the objectβs acceleration then (in m/s2)?
58. In exercise 51c, you used the power rule (plus a linearity property of the derivative)
to show that if π΄ = ππ2, then ππ΄/ππ = 2ππ. By thinking geometrically, youβll be
able to understand this striking result better. If a circleβs radius π increases by an
infinitesimal amount ππ (which is necessarily exaggerated in the figure), then its
area increases by an infinitesimal amount, ππ΄. In the figure, ππ΄ is the area of the
infinitesimally thin ring bounded by the two circles. Given that the outer circleβs
radius is (π + ππ), express ππ΄ in terms of π. Then divide by ππ, and verify that the
derivative ππ΄/ππ is indeed 2ππ.
59. Recall that a sphere of radius π has volume π = (4 3β )ππ3. The power rule tells us that ππ ππβ = 4ππ2, which
is β as the previous problem might lead you to expect β the sphereβs surface area. Verify this geometrically as
in the previous problem.
60. A square of side length π₯ has area π΄ = π₯2 and perimeter π = 4π₯. Contrary to what one
might expect from the previous two problems, the power rule shows at once that
ππ΄/ππ₯ β π. To understand this geometrically, consider the figure at right. As with all
such schematic depictions of infinitesimals, you must imagine the ππ₯βs as being
incomparably tinier than they appear. If π΄ represents the original squareβs area (before
the infinitesimal change to π₯), to what part of the figure does ππ΄ correspond? Compute
the value of ππ΄ and use it to verify geometrically that ππ΄/ππ₯ is indeed 2π₯, as given by
the power rule.
61. In the previous problem, if we had drawn the figure so that the original square
of side π₯ was centered in the enlarged square of side (π₯ + ππ₯), then ππ΄ would
still have been 2π₯ππ₯. Verify that this is so. The moral of the story is that how the
quantity π₯ changes by ππ₯ doesnβt matter; all that matters to the derivative is
that it does change by ππ₯. You might wish to further convince yourself of this by
playing with a third figure whose sides increase by (1 3β )ππ₯ at one end and
(2 3β )ππ₯ at the other.