Mathematics in Context: Calculus Overview
Newton calculated the fluxion (today, called the derivative) of a quantity moving (i.e., “in flux”)
according to the equation y = x2 by first finding the slope of the secant line between the two points
(x, x2 ) and (x + o, (x + o)2 ) as shown in the diagram below.
To get the slope of the tangent line, Newton
thought of the point x + o, as being an “infinitesimal” distance away from the point x.
Bishop Berkeley was uncomfortable with such
vague language. He was especially upset when
mathematicians, who also used such language,
criticized people of faith for being imprecise
when they spoke about God. Berkeley thought
the logic mathematicians used in “expunging”
terms that had the symbol o in them was suspect. The following example illustrates the use
of such logic.
(x + o, (x + o)2 )
The secant line
(whose slope is 2x + o)
Rise = (x + o)2 − x2
The tangent line
(whose slope is 2x)
(x, x2 )
Run = (x + o) − x
x
x+o
The slope of the secant line between the points (x, x2 ) and (x + o, (x + o)2 ) is rise
run , which equals
(x + o)2 − x2
(x2 + 2xo + x2 ) − x2
=
(x + o) − x
o
2
2xo + o
=
o
o(2x + o)
=
o
= 2x + o.
Newton reasoned that the slope of the tangent line was equal to 2x. He arrived at this value by
“expunging” the last o in the above equation, as it was “infinitesimally” small. Berkeley had strong,
and valid, objections to this approach. Today mathematicians justify their calculations with the
notion of a limit. Rather than expunging the o, the question is asked, “Is there some number to
which the quantity 2x + o can be made arbitrarily close by taking o to be sufficiently small?” If
there is such a number, it is called the limit of the expression 2x + o as o approaches zero. The
fluxion is then defined to be this limit. For the expression 2x + o, the limit is simply 2x, and we
write lim (2x + o) = 2x. Because the symbol o can be confused with the number zero, frequently a
o→0
different symbol, such as h, is used, and we write lim (2x + h) = 2x. The last equation is read as,
h→0
“The limit as h approaches zero of the quantity 2x + h equals 2x.”
Isaac Newton and Gottfried Leibniz are considered to be the founders of Calculus. Leibniz published
his ideas in October 1684 (Acta eruditorum his Nova methodus pro maximis et minimis, itemque
tangentibus, quae nec fractas nec irrationales quantitates moratur, et singulare pro illis calculi
genus). Newton did not publish his ideas until 1687 (Methodus fluxionum et serierum infinitarum
and the De quadratura curvarum), although these pieces were written, respectively, in 1671 and in
1676. Unfortunately, there was quite a bit of jealousy between the students of Newton and Leibniz,
each claiming their teacher founded Calculus first. Their approaches are different, however, which
makes plausible the belief that they developed their ideas independently.
The term mathematicians use to denote the slope of the tangent line is the derivative, as it is
dy
derived from a given equation. The notation they often use to designate a derivative is dx
, or
d
d
(y).
It
is
important
to
understand
that
the
symbol
is
not
a
fraction.
It
is
just
notation
that
dx
dx
stands for “the derivative of” For example, if y = x2 then the slope of the tangent line is 2x, so
d
d
mathematicians write dx
(y) = 2x, or dx
(x2 ) = 2x. This notation is due to Leibniz. It is very
convenient for more advanced calculations, which we will not get into during this course.
With y = 2x2 − 3x + 5, the following illustrates the calculation of
d
dx (y)
using limits.
d 2
d
(y) =
(x − 3x + 5)
dx
dx
[2(x + h)2 − 3(x + h) + 5] − [2x2 − 3x + 5]
= lim
h→0
(x + h) − x
[2(x2 + 2xh + h2 ) − 3(x + h) + 5] − [2x2 − 3x + 5]
= lim
h→0
h
2
2
[2x + 4xh + 2h ) − (3x + 3h) + 5] − [2x2 − 3x + 5]
= lim
h→0
h
2
2
2x + 4xh + 2h − 3x − 3h + 5 − 2x2 + 3x − 5
= lim
h→0
h
4xh + 2h2 − 3h
= lim
h→0
h
h(4x + 2h − 3)
= lim
h→0
h
= lim 4x + 2h − 3
h→0
= 4x − 3.
Page 2
The limit process given on the previous page is somewhat tedious, so mathematicians have developed rules for calculating derivatives.
d
Rule 1: If c is any number and we have a line given by the equation y = c, then dx
(y) =
In other words, the line that is tangent to a horizontal line has zero for its slope.
d
dx (c)
= 0.
d
(xn ) = nxn−1 .
Rule 2: If n is any number (even a fraction, even a negative number) then dx
d
Note: On the previous page we saw that, indeed, dx (x2 ) = 2x2−1 = 2x1 = 2x.
Rule 3: The derivative of a sum (or difference) is the sum (or difference) of the derivatives.
d
d
Example: We have shown in class that dx
(x2 ) = 2x, and Rule 2 gives dx
(x3 ) = 3x3 (you will also
d
d
show for homework that dx
(x3 ) = 3x3 using the limit process). It follows, then, that dx
(x2 + x3 ) =
d
d
d
d
d
2
3
2
2
3
2
3
2
dx (x ) + dx (x ) = 2x + 3x . Likewise, it follows that dx (x − x ) = dx (x ) − dx (x ) = 2x − 3x .
d
d
Rule 4: If c is a constant, then dx
(cxn ) = c dx
(xn ). In other words, the derivative of a constant
times a function is the constant times the derivative of the function.
d
d
(5x2 ) = 5 dx
(x2 ) = 5(2x) = 10x.
Example: dx
Let’s compute the derivative of y = 2x2 − 3x + 5 again, but this time using derivative rules.
d
dx (y)
=
d
2
dx (2x
− 3x + 5)
=
d
2
dx (2x
− 3x1 + 5)
=
d
2
dx (2x )
−
d
1
dx (3x )
+
=
d
2
dx (2x )
−
d
1
dx (3x )
+0
=
d
d
(x2 ) − 3 dx
(x1 )
2 dx
(Rule 4)
=
2(2x2−1 ) − 3(1x1−1 )
(Rule 2)
=
2(2x) − 3(x0 )
(re-writing)
=
4x − 3(1)
(re-writing, with x0 = 1)
=
4x − 3
Q.E.D.
(re-writing the expression)
d
dx (5)
(Rule 3)
(Rule 1)
Page 3
Assignment
1. Use the limit approach to obtain derivatives of
a. y = x2 + 7x;
b. y = x3 .
2. Use derivative rules to find derivatives for the following.
a. y = x3 + x2 + 7x.
b. y = x2 + 3.
c. y = 7x2 − 9x + 10.
√
d. y = x + x3 − x12 .
Page 4