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Zhenya Rock
Professor Michael Wilson
Math 161
4 May 2016
The Product Rule
Abstract.
This paper outlines the idea that the development of mathematical concepts is not a
smooth and error free process despite its appearance as such in textbooks, detailing Leibniz’s
incorrect first attempt at developing the product rule for finding the derivative of the product of
two functions as well as his subsequent correction.
When studying mathematics, it may appear that the development of the fundamental
mathematical concepts flowed naturally from one realization or discovery to the next over time,
as if there were no difficulties or uncertainties involved in the process. Textbooks generally
outline only the correct results and their proofs, omitting mention of any struggles that
mathematicians may have faced or errors that they may have committed in developing these
methods. Many students struggle in their first calculus class, finding the ideas presented to be
foreign and unusual in comparison to the algebra that they are accustomed to. However, incorrect
hypotheses are typical of the scientific process, and even great mathematicians have committed
prominent errors in their work at times. One such example would be Gottfried Wilhelm
Leibniz’s original speculation of what would be called the product rule, which is used in taking
the derivative of the product of two functions. At first, Leibniz believed that this derivative
would be equal to the product of the derivatives of the functions (Cirillo), which is incorrect.
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This is a common assumption of many Calculus I students when they are first presented with a
problem that involves taking the derivative of the product of two functions. This instance clearly
demonstrates the idea that our current understanding of mathematics did not develop smoothly,
as textbooks seem to imply.
In a typical calculus class, students first spend time developing an understanding of the
definition of a derivative, as shown below (Banchoff):
𝑑
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
(𝑓(𝑥)) = lim
ℎ→0
𝑑𝑥
ℎ
Once students adequately comprehend and are able to use this definition, they begin learning the
numerous basic derivative rules, beginning with the simplest ones, which are the rules associated
with finding the derivative of a constant function and that of a constant multiple of a function.
The derivative of a constant is always zero, and the derivative of the constant multiple of a
function is the constant multiple multiplied by the derivative of the function (Banchoff). The
next logical step is to introduce the summation and difference derivative formulae, where the
derivative of the sum of two functions equals the sum of the derivatives of each of the two
functions, and the derivative of the difference of two functions equals the difference of their
respective derivatives (Banchoff), as follows:
𝑑
𝑑
𝑑
(𝑓(𝑥) ± 𝑔(𝑥)) =
𝑓(𝑥) ±
(𝑔(𝑥))
𝑑𝑥
𝑑𝑥
𝑑𝑥
So far, the forming of these formulae is fairly intuitive, and their proofs are not difficult to
follow. For example, proving that the derivative of a summation of two functions is the
summation of the derivatives of the functions may be carried out by simply applying the
definition of the derivative and carrying out some basic algebra (Banchoff). As in Thomas
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Banchoff and Steven Miller’s page entitled Basic Derivative Rules, one must first use the
definition of the derivative, as follows:
𝑑
(𝑓(𝑥 + ℎ) + 𝑔(𝑥 + ℎ)) − (𝑓(𝑥) + 𝑔(𝑥))
(𝑓(𝑥) + 𝑔(𝑥)) = lim
ℎ→0
𝑑𝑥
ℎ
Next, the terms in the numerator should be altered using the distributive property and grouping
rules, as shown below:
(𝑓(𝑥 + ℎ) − 𝑓(𝑥)) + (𝑔(𝑥 + ℎ) − 𝑔(𝑥))
ℎ→0
ℎ
= lim
Then, this may be separated into two different limits:
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
+ lim
ℎ→0
ℎ→0
ℎ
ℎ
= lim
Finally, this may be converted to a summation of derivatives, yielding the desired result:
=
𝑑
𝑑
(𝑓(𝑥)) +
(𝑔(𝑥))
𝑑𝑥
𝑑𝑥
By the transitive property of equality, then:
𝑑
𝑑
𝑑
(𝑓(𝑥) + 𝑔(𝑥)) =
(𝑓(𝑥)) +
(𝑔(𝑥))
𝑑𝑥
𝑑𝑥
𝑑𝑥
The difference formula is proven with much the same process.
With this in mind, a student of calculus might infer that the product rule for the derivative
of two multiplied functions would be similar, as follows (Cirillo):
𝑑
𝑑
𝑑
(𝑓(𝑥) ∙ 𝑔(𝑥)) =
(𝑓(𝑥)) ∙
(𝑔(𝑥))
𝑑𝑥
𝑑𝑥
𝑑𝑥
Even though this seems to follow the same logic as the summation and difference formulae, it
would be incorrect. However, there are certain situations where this method would yield the
correct answer. Clearly, this would be so if either function were the zero function, or if both
functions were constants, as a constant multiplied by a constant would equal a constant, and
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therefore of the derivatives in the above equation would equal zero regardless (Cirillo). In
Michelle Cirillo’s textbook entitled Humanizing Calculus, she explains that this incorrect
product rule would work for any two functions of the following forms (Cirillo):
𝑓(𝑥) = 𝐶(𝑛 − 𝑥)−𝑛
𝑔(𝑥) = 𝑥 𝑛
For example, taking C equal to seven and n equal to two and using the quotient rule, we see that:
(2 − 𝑥)2 (14𝑥) − 7𝑥 2 ∙ 2(2 − 𝑥)(−1)
𝑑
𝑑
7𝑥 2
(𝑓(𝑥) ∙ 𝑔(𝑥)) =
(
)
=
(2 − 𝑥)4
𝑑𝑥
𝑑𝑥 (2 − 𝑥)2
This solution simplifies to:
=−
28𝑥
(−2 + 𝑥)3
Then, taking the product of the derivatives of each function:
𝑑
𝑑
28𝑥
(𝑓(𝑥)) ∙
(𝑔(𝑥)) = (−14(2 − 𝑥)−3 (−1)) ∙ 2𝑥 = −
𝑑𝑥
𝑑𝑥
(−2 + 𝑥)3
Therefore, by the transitive property of equality this incorrect product rule holds for this case.
However, for any other case, this does not work. For example, taking the functions:
𝑓(𝑥) = 𝑥 2
(1)
𝑔(𝑥) = 4𝑥
(2)
The derivative of the product of the functions is as follows:
𝑑
𝑑
(4𝑥 3 ) = 12𝑥 2
(𝑓(𝑥) ∙ 𝑔(𝑥)) =
𝑑𝑥
𝑑𝑥
(3)
However, this does not equal the product of the derivatives:
𝑑
𝑑
(𝑓(𝑥)) ∙
(𝑔(𝑥)) = 2𝑥 ∙ 4 = 8𝑥
𝑑𝑥
𝑑𝑥
(4)
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Despite the fact that there are infinitely many instances where the derivative of the
product of two functions would in fact equal the product of the derivatives of the two functions,
there are also infinitely many scenarios where it would not (Cirillo). Given that this inference
does not work for every possible situation, it cannot be considered a rule. As obvious as it may
seem to a seasoned mathematician or upper level student of mathematics now that this would be
an incorrect assumption, people new to the concept may not expect as such. In fact, Leibniz
himself originally believed that this would be the correct formula to use when he began working
on derivatives in his manuscripts (Cirillo).
After realizing his mistake, Leibniz determined the correct formula to use in order to find
the derivative of the product of two functions, which is as follows (Cirillo):
𝑑
𝑑
𝑑
(𝑓(𝑥) ∙ 𝑔(𝑥)) = 𝑓(𝑥) ∙
(𝑔(𝑥)) +
(𝑓(𝑥)) ∙ 𝑔(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
Once he had figured this out, Leibniz proceeded to provide a proof that would show how this
new formula would work to find the derivative of the product of any two equations. This proof is
extremely similar to the one developed by Leonhard Euler, and uses the idea of the infinitesimal,
which is the term used to describe a value that is less than any given positive quantity and that
will not change the value of a number if added to it (Ferzola). One main fact used in this proof is
that infinitesimals of order higher than one should be considered negligible, meaning that an
infinitesimal plus a higher order of the same infinitesimal would equal the original first order
infinitesimal (Ferzola). Euler’s reasoning for this is as follows (Ferzola):
𝑑𝑥 + (𝑑𝑥)2
= 1 + 𝑑𝑥 = 1
𝑑𝑥
The fact that this quotient is equal to one implies that the values in the denominator and
numerator are equal, showing that the square of dx must be equal to zero.
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From the textbook entitled Euler and Differentials by Anthony Ferzola, the details of
Euler’s proof of the product rule that uses infinitesimals are the following, where f and g are
functions and df and dg denote infinitesimals (Ferzola):
𝑑(𝑓 ∙ 𝑔)) = (𝑓 + 𝑑𝑓)(𝑔 + 𝑑𝑔) − (𝑓 ∙ 𝑔)
= (𝑓 ∙ 𝑑𝑔) + (𝑑𝑓 ∙ 𝑔) + (𝑑𝑓 ∙ 𝑑𝑔)
= (𝑓 ∙ 𝑑𝑔) + (𝑑𝑓 ∙ 𝑔)
[𝑇ℎ𝑒 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑅𝑢𝑙𝑒]
The logic allowing the movement from the second line to the third employs the knowledge of the
fact that higher orders of infinitesimals are negligible (Ferzola). This lets us ignore the third term
of the second line, as seen above, eventually resulting in the proposed product rule, written out in
the third line.
The more modern proof uses the definition of the derivative rather than infinitesimals to
prove that the derivative of the product of two functions is actually equal to the sum of the first
function times the derivative of the second and the second function times the derivative of the
first (Banchoff). The first step of this proof is to use the definition of the derivative to rewrite the
derivative of the product of the two functions (Banchoff):
𝑑
𝑓(𝑥 + ℎ)𝑔(𝑥 + ℎ) − 𝑓(𝑥)𝑔(𝑥)
(𝑓(𝑥)𝑔(𝑥)) = lim
ℎ→0
𝑑𝑥
ℎ
Then, one may add and subtract f(x+h)g(x), as the sum of these two values is zero, and therefore
they have no effect on the end value of the function (Banchoff):
𝑓(𝑥 + ℎ)𝑔(𝑥 + ℎ) + 𝑓(𝑥 + ℎ)𝑔(𝑥) − 𝑓(𝑥 + ℎ)𝑔(𝑥) − 𝑓(𝑥)𝑔(𝑥)
ℎ→0
ℎ
= lim
The terms in the numerator can be rearranged and factored into (Banchoff):
= lim (𝑓(𝑥 + ℎ)
ℎ→0
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
+ 𝑔(𝑥)
)
ℎ
ℎ
When the value of this limit is evaluated, the final result is that (Bandchoff):
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= 𝑓(𝑥) ∙
𝑑
𝑑
(𝑔(𝑥)) +
(𝑓(𝑥)) ∙ 𝑔(𝑥)
𝑑𝑥
𝑑𝑥
This is the desired answer, proving the product rule. Therefore, this should give the correct result
for the derivative of the two functions previously mentioned, (1) and (2), which was not the case
with the originally assumed product rule formula. This correct product rule formula should then
produce the same value as (3). Employing the product rule, we see that:
𝑓(𝑥) ∙
𝑑
𝑑
(𝑔(𝑥)) +
(𝑓(𝑥)) ∙ 𝑔(𝑥) = 𝑥 2 (4) + 2𝑥(4𝑥) = 12𝑥 2
𝑑𝑥
𝑑𝑥
This does indeed match (3).
One may be inclined to question why the incorrect formula works in any case at all. Why
only sometimes? When would one be able to calculate the derivative of the product of two
functions with either method and produce the same result? The answer to this question would be
that there are particular pairs of functions where the following is true:
𝑑
𝑑
𝑑
𝑑
(𝑓(𝑥)) (𝑔(𝑥)) = 𝑓(𝑥) ∙
(𝑔(𝑥)) +
(𝑓(𝑥)) ∙ 𝑔(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
Any such equation pair would yield the correct result despite the fact that the wrong method was
used. Clearly, if either function is zero, then its derivative is also zero, and therefore the left and
right side of the equation are equal (Cirillo). In addition, both sides of the equation also equal
zero if both of the given functions are constants. This is also true for any equation pairs that fit
the general form from Michelle Cirillo’s textbook entitled Humanizing Calculus (Cirillo).
Although Leibniz had thought that he had proven his original, incorrect formula, he notes
in his manuscript that he was actually incorrect, even though he had “stated that [his original
assumption] was the case, and it appeared to be proved” (Cirillo 25) and that it was “a difficult
point” (Cirillo 25). This is one example of when the path of mathematical development became
turbulent. This same misconception arose in the original attempts to devise a rule to take the
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derivative of the quotient of two functions. The correct formulae for these derivatives are not
generally intuitive, despite how a math textbook might make them out to be. The development of
mathematics is often slow and rigorous, requiring hours of thought and logical reasoning to
devise new strategies and ways of thinking.
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Works Cited
Banchoff, Thomas, and Steven Miller. "Basic Derivative Rules." Basic Derivative Rules.
Brown University, 06 Aug. 2008. Web. 01 May 2016.
<http://www.math.brown.edu/UTRA/derivrules.html>.
Cirillo, Michelle. “Humanizing Calculus”. The Mathematics Teacher 101.1 (2007): 23–
27. Web...
Ferzola, Anthony P.. “Euler and Differentials”. The College Mathematics Journal 25.2
(1994): 102–111. Web...