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paper
Trig
Derivatives found without the old Calculus
now
including a disproof the chain rule
by Miles
Mathis
First
written October 2006, extended greatly June 2015
Be
sure to read the addendum, where I show the current proof for the
derivative of (sinx)^{2
}is
fudged. I show that the mainstream long ago pushed the proof via
a misuse of the chain rule.
Since I first
published my very long
paper on the calculus several years ago, I have gotten many
angry emails like this one:
You
are wrong! Mathematics is a science about numbers. Graphs, plots
are for illustration—you can prove nothing from them.
y
= sin x
y'
= cos x
How
do you prove that by your method?
Yuri
Even
my mother, who is a professional mathematician, has failed to see
how I can incorporate my table into an analysis of all functions.
She never got angry, but she has used my silence as proof that my
method has limited use.
Now, I said
in a footnote to that paper that my method applied to all of
calculus and all functions, not just differentials or
polynomials. It applies to trig functions, logarithms, integrals,
and so on. I assumed that anyone who understood my argument would
see that immediately. I didn't even bother to write a followup
paper on integration, it seemed so clear to me that anyone could
just read up the tables instead of down. I was busy with other
important problems and decided to let that paper hang, along with
any paper that specifically addressed trig functions. Frankly I
had hoped that someone might come along and see my point, and
that they would do the dirty work of advancing my theory into
these other alleys. Once I have solved a problem, I tend to get
bored, and stating the obvious does not really inspire me to
write.
However, I
now see, years later, that I was mistaken in assuming that my
initial paper would penetrate into the mathematical community. It
has been turned down for publication in all the top forums, for
what I think are political reasons. So I have recently gone back
and simplified my argument, selfpublishing a shorter
and simpler paper, argued in what I consider to be an
extremely transparent manner and language. I hope that this paper
may eventually make some headway in the mainstream, even if I
continue to be blocked by the higherups.
Beyond that,
I have decided to publicly solve Yuri's trig problem for him,
knowing full well that it won't be the further miracle anyone
needs. No matter what I do or how I do it, I now expect most of
the status
quo to find
a way to dismiss it out of hand. They weren't bothered by the
fact that the current equation has been hanging from skyhooks for
350 years, and so they won't be impressed to see the equation
finally grounded. Anyone who studies my table and doesn't undergo
an epiphany is someone who is pretty much unreachable, and
solving this trig problem with the table won't reach them either.
But here goes.
So, Yuri,
watch closely, my friend. I will do it so quickly and so easily,
you will no doubt think it is nothing. I will show you how to do
it without limits, without going to zero, without infinite
series, and without the current derivation of the calculus. I
will do it using only my table of exponents and the constant
differential.
We
start with this equation, which is one of the defining trig
equations:
y
= sinx
= ±√(1  cos^{2}x)
Notice that
we are still dealing here with exponents. The cosine is squared
and that is the important fact here, not the fact that we are
dealing with trig functions. From a rate of change perspective,
the trig function is meaningless. A sine or cosine is just a
number, like any other. It is written as function of an angle,
but that does not affect the rateofchange math at all. The
cosine of x
is a single variable, and we
could rewrite it as b
if we wanted to, to simplify
the variable for the rate of change math. Likewise, we could
rewrite sin x
as a,
if we desire. All we have to do is make sure we don't confuse
sine and cosine, since they vary in different ways, but we can
mark them anyway we want.
Let a
= sin x
b
= cos x
Therefore, we
could rewrite the equation as
y
= a
= ±√(1 –
b^{2})
Square both
sides
y^{2}
= 1 – b^{2}
Since sine
and cosine are codependent, we can differentiate either side, or
both sides, starting with either side we like.
Let z
= 1 – b^{2}
z
= y^{2}
Δz
= z'
= 2y
(from my table of integer
exponents)
Δ(1 
b^{2})
= 2y
Now switch
sides and differentiate again
2y
= Δ(1 – b^{2})
2Δy
= 2y'
= 2b
(once again, straight from the
table of exponents)
y'
= b
= cos x
[For a
slightly more expanded proof in the same vein, you may wish to
consult my new paper
on the derivative of sin5x.
It may clarify some of the steps above for some readers.]
You will say
that I just followed normal procedure (a kind of chain rule), but
I didn't, since whenever I use the equation nz^{n1}
= Δz^{n}
I pull it from my
table of exponents and constant differentials, not from
current sources, which I have shown are all faulty. I prove this
equation using a constant differential, not a diminishing
differential or a method using limits. My table shows that with
the exponent 2, you only have to go to a third subchange in the
rate of change chart in order to find a straight line, or a
constant rate of change. This means that you aren't anywhere near
zero, and aren't anywhere near an infinite series of any kind.
You are two steps below the given rate of change for this problem
(which is an acceleration or its pure math equivalent) and two
steps is two steps, not an infinite number of steps. In any rate
of change problem, we simply aren't dealing with infinite series,
points, or limits. We are dealing with subchanges, and we are
seeking a line of constant differentials. Not a point, a line.
This is why my method is so important.
It does not
matter in this problem that the curve was created by sine or
cosine. The way the curve was created does not concern us in
calculus. All we need is at least one dependence. If we have that
dependence then we can use the definition of exponent and integer
to create the table, and that table will straighten our curve out
in a definite and finite number of steps—the number of
steps being absolutely determined by the exponent itself.
An infinite
series is only created by an infinite exponent. But an exponent
signifies a change, and a change requires time, so that an
infinite exponent would imply infinite time. We do not need to
solve equations concerning infinite time, not in physics and not
in mathematics. Therefore we have no need of infinite series in
rate of change problems.
Now to briefly answer Yuri's
other concern (from the first paragraph). If you will remember,
he ridiculed me for studying graphs and realworld situations in
trying to understand math. Hopefully you know see that my
analysis of the Cartesian graph (in my long paper) was necessary
to discover the problems within the calculus, and that therefore
this analysis certainly transcends "illustration". I
never claimed that calculus was all about graphs, or implied that
the graph was the central feature of either calculus or of my
argument. But I would never have discovered what I did without an
indepth analysis of the graph and the way the curve is created
there, and I could never fully explain my method without using
the graph to make it clear. Trying to understand the calculus
without looking at graphs and real motions would be like trying
to understand trigonometry without looking at triangles.
[Addendum,
June 8, 2015: I just got a complaint from a person claiming to be
a highschool student, accusing me of an error here. I don't
believe the complaint was really from a highschool student, since
she seemed entirely too sure of herself ( and nasty) for someone
that age. But I will address the complaint nonetheless. The
complaint was that I seem to be implying that the derivative of
(sinx)^{2}
is 2sinx,
when of course it isn't. In an intermediary step, I differentiate
y^{2}
to find 2y, and y^{2}
is supposed to be equal to
sinx.
Although the mainstream currently does exactly
the same thing in the first
part of the chain rule (see below), she told me I have broken a
rule of math. I tried to explain to her the difference between
differentiating and finding a derivative, but she wasn't
listening. She was so sure I was wrong, I have to wonder why she
wrote me to ask for my response. She didn't want a response, so
why write? But I will see if anyone else out there can comprehend
the difference. I expect that few will, since the difference is
not taught in any school I know of.
To start
with, I send you to the current and mainstream proof for the
derivative of (sinx)^{2}.
To find this derivative, one normally uses the chain rule. To
apply the chain rule, you are instructed to first “take the
derivative” in the disallowed way. In other words, you just
drop the 2 down in front to find 2sinx
(which I was told was
disallowed above, in my proof). Next, you find the derivative
inside
the parentheses. Since the
derivative of sinx
is known to be cosx,
we multiply that by what we already have, obtaining the final
result 2sinxcosx.
Curious, isn't it, that the mainstream can use multiple steps in
its chain rule, but I can't use multiple steps in my own method?
What is stopping me from making the same complaint against this
proof that was made against mine above? Why can't I just pause
the mainstream proof after step one and complain that the
derivative of (sinx)^{2
}isn't
2sinx?
To avoid that complaint, the mainstream normally (or often)
doesn't
call that first step taking a derivative. Although it is
exactly the same manipulation, they just call it the first step
in the chain rule. Because they have given it a different title,
you aren't expected to notice it is exactly the same
manipulation.
But back to
my proof of the derivative of sinx.
Notice that to “find a derivative” here, we have to
“differentiate” twice, in both directions. That
should seem odd, since normally when we have a function to the
power of two, we only need to differentiate once
to find a derivative. We only
have to look at one rate of change, right? Well, you aren't
taught it that way, but that is what is happening. But in this
functional equation we are looking at, we have both
sine and cosine, and both are
to the power of 2. We don't just have one function, we have two
interdependent or codependent functions, and both are squared.
And that is why we have to look at rates of change on both sides.
Now,
admittedly, that is not the way these problems are normally
solved. But I am not claiming to solve these problems in the
normal ways, am I? I am claiming to solve them in a new way. So
complaining that I do not follow old rules is sort of missing the
point. I am showing that the old rules are sometimes wrong and
sometimes unnecessary. Yes, sometimes they are necessary,
especially if you don't know what is going on with the rates of
change. In that case, you are best to stick to the old rules. But
if you know the difference between slopes, derivatives,
differentiation, and rates of change, you can simplify the math
considerably.
To see this
in another way, notice that I never claim that the derivative of
(sinx)^{2}
is 2sinx.
I only claim that you can differentiate
y^{2}
to find 2y, even while y =
sinx,
provided that
do not stop there.
You must also differentiate in the other direction, because if
you don't you will not have related the two rates of change to
one another at the same time.
In other
words, I am not claiming that the derivative of (sinx)^{2}
is 2sinx.
I am claiming that the rate of change of (sinx)^{2},
taken alone
or in isolation,
can be written as 2sinx.
But since sinx
is, in fact, never found in
isolation, to find a derivative you have to keep going. You have
to find the rate of change of cosine simultaneously, you see.
Which is why I differentiate in both directions.
What my
critic is doing is stopping the proof between the two
differentiations, and saying that because my first
differentiation didn't achieve the known derivative, I have
cheated. But that is to completely miss my point. Of
course the
first differentiation wouldn't find any known derivative, since
it is only half a manipulation or half a relationship.
Let me
continue. My critic might say, “Well, if that is so, why
can't you just insert (sinx)^{2}
into the basic derivative
equation to find the real derivative? Why exactly is it allowed
in this interior differentiation, but not allowed in that direct
way?” Because if we are seeking the
derivative of
(sinx)^{2},
we aren't just seeking the rate
of change of
sinx
by itself. We can only drop
the 2 down like that in cases where we are dealing with one rate
of change. But as we have seen, sine and cosine are fatally
linked at all times. We saw it in the very first equation, sin^{2}x
= (1 – cos^{2}x).
And we see it after
we find the derivative of
sin^{2}x,
which is 2cosxsinx.
That just tells us that whether you are given both or only one of
the two, you have to monitor both
rates of change. If you were
to just drop down the 2, you wouldn't be monitoring both rates of
change, and you would get the wrong derivative. But I am not
doing that in my proof, am I? No, I am explicitly
monitoring the rate of change
in both directions (albeit in a somewhat compressed manner).
Which is why I get the correct derivative at the end (or the
absolute value, at any rate—the negative sign can be found
by monitoring the relative directions of change)*.
This analysis
is borne out if we look at the
current proof that the derivative of sinx
is cosx.
Unlike the proof for (sinx)^{2},
the proof for sinx
cannot be found by any chain
rule, obviously. The proof is actually extremely long and
unwieldy, as you see if you take that link or study any other
similar site. Strictly, it is exactly as complex as the proof of
the calculus itself, since it relies on the same basic
“identity”—an identity that itself relies on
the infinitesimal h.
It is precisely this reliance on infinitesimals that I am trying
to skirt with my new method. At any rate, in this proof you can
see that sinx
can never be monitored by
itself. Every proof of sinx
must include cosx,
and the reverse. They aren't just dependent functions, they are
interdependent
functions. Both have to be
monitored at the same time, because the slope of one is always
determined by both functions.** Which is why I started with the
equation stating their interdependence, and why any
proof of the derivative of
either function must start with an equation that includes both.
I will argue
that the chain rule (used in the current way) is the fudge, and
that my method is the correct one. I hope you can see that
applying a chain rule in this way to trig functions is actually
very slippery. You are being to taught to basically make two
different manipulations on the same entity, but not being told
why. There is no chain here. There is only one link, that being
sinx.
Or, if you wish, we have two links, but both
of them are sinx.
We see that if we write the square out in long form:
(sinx)(sinx)
Why would you
take the derivative of one of those terms, but not the other? It
appears to me that we are actually taking the derivative of one
of them, then multiplying by a term that is not really the
derivative of both together. Why on earth would we wish to do
that?
To see what I
mean, notice again that when I solve these trig functions for
derivatives, I make sure to manipulate both sine and cosine at
the same time. But when they apply the chain rule to (sinx)^{2},
they aren't doing that, are they? They aren't looking at cosine
at all. They are trying to differentiate sinx twice, in
two different ways, which is why I could immediately see they
were pushing the proof.
We see that
again by the fact the chain rule was meant to apply to a
functional relationship inside a functional relationship,
as in the equation
Or the
equation
The
derivatives of simple trig functions don't fit that form, as we
can see even more clearly when they
try to use the chain rule on a term like sin5x.
They write it as sin(5x)
and then try to use those imported parentheses to convince you
they suddenly have a chain of functions. They tell you g(x)
is 5x
and sin(x)
is h(x).
But that is an obvious push, since to split the term that way
implies sin5x
can
be written as two numbers. It
can't. The
sin5x
is
only one
number
or one variable. One number can't give you two functional
relationships. There is no real “interior” and
“exterior” here, since the sin signifier as written
is neither a power signifier nor any other sort of multiplicative
signifier. Sin with its variable is one term, and it cannot be
split into two subfunctions. So the proof is just a fudge from
the getgo.
Actually, the
only reason you are being taught this faked mainstream proof is
that it works. It is a push that gives you the correct result.
And they have to push the proof because they don't understand
what is really happening.
I will now
tell you what is really happening, and it has nothing to do with
a chain rule (or not the given chain rule, at any rate). To
discover it, we have to go back to my proof above, starting with
the equation
(sinx)^{2
}=
1 – (cosx)^{2
}
Since
we are seeking the derivative of (sinx)^{2
}instead
of sinx,
we need to start by differentiating the right side instead of the
left. That gives us
Δ(sinx)^{2
}=
2cosx
But, as we
have seen, that can't be the full manipulation, since we have
found a rate of change of cosx
but not of sinx.
We have monitored cosine for change but not sine. So what we need
is the rate of change of (sinx)^{2
}with
respect to sinx.
Or, to say it another way, we need to compare Δsin^{2}
to sin. That is a sort of
chain rule, but it is a very different chain rule than the one
you are taught. To do that, we don't even have to differentiate:
we simply divide the left side by sine, which gives us this:
Δ(sinx)^{2}/sinx
= 2cosx
That
fraction, by itself, tells us how one term changes
relative to the other. That is what a fraction is, in this case.
You may think of it as one of those “with respect to”
fractions if it helps you, but even that isn't really necessary.
We
can then move the sinx over to obtain this:
Δ(sinx)^{2}
= 2cosxsinx
Since we know
the rates of change are opposing*, we can use that knowledge to
drop the minus sign, giving us
Δ(sinx)^{2}
= 2cosxsinx
Since in this
case, my Δ
is indicating the same thing
as the current derivative (both are indicating a rate of change),
my proof is complete. I have matched the finding of the
mainstream without using their fudged chain rule.
Also notice
that I have explained each step. When have you ever been given a
reasonable explanation of the chain rule as applied to this
proof? There can be no reasonable explanation of the current
chain rule as applied to this problem, because, as we have just
seen, they totally ignore cosine in the proof. They manipulate
sinx
twice, in completely illogical
ways. Cosine comes into the current proof only as a manipulation
upon sine. In other words, they don't currently monitor any slope
of cosine in order to solve. That cannot work, since to solve any
trig derivative or slope, you have to monitor both sine and
cosine.** In my proof, I monitor the change of sin^{2}
against both cosine and sine,
as you see.
I also draw
your attention to the fact that the 2 actually enters the last
equation above with the cosx,
not the sinx.
Which again proves the chain rule was a fudge in this case. We
also see that the sinx
in the equation isn't the
derivative of anything. It wasn't found by taking a derivative or
by differentiating. It is just a straight relationship, as in any
other fraction. You could say it that it is found by a form of
calculus, since the relationship of sin^{2}
to sine here is a “with
respect to” relationship. But what I mean by saying it
isn't the derivative of anything is that we don't have to
manipulate powers in any way. We just write the relationship as a
fraction.
[You may
consult my next paper
for clarification of the above proof. There, I make clear what
some may already comprehend: I am basically scaling my solution
to sinx.
So although it may look like I am illegally dividing only the
left side by sinx,
that isn't what is happening. I am scaling a completed equality
to an external function.]
So while it
may seem I have broken some great rules in my first proof of
sinx,
I actually haven't. I have just used the given and longstanding
manipulations in a more direct manner. In doing so, I have
greatly simplified the old proof. Ironically, the ones who have
broken rules are those who try to insert sinx
here in the middle of my
manipulation, claiming I have found the wrong derivative for its
square. But since I am finding the rate of change of sinx
in isolation in that step, not
its derivative, this entire
argument against me is another strawman argument. As we have
seen, they manufacture that strawman to use against me, but do
not use it against their own chain rule.
More
importantly, we have found that the mainstream is making up fake
chain rules to suit themselves, and that they have been doing it
for centuries. Although I am not making a blanket argument
against the chain rule here, I have certainly shown the rule is
fudged in this particular proof. [We will see it is fudged in
other trig functions in upcoming papers, so this is not an
isolated case.] A bastardized form of the chain rule was used
here to push this derivative of (sinx)^{2}
to
the known number. But, as we have seen, both manipulations in the
chain rule were manufactured. In neither step were the proper
relationships between functions discovered, and the 2 was
completely misassigned.
It
would be interesting to see who first fudged this proof of
(sinx)^{2},
and how long ago. I suspect it has been sitting out in the open
since the time of Euler, but I could be wrong. Someone else will
have to do the research in this case, since I have other more
pressing things to attend to. But it is just more proof that both
math and physics have been unmoored for a very long time. What we
had been sold as bedrock is actually just a cardboard foundation
cast over a vast abyss.
To
sort of bookend this addendum, I will answer one last question. I
will be asked why I can differentiate (sinx)^{2}
to
find 2sinx
(as
in my first proof), but the mainstream can't. Haven't I claimed
their method was a fudge, and that it was a fudge due to the fact
that the chain rule requires a chain of correct derivatives? No,
I never said that. I said the chain rule was misused with this
trig function, but not for that reason. It is fudged not because
that is a fake derivative, but for several other reasons,
including the fact that the mainstream is creating a fake chain
by misreading the parentheses, and the fact that they are failing
to monitor both sine and cosine.
I suppose I
should be grateful for this latest email, since it gave me an
idea of the many ways I am being misunderstood. It appears that
some readers are coming to this paper without making an effort to
read and understand my longer paper on the foundations of the
calculus. That doesn't really surprise me, since I get these
isolated strawman arguments against many of my ideas, always from
people who have made no effort to comprehend what I am trying to
do. They tell me I am breaking their teachers' rules, expecting
me to be chastened, but that just means they haven't even
comprehended my titles or my basic intent: that
is exactly what I am trying to do.
I am breaking all the old rules on purpose, because I have proved
the rules are false or faulty. I don't really understand why
those who are content with all the current rules would even click
on my papers. If they are so confident in what they have, they
are welcome to it. It isn't working worth a damn in the real
world (see physics), but they have never been too concerned with
that, as we know.
More than
that, I should be thankful for this latest email, because it made
me angry enough to return to this paper and extend it into an
even stronger argument against the status
quo. This
pulling apart of the proof of (sinx)^{2
}was
long overdue, and I might never have done it without the prodding
of this clueless schoolgirl.]
*Some
won't understand what I am doing there, so I will clarify it with
a simple diagram.
Cosine is
defined by that adjacent leg of the triangle, while sine is
defined by the opposite leg (both relative to the hypotenuse, of
course). If we draw the hypotenuse flat to the horizon, then we
can “see” the relative slopes of sine and cosine. If
we think of those upper legs like a roof, one slopes up while the
other slopes down, right? You may ask, how do we know which is
which? Aren't “up” and “down” relative
terms? Well, we have to put the triangle in some sort of
Cartesian graph, which specifies a direction. Once we do that,
with right indicating positive direction, say, then if one leg
slopes up, the other slopes down. Although this visualization is,
as usual, a simplification, it is useful when trying to
understand sine's relationship to cosine.
**Now, to
find the slope of the opposite leg, say, rather than its length,
you can't just monitor sine alone. You can't build a triangle
with two sides, as the old saying goes. You also have to monitor
cosine. Why? Because if you are just given a term like sinx
and asked to differentiate,
you don't actually have enough information to solve. And I don't
mean solve as in find a specific number. I mean you can't solve
because you don't have enough information to build a real
derivative equation or find a general slope equation. You don't,
because you aren't given the length of the hypotenuse, for
instance, or even a variable for it. Without information about
the length of the hypotenuse, one leg of the triangle with its
angle isn't enough to tell you any possible slope of sine or
cosine. Just consult the image above. Say you are given sinx.
That gives you possible values for x
and for the length of the
opposite leg, right? Now, if you also
have a possible value for
the length of the hypotenuse, you could write an equation for the
slope of sine, since you then have enough information to build
your triangle. Given all that information, there is only one way
you could draw in the adjacent leg to fit. In that case, for each
value of x
and sinx,
there is only one value of cosx.
But if you are not given any possible value for the hypotenuse,
you can't solve, since the adjacent leg could be any length. And
its length will then determine the slope of sine.
That is why
I started with the equation sinx
= ±√(1
 cos^{2}x)
in this paper (in both proofs), and why any proof of the slope or
derivative of either sine or cosine must start from an equation
that includes both.
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