The current and historical method for differentiating is a mess. In Lagrange's derivation of the virial^{1}, we find him differentiating x^{2} with respect to t, to find 2xdx/dt. He then lets dx/dt equal v, so that dx^{2}/dt is equal to 2xv. I found this astonishing, but then was more astonished to find that it is done all the time, to this day. No one calls Lagrange on this cheat because they want to use it themselves. It is the same reason the Democrats never called the Republicans for stealing elections with voting machines. The Democrats wanted their turn with the machines, and got it.

We could continue this cheat of Lagrange to find that 2xv = 2x^{2}/t, which would mean that the derivative of x^{2} with respect to t is 2x^{2}/t.

Does any of this make any sense? Does anyone ever bother to check to see if these equations are physically true? Apparently not.

If the derivative of x^{2} with respect to t is 2xv, then what velocity is that? I will be told it is the velocity at (x,t), but 2x is already the velocity at (x,t). Because our power is 2 here, we have a curve equation. The derivative of a curve at a point is a velocity. It is supposed to be the instantaneous velocity at that point on the curve. The derivative is the tangent to the curve, remember, and it is also the velocity at that point. So by this finding of 2xv, we appear to have two simultaneous velocities at a single point, multiplied together. The value 2x is the velocity, and v is also the velocity, so we actually have v^{2}. The derivative of a power 2 curve equation is a velocity squared?

The derivative of x^{2} with respect to t cannot be 2xv, since the derivative is the rate of change. The rate of change of x^{2} with respect to t cannot be 2xv, unless v=1. Historically, Newton used the notation dx/dt (or the equivalent) here simply as a reminder of the relationship. It is like a listing of physical dimensions in an equation, not a continuation of variables. For example, if we find the derivative of x^{2} with respect to x, we find 2x, which could be written 2x(dx/dx). That would be read, “The rate of change of x^{2} is 2x(x with respect to x).” But since it is clear that dx/dx=1, we drop the notation.

When we find the derivative of x^{2} with respect to t, we are doing the same thing. We can write that as 2x(dx/dt), but only if we remember that dx/dt is just notation. In that equation, dx/dt is reducible to 1, so it cannot be written later as a velocity *variable*.

In fact, in this notation, dx/dt IS the dimensional notation. It does not stand for a velocity variable, it stands for meters/second, or something like that. If we find the derivative of x^{2} at x=4, the answer is 8, and the dx/dt only stands for the dimensions. We need dimensions, since 8 is not a physical answer. Since x^{2} is an acceleration, the rate of change of that curve at a given x must be a velocity. The derivative of an acceleration is a velocity. So dx/dt is giving us the velocity dimensions of length over time. The ratio dx/dt is not a velocity variable, it is dimensional constant.

The ratio dx/dt is reducible to 1 simply because, physically, distance and time cannot be changing at different rates, as infinitesimals or at the limit. You will say that if our velocity is 2, our distance is changing at a rate of 2 while our time is changing at a rate of 1, but in the calculus, first order changes like that are ignored. The derivative of 2 is 0 and the derivative of 1 is 0. That is why the ratio in the notation is written dx/dt. It is not written x/t. The letter d here does not mean delta, it means fluxion or infinitesimal change. At the limit, all first order changes are equal, so dx/dt is vanishing in the same sense that 1 is vanishing in equations. It is vanishing in the sense that you can ignore it. You can cross it out of the equation. If you were Newton, you would think of it as a dot over a dot, and let it vanish for that reason. If you are with me, you think of this as 1/1, which is “vanishing” for a different but mostly equivalent reason.

That notation dx/dt here is not a representation of how x and t are changing relative to each other in any *specific* problem, it is a representation of how x and t are changing relative to each other in the physical field or space, absolutely. It is analogous to dx/dy. If we study space itself, does x change at a different rate than y? Do we measure the width of space differently than we measure the depth of space? Does space itself get wider faster than it gets longer? No. In physics, dx/dy always equals 1, just as dx/dx always equals 1. For the exact same reason, dx/dt always equals 1 *in this sort of notation*. Since time is defined relative to distance, they cannot be changing at different rates. If x is changing at a rate of 1, t must be changing at a rate of 1, and this has nothing to do with the velocity of any real object in any real measurement.

To show what I meant when I said that this notation is equivalent to dimensional notation, we can look at some common equation in physics. Say,

G = 6.67 x 10^{-11} m^{3}/kgs^{2}

You can't rewrite that as

G = 6.67 x 10^{-11} v^{2}m/kg

And then start inserting different values for v into that equation. That second equation is true only in the case that v=1. v is not a variable there, it is just a dimensional analysis.

It is the same with the notation dx/dt. The ratio dx/dt can be written as a velocity only if you remember that v=1. In that case, v is not a variable, it is just a constant dimension, reminding us of a prior relationship. In Newton's and Leibniz's notation, they used dx/dt (or its equivalent) only in this way.

Therefore, Lagrange's math is just a cheat. Any time you see dx/dt written as a velocity, in a situation like this, a big red flag should pop up.

You will say, “C'mon, we all know that dx/dt is a velocity. What are you talking about? Velocity is *defined* as dx/dt, for heaven's sake!” Yes, in many situations, it is. I am not denying that dx/dt is a velocity, as long as it is used correctly. What I am denying is that dx/dt *in this particular case* is equivalent to the definitional notation of velocity. I am pointing out something fundamental and of great importance, and you better open your eyes to it. The notation of calculus has always been convoluted and sloppy, and this sloppiness had already reached epidemic proportions by the time of Lagrange. If you differentiate x^{2}, finding 2xdx/dt, the dx/dt in that notation is not a velocity. The notation is telling us that we are finding the rate of change of x^{2} with respect to the given rate of change of time. Since the given rate of change of time is always 1, dx/dt must also be one.

Let me put it another way. I try to explain this in as many ways as possible. In physics, does time ever change at any other rate than 1? Can time itself be accelerated or dilated? No. Not even in Relativity. In Relativity, *measurements* of time can be dilated or compressed, but time itself is invariable. We never find time to any power. We never find time going at any rate above 1 or below 1. Time is always ticking at a rate of 1, by definition. The rate of change of time is and must be 1. Therefore, if we are given the ratio dx/dt, and we know that dx is 1, then dx/dt must be 1. Well, if we differentiate some power of x with respect to t, then dx can hardly be anything but 1, can it? Given x^{2}, we assume dx=1. If we differentiate x^{2} with respect to x, we assume dx=1, don't we? If we didn't, then we couldn't find that dx^{2}/dx=2x. The denominator dx has to equal 1 or the equality is ruined. In the same way, when we differentiate with respect to t, dx is still changing at a rate of 1. Therefore, dx/dt=1. It can be dropped.

If we differentiate x^{2} with respect to t, the answer is 2x, not 2xdx/dt. This sloppy and confusing notation should be abandoned. The ratio dx/dt should mean one thing in calculus and one thing only. It should not be used differently in different places, since that only encourages this sort of cheating.

I have also seen x^{2} differentiated with respect to t to get 2xx'. That is unnecessary for the same reason. It just gives cheaters another variable to play with, to fudge later if they want. The term x' just means the derivative of x, and of course the derivative of x is just 1. So there is no reason to write it. The only reason to write it is so that you can use the Lagrange fudge later, claiming that x' = v, and push your equations that way. I have seen “real” mathematicians defending this manipulation, claiming that whenever you differentiate a length, you get a velocity. But that is the upside-down calculus I have talked about elsewhere^{2}. Many mathematicians and physicists actually believe they can differentiate any length they like into a velocity. But they can't, as I have shown with Lagrange and his virial proof. They have to be given an acceleration, and every length is not part of an acceleration. Because they have misunderstood calculus from day one in high-school, they think that when they differentiate some power like x^{2}, they are differentiating a length. But they aren't. When you differentiate x^{2}, you are differentiating a curve, which is an acceleration. The term x^{2} is *already* an acceleration, even without a ratio, a denominator, or a “with respect to.” The term x^{2} stands for this accelerating series of numbers:

x^{2} : 1, 4, 9, 16, 25, 36, 49, 64, 81

Therefore, you differentiate an acceleration into a velocity. You do not differentiate a length into a velocity. You differentiate down, not up.

But as it is, mathematicians think they can differentiate both up and down. They can differentiate x^{2} into a velocity, because 2x is a velocity at the point x; and they can also differentiate x into a velocity, since x' = v. But their second manipulation here is illegal, since they have just differentiated *up*, with no given acceleration or curve. If you are given x, you are not given a curve or any variable change. You are given this series of numbers:

x : 1, 2, 3, 4, 5, 6, 7, 8

The derivative of that series of numbers is 1, because the rate of change of that line is 1. There is a difference of 1 between each and every term. If you assign a velocity to that series, it is also 1. You cannot differentiate x into a velocity *variable*. You can only differentiate x into the velocity 1.

Therefore, the calculus has been corrupt and almost infinitely fudgable, at least since the time of Lagrange.

For more big cheats of Lagrange, you may visit my paper on Lagrange Points.

^{1} Mathis, Miles. *The Virial Theorem is False.* 2010.

^{2} Mathis, Miles. *A Study of Variable Acceleration.* 2009.

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