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by Miles Mathis

Abstract: Following the simple method of previous papers, I will show that the current derivative of log(x) is wrong. I will correct it.

Like the other papers I have put up in the last week, this one has also had to be extended and corrected. My problems with ax forced me to continue studying these solutions, until they all matched one another and matched the correction to the calculus that I knew was needed. It took a bit of work over the holidays, but I believe it has paid off.

Not only have I not backed off, I have found a way to advance even further. In my next paper3, I will show that even the derivative equation for powers is proved in a faulty way. Yes, although I have confirmed the equation y' = nxn-1 in my long paper, it turns out even that proof will fall. It gets the correct slope and velocity and so on, but it is not absolutely correct as a matter of math or physics.

This is a table of the actual differentials of log(x):

log(x) 0. .301, .477, .602, .699, .778, .845, .903, .954
Δlog(x) .301, .176, .125, .097, .079, .067, .051
ΔΔlog(x) .125, .051, .028, .018, .012, .016
ΔΔΔlog(x) .074, .023, .01, .006, .004
ΔΔΔΔlog(x) .051, .013, .004, .002

The following are the real differential equations for log(x), finding line 3 from line 1. According the current definition of the derivative, line 3 is the derivative of line 1. Line 2 is the general rate of change of the curve log(x), and line 3 is the rate of change at a given interval x.

ΔΔlog(x) = Δlog(x ) Δlog(x + 1)
Δlog(x) = log(x + 1) log(x)
Δlog(x + 1)) = log(x + 2) log(x + 1)
ΔΔlog(x) = log(x + 1) log(x) log(x + 2) + log(x + 1)
log(x)/dx = 2log(x + 1) log(x) log(x + 2)

In the differential table, each line is a factor of 2 separated from the previous line. What I mean is, the first real differential in line 1 is the log of 2. So we are starting with 2. If we want a slope, we have to shift the entire table by a factor of 2. Since we are finding line 3 from line 1, we must shift or multiply by 22=4.

rate of change = 4[2log(x + 1) log(x) log(x + 2)]

But, as before, that isn't the slope. The slope is found most easily and perfectly by this equation, as I discovered today while chewing for the third day in a row on the derivative of ax.

slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2

slope x=3 is .1505, not .145
slope x=4 is .111, not .109
slope x=5 is .088, not .087
slope x=6 is .073, not .0725
slope x=7 is .0625, not .0621
slope x=8 is .0545, not .0543
slope x=10 is .0437, not .0435

Once again, my critics have to be sweating. Those numbers are astonishing, and there is no way around it. To see a full explanation of why averaging forward slope and backward slope is actually better than going to zero, you will have to read the extended analysis and question answering in my earlier paper on the derivative for exponents2.

Now let us look at the current proof for log(x)/dx. This log is proved straight from the natural log using the change of base rule.

Assume (lnx)' = 1/x
(logx)' = d(lnx)/dx(lnb) = 1/x(lnb) = 1/(2.3x)

The problem there is that the assumption is false. I have proven1 that 1/x is not the derivative of ln(x). Therefore this proof falls with several others.

Conclusion: As with my other recent papers, this solution remains incomplete. I am satisfied neither by the current proofs nor by my own. This averaging of forward and backward slopes works very well here, but I have not proved it is correct while the current numbers are not. I need to connect my two solutions here, the rate of change taken from the table and the slope taken from the graph. All derivatives and slopes should be provable without going to zero.


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