return to homepage THE DERIVATIVES OF THE NATURAL LOG
Abstract: I will show that the current derivative of the natural log and the current derivative of 1/x are both wrong. In doing so, I will show the magnificent cheat in the current derivation of dln(x)/dx, embarrassing every living mathematician.
The derivative for the natural log is currently found by this method:
But is that the slope? No. As with the exponential functions, to find a slope we just find an average of the forward slope and the backward slope, like this: slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2 The slope at x=2 is .5495, not .5. The slope at x=3 is .3464, not .3333. The slope at x=4 is .255, not .25. The slope at x=5 is .203, not .2. The slope at x=6 is .1685, not .1667. The slope at x=7 is .1435, not .1429. You can see that these new slopes are very close to the current ones. To see a full argument for why averaging forward slope and backward slope is actually better than going to zero, you will have to read the full analysis in my paper on the exponential functions. Once I made these tables, I could see that ln(x) wasn't acting like a power curve. It wasn't straightening out as we differentiated. The first solution I considered was integrating. Because our curve is increasing instead of decreasing, we can try integrating instead of differentiating. To find a velocity, you need to straighten out the curve, and differentiating is further curving this curve, you see. So reversing the process might have helped us. We can seek an integral or anti-derivative of ln(x) just by reversing our chart, like this: Δln(x) .6931, .4055, .2877, .2231, .1823, .1542, .1335, .1178 ln(x) 0, .6931, 1.099, 1.386, 1.609, 1.792, 1.946, 2.079, 2.197 ∫ln(x) .6931, 1.792, 3.178, 4.787, 6.579, 8.525, 10.60, 12.80 ∫∫ln(x) .6931, 2.48, 5.663, 10.45, 17.03, 25.55, 36.16, 48.95 All I have done is turn the table upside down, then try to find more lines of differentials. But since we are going the opposite direction of what has historically been called differentiation, I call these lines integrals. They are found by adding instead of subtracting, and that is what integration originally meant. But, as you now see, we find the same problem in this direction in the table. Our curve is again increasing. It switched directions, and then immediately began increasing again. So we can't find a straight line, or what has been called the velocity, by either the old-fashioned differentiation or integration. Calculus was invented in relation to the power functions, and ln(x) is not a power function. The power functions could be related to one another, because the rate of change of a higher power was equal to the rate of change of a lower power. As we found the derivative, we took higher powers to lower powers, thereby straightening them out. Here, we can't do that. Both up and down the table, we get more curvature. It is also very important to notice that with this function, differentiating never lowers the rate of change. With power series, finding a derivative always meant decreasing the curve of the function, which means the rate of change of the curve was decreasing. But with ln(x), neither differentiating nor integrating decrease the curvature. Both increase it. In problem solving, this must be important. The current calculus has solved this dilemma by 1) Ignoring it. They pretend these simple tables don't exist; 2) Fudging the equations to match the slope they know exists, from studying the lines on real graphs. If the slope looks like the slope of 1/x, they assume that it is, and they push their math to find 1/x, using chain rules and finessed limits and fake substitutions and whatever else they need to hammer the solution down. The current method has found a fair approximation with 1/x, but my method finds the right answer with no approximation. I never go to zero and don't use infinitesimals. My rates of change are absolutely correct relative to one another, and there is no margin of error. My method is vastly superior to the current method in both operation and answer. My operation is simple and transparent, with no fudgy manipulations. My notation is much simpler than the current method. And my answer is correct, whereas the current answer is simply incorrect. The current answer is nothing but a poor estimate of my answer, and I have explained why in my long paper on the derivative and its history.
Now let us look at the derivative of 1/x. The current calculus treats this function as equivalent to x
As with the exponential functions, these two functions behave in strange ways. I remind you, for example, that ln(x) increases its curvature with both differentiation and integration. I said that this must affect solutions in physics, and what I meant is that in physics we find a velocity from a curve by straightening out the curve. By the modern interpretation, the velocity is always equal to the tangent, which is always equal to the derivative, but that isn't true even with power functions. It is true only with the power 2. I showed in a recent paper on cubed acceleration
Conclusion: By ignoring the differential tables themselves, the calculus has vastly overcomplicated the methods and proofs. In doing so, they have actually gotten the wrong answer for many derivatives, as with these two. Moreover, they have been forced to fudge many proofs. The fudge above with ln(x) is not what one would call subtle, and the modern calculus is full of fudges like that. For another, see my explosion of the proof of d If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |