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I was reasoning as follows:

  • Any function can be decomposed as the product between the rate of change of output over input (e.g.: v = ∆r/∆t) and the interval along which input ha changed (∆t).

  • Thus the functions r = 0, r = t, r = t2, r = t3 can be decomposed as follows:

    r = velocity (0 m/s) * time elapsed (∆t s) = 0 m (static object)

    r = velocity (1 m/s) * time elapsed (∆t s) = ∆t m (constant velocity)

    r = velocity (∆t m/s) * time elapsed (∆t s) = ∆t2 m (constant acceleration)

    r = velocity (∆t2 m/s) * time elapsed (∆t s) = ∆t3 m (constant “super-acceleration”)

  • One could say that the in each type of motion there is a “number of rates”. A static object has 0 rates, one moving with constant velocity has 1 rate, one where also velocity is uniformly increasing has 2 rates, one with constant re-acceleration has 3 rates and so on.

  • If you divide by time elapsed, you extract average velocity. This is equivalent to the first sub-rule of the power rule = “reduce the exponent by 1”.

  • We still need to shift from average to instantaneous velocity. At a given time, if velocity has been increasing, the instantaneous velocity will always be higher than the average velocity. So this calls or a coefficient. The pattern points clearly at this choice: multiply by the “number of rates”. This is equivalent to the second sub-rule: “the original exponent is placed as coefficient”.

  • To obtain the antiderivative and return to the original function, you just undo both operations: go back to average velocity (by dividing by the number of rates) and multiply by time elapsed (to obtain displacement).

I understand that by taking the limit as ∆t tends to 0, you come across this pattern as well, comprising both sub-rules. But I was wondering if there is any simple logical reason explaining only the second sub-rule, i.e. explaining why, to shift from average to instantaneous velocity, you must multiply by the “number of rates”.

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  • $\begingroup$ In my answer I gave the simplest logical reason explaining the differentiation fact you are interested in. One cannot split it into 'sub-rules'. However, here is a related fact. If you expand a square by dragging one corner slightly, the change in area is approximately $2$ times the side length times the side length increase. Similarly if you expand a cube, the change in volume is approximately $3$ times the face area times the side length increase. For an $n$-dimensional hypercube, observe that exactly $n$ of the hypercube's $n-1$-dimensional hyperfaces move in the expansion. $\endgroup$ – user21820 Aug 1 '18 at 3:33
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    $\begingroup$ Let me emphasize that my preceding comment cannot serve as a proof, for many reasons. Most crucially, one needs to show that all the little error bits are insignificant and so do not contribute to the ratio of rates of change as the side length increase tends to zero. This is in fact very troublesome, and that is why the product rule plus induction is a far better way to go. $\endgroup$ – user21820 Aug 1 '18 at 3:36
  • $\begingroup$ @user21820 Before going to bed I was thinking of the same analogy, so I was glad when I woke up and saw that you mentioned it. But based on morning thoughts, unlike you, I think that the analogy is perfect as it is. $\endgroup$ – Sierra Aug 1 '18 at 9:33
  • $\begingroup$ The square being stretched is exactly scenario c) in my post. The area of the square plays the part of “displacement”; one side length is “time elapsed”, the other is “average velocity” and 2 * side length is “instantaneous velocity”. And the logical thread joining both situations is that the number of “rates of change” (constant velocity, constant acceleration, constant super-acceleration…) is akin to number of “dimensions”. $\endgroup$ – Sierra Aug 1 '18 at 9:33
  • $\begingroup$ I repeat that what I wrote in my comments is not logic, and you have no idea how much I swept under the carpet. Any proper proof along those lines will be about 10 times the length of the proof following the approach I gave you in my answer. Your answer is mathematically very wrong, and so please first understand my answer completely. $\endgroup$ – user21820 Aug 1 '18 at 10:27
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Honestly, your reasoning does not make sense. In mathematics we simply cannot use ad-hoc handwaving or pattern-matching. But your inquiry is a good one, and I've left a comment that may give some intuition along the lines of your attempted approach, even though it isn't a proof. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

If you want to know the real reason for this fact:

$\lfrac{d(x^n)}{dx} = n·x^{n-1}$ for any real variable $x$ and natural number $n$.

Then there are a couple of ways to do it. The easiest way is by the product rule and induction:

$\lfrac{d(x^0)}{dx} = \lfrac{d(1)}{dx} = 0 = 0·x^{0-1}$.

Given any natural $n$ such that $\lfrac{d(x^n)}{dx} = n·x^{n-1}$:

  $\lfrac{d(x^{n+1})}{dx} = \lfrac{d(x^n·x)}{dx} = \lfrac{d(x^n)}{dx}·x + x^n·\lfrac{dx}{dx} = n·x^{n-1}·x + x^n·1 = (n+1)·x^n$.

Therefore by induction $\lfrac{d(x^n)}{dx} = n·x^{n-1}$ for every natural $n$.

You can see this post for a summary of common properties of differentiation (the product rule is the 5th point in the list).


There are in fact further generalizations:

$\lfrac{d(x^q)}{dx} = q·x^{q-1}$ for any real variable $x > 0$ and rational $q$.

$\lfrac{d(x^r)}{dx} = r·x^{r-1}$ for any real variable $x > 0$ and real $r$.

The first can be proven using implicit differentiation. The second is highly non-trivial to prove (rigorously).

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[Warning: this does not intend to be a rigorous proof, expressed in technical terms, it is only an intuitive mnemonic help to better understand the power rule.]

In the end the answer to my specific question was more or less obvious. If what drives instantaneous velocity away from average velocity is “something with a number”, it is only logical that this same number provides the rule for quantifying the said separation. With a linear or “t alone” function, average and instantaneous growth rates go together. With a quadratic one like "t2", average lags behind and with a cubic one like “t3”, average lags even more. Hence it seems to follow that the degree or exponent of the function is the coefficient that must be applied onto an average rate to convert it into the instantaneous one. If we look at the reverse operation, the same logic applies: if the instantaneous rate has rocketed in proportion to the exponent, in order to return to the average rate you should “spread” the bigger rate among (i.e., divide it by) the number of causes that made it grow.

For a visual confirmation of this preliminary conclusion, I think that the analogy mentioned in the comment above is still valid: the exponent plays that role because it makes the dependent variable grow as the space occupied (area, volume and so on) by an expanding geometric shape. Let us develop this.

The function for finding displacement of an object, up to third degree rate of change (officially called “jerk”), is:

∆r = r o + v o∆t + a∆t2/2 + j∆t3/6

This is like a sum or superposition of functions, as if the object were undergoing three cumulative “states of motion”: it is static at an initial position, it is moving with an initial velocity, it is subject to a given acceleration and to a given jerk. But it is when you focus on a specific term, that you can see that its exponent acts as indicated above. For example, focusing on the 3rd degree term (“jerk”), we could represent the displacement caused by it as the volume of a 3D figure whose sides grow by 1 with every second (just read the output as “linear meters” instead of “cubic meters”).

The specific geometric figure to be taken as example depends on the “a” (acceleration) or “j” coefficients. If I am not mistaken, with an acceleration of modulus = 1, we would be in face of a right triangle with area growing as t2/2; we need acceleration = 2 to have an expanding square of growth = t2; similarly we need jerk = 6 to be in front of an expanding cube growing as per t3. But no matter the specific shape, what is important for the purpose of the analogy is that we are always in front of growth in several equal rates corresponding to the figure’s number of dimensions. Just like an expanding cube grows in 3 directions as per three equal rates, an object subject to jerk is gaining a “volume” of displacement also as per three equal rates.

Given so, how to visually guess the instantaneous growth rate? I would note in this sense that, when you make a geometric shape expand, what you do is sort of pulling it by a skin comprising each of its dimensions, nothing less, nothing more: a sort of L form (= a half-perimeter = 2t = 2 sides) in the case of 2D figures and an L of surfaces with an additional lateral wall in 3D figures (3t2), always of width = 0. Translating this image into algebra, the power rule can be formulated as follows: pull from the skin of the shape = descend 1 dimension = divide by the time elapsed and thus descend to the level of average rate = reduce the power by 1, but pull in all of the shape’s growth directions = multiply by the number of dimensions = make the average rate become instantaneous = multiply by the power.

PS: There is a good discussion here showing why 2L is the derivative of an expanding square. There they show it by pointing out that the area of the form that is added to a square when you stretch it is L∆l + L∆l (the fringes added to each side) + ∆l2 (a little square between those two fringes). With an area we are still at the level of the function (i.e. "displacement" in the motion case). To move to "average velocity" you must divide by ∆l and thus arrive at the well-known expression that appears in the difference quotient: 2L + ∆l. Final step is just stipulating that ∆l tends to 0 and thus you end up with the instantaneous rate or derivative, 2L.

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  • $\begingroup$ I understand that the answer may not be very technical, but why the downvote, only for that reason? Could the downvoter point out math errors or flaws in the reasoning if any? $\endgroup$ – Sierra Aug 14 '18 at 10:31

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