# Can you deduce the power rule of differentiation just by logic?

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”.

• 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. Commented Aug 1, 2018 at 3:33
• 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. Commented Aug 1, 2018 at 3:36
• @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. Commented Aug 1, 2018 at 9:33
• 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”. Commented Aug 1, 2018 at 9:33
• 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. Commented Aug 1, 2018 at 10:27

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).