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

Question

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”.

Answer 1

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).

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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).