power rule scenic route

6:30 am. Jackson Hole. I’m in a coffee shop sipping a cappuccino, waking up and killing time before meeting up with climbing partner/middle school math teacher S. Suddenly, a blinding insight occurs to me: \(x^n\) is just a product of \(n\) \(x\)’s multiplied together… so why not derive the power rule by using the product rule? Or rather, using the product rule’s bigger badder heavy-equipment off-road version? Digging a garden using a backhoe.

In its normal form, the product rule is: \[\Big(\,\, f(x)g(x) \,\,\Big)' = f'(x)g(x) + f(x)g'(x)\] But if we have not two functions but a bunch of functions, the product rule becomes: \[\Big(\,\, f_1(x)f_2(x)\cdots f_n(x)\,\,\Big)' \quad=\quad \begin{matrix} {\color{red} f_1'(x)}\cdot f_2(x)\cdot f_3(x)\cdot \cdots f_n(x) \\ + \\ f_1(x)\cdot {\color{red} f_2'(x)}\cdot f_3(x)\cdot \cdots f_n(x) \\ + \\ f_1(x)\cdot f_2(x)\cdot {\color{red} f_3'(x)}\cdots f_n(x) \\ + \\ \vdots \\ +\\ f_1(x)\cdot f_2(x)\cdot f_3(x)\cdot \cdots{\color{red} f_n'(x) } \end{matrix}\] Maybe it’s clearer if we just drop all the ``of \(x\)’’ \((x)\)’s: \[\Big(\,\, f_1f_2\cdots f_n\,\,\Big)' \quad=\quad \begin{matrix} {\color{red} f_1'}\cdot f_2\cdot f_3\cdot \cdots f_n \\ + \\ f_1\cdot {\color{red} f_2'}\cdot f_3\cdot \cdots f_n \\ + \\ f_1\cdot f_2\cdot {\color{red} f_3'}\cdots f_n \\ + \\ \vdots \\ +\\ f_1\cdot f_2\cdot f_3\cdot \cdots{\color{red} f_n' } \end{matrix}\] If we write it more compactly and intimidatingly with \(\prod\) and \(\sum\) notation (I’ll keep the red color-coding to make comparison easier), it’s: \[\left(\,\, \prod_{i=1}^{i=n}f_i(x) \,\,\right)' \,\,=\,\, \sum_{i=1}^{i=n} \left(\, {\color{red} f_j'(x) } \prod_{\substack{i=1 \\ i\neq j}}^{i=n}f_i(x) \,\right)\] Or again sans \((x)\)’s: \[\left(\,\, \prod_{i=1}^{i=n}f_i \,\,\right)' \,\,=\,\, \sum_{i=1}^{i=n} \left(\, {\color{red} f_j'} \prod_{\substack{i=1 \\ i\neq j}}^{i=n}f_i \,\right)\]

Anyway, we can use this to prove the power rule!!! The power rule is the theorem that tells us that the derivative of any monomial \(x^n\) is just \(nx^{n-1}\): \[\boxed{\Large \,\, \left(x^n\right)' = nx^{n-1} \,\,}\] After all, \(x^n\) is just \(n\) \(x\)’s multiplied together! So we have:

\[\begin{align*} {\Large \left(x^n\right)'} \quad&=\quad \left(\,\underbrace{x \cdot x \cdot x \cdots}_{n \text{ times}} \right)' \\ \\ &=\,\, \left. \begin{matrix} {\color{red} x '}\cdot x\cdot x \cdots x \\ + \\ x\cdot {\color{red} x '}\cdot x \cdots x \\ + \\ x\cdot x\cdot {\color{red} x '}\cdots x \\ + \\ \vdots \\ +\\ \underbrace{ x\cdot x\cdot x \cdots{\color{red} x ' }}_{\text{also $n$ times}} \end{matrix} \,\,\right\} \text{$n$ times} \\ \\ \\ &=\,\, \left. \begin{matrix} {\color{red} 1}\cdot x\cdot x \cdots x \\ + \\ x\cdot {\color{red} 1}\cdot x \cdots x \\ + \\ x\cdot x\cdot {\color{red} 1}\cdots x \\ + \\ \vdots \\ +\\ \underbrace{ x\cdot x\cdot x \cdots{\color{red} 1 }}_{\text{also $n$ times}} \end{matrix} \,\,\right\} \text{$n$ times} \\ \\ \\ &=\,\, \left. \begin{matrix} x\cdot x \cdots x \\ + \\ x\cdot x \cdots x \\ + \\ x\cdot x \cdots x \\ + \\ \vdots \\ +\\ \underbrace{ x\cdot x\cdot x\cdots}_{\substack{\text{now $n-1$ times}\\\text{'cause we zonked off the $1$s}}} \end{matrix} \,\, \right\} {\text{$n$ times} } \\ \\ \\ &=\,\, \left. \begin{matrix} x^{n-1} \\ + \\ x^{n-1} \\ + \\ x^{n-1} \\ + \\ \vdots \\ +\\ x^{n-1} \\ \end{matrix} \,\,\right\} \text{$n$ times} \\ \\ \\ &=\,\, {\large nx^{n-1} } \end{align*}\] Ta-da!!!

If we want to be intimidating and terse, we can write this derivation using \(\prod\)- and \(\sum\)-notation! We have: \[\begin{align*} {\Large \left(\,x^n\,\right)'} \,\, &= \left(\,\underbrace{x \cdot x \cdot x \cdots}_{n \text{ times}} \right)' \end{align*}\] These \(x\)’s are all identical but let’s label them individually anyway: \[\begin{align*} &= \left(\,\underbrace{x_1 \cdot x_2 \cdot x_3 \cdots}_{n \text{ times}} \right)' \\ \\ &= \left(\,\prod_{k=1}^{k=n}x_k \right)' \end{align*}\] So then by the generalized product rule, we have: \[\begin{align*} &= \sum_{i=1}^{i=n} {\color{red} \left( x_j\right)'} \prod_{\substack{k=1 \\ k\neq j}}^{k=n}x_k \\ \\ &= \sum_{i=1}^{i=n} {\color{red} 1}\cdot \prod_{\substack{k=1 \\ k\neq i}}^{k=n}x_k \\ \\ &= \sum_{i=1}^{i=n} {\color{red} 1}\cdot \left(\,\underbrace{x \cdot x \cdot x \cdots}_{(n-1) \text{ times}} \right) \\ \\ &= \sum_{i=1}^{i=n} \left(\,\underbrace{x \cdot x \cdot x \cdots}_{(n-1) \text{ times}} \right) \\ \\ &= \sum_{i=1}^{i=n} x^{n-1} \\ \\ &= \underbrace{ x^{n-1} + x^{n-1} + x^{n-1} + \cdots}_{n \text{ times} } \\ \\ &= {\Large nx^{n-1}} \end{align*}\] Of course if you aren’t already deeply comfortable with the generalized product rule, this is absurd and intimidating! This is a description of why the power rule is true, using these tools, but if you’re not already very comfortable with the tools, it’s a description that might confuse more than clarify. In this case that’s deliberate irony—hence the title “scenic route”—but often math textbooks are written by people who struggle with that distinction. Describing things you already know can be very different than helping others come to that same understanding.

Problem

This proof, delightful and absurd as it is, has the same problem that our binomial theorem-based proof had: it’s only true for \(n\) being a positive integer. Can you improve it? Can you use these ideas to prove a stronger version of the theorem (one that’s true for \(n\) also being a negative integer? or a rational? or any real number???)