power rule scenic route

At 6:30 in the morning in a coffee shop in Jackson Hole, this delightful, absurd, using-a-backhoe-to-dig-a-garden derivation of the power rule ocurred to me.

The product rule, in its normal form, 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: \[\Big(\,\, f_1(x)f_2(x)\cdots f_n(x)\,\,\Big)' \,\, = \,\, \begin{matrix} {\color{red} f_1'(x)}\cdot f_2(x)\cdot f_3(x)\cdot \cdots f_n(x) \\ + \\ f_1(x)\cdot {\color{red} f_2'(x)}\cdot f_3(x)\cdot \cdots f_n(x) \\ + \\ f_1(x)\cdot f_2(x)\cdot {\color{red} f_3'(x)}\cdots f_n(x) \\ + \\ \vdots \\ +\\ f_1(x)\cdot f_2(x)\cdot f_3(x)\cdot \cdots{\color{red} f_n'(x) } \end{matrix}\] Maybe it’s clearer if we just drop all the ``of \(x\)’’ \((x)\)’s: \[\Big(\,\, f_1f_2\cdots f_n\,\,\Big)' \,\, = \,\, \begin{matrix} {\color{red} f_1'}\cdot f_2\cdot f_3\cdot \cdots f_n \\ + \\ f_1\cdot {\color{red} f_2'}\cdot f_3\cdot \cdots f_n \\ + \\ f_1\cdot f_2\cdot {\color{red} f_3'}\cdots f_n \\ + \\ \vdots \\ +\\ f_1\cdot f_2\cdot f_3\cdot \cdots{\color{red} f_n' } \end{matrix}\] If we write it more compactly and intimidatingly with \(\prod\) and \(\sum\) notation, it’s: \[\left(\,\, \prod_{i=1}^{i=n}f_i(x) \,\,\right)' \,\,=\,\, \sum_{i=1}^{i=n} \left(\, f_j'(x) \prod_{\substack{i=1 \\ i\neq j}}^{i=n}f_i(x) \,\right)\] Or again sans \((x)\)’s: \[\left(\,\, \prod_{i=1}^{i=n}f_i \,\,\right)' \,\,=\,\, \sum_{i=1}^{i=n} \left(\, f_j' \prod_{\substack{i=1 \\ i\neq j}}^{i=n}f_i \,\right)\]

Anyway, we can use this to prove the power rule!!! \[\boxed{\Large \,\, \left(x^n\right)' = nx^{n-1} \,\,}\] After all, \(x^n\) is just \(n\) \(x\)’s multiplied together! So we have: \[\begin{align*} {\Large \left(x^n\right)'} \quad&=\quad \left(\,\underbrace{x \cdot x \cdot x \cdots}_{n \text{ times}} \right)' \\ \\ &=\,\, \left. \begin{matrix} {\color{red} x '}\cdot x\cdot x\cdot \cdots x \\ + \\ x\cdot {\color{red} x '}\cdot x\cdot \cdots x \\ + \\ x\cdot x\cdot {\color{red} x '}\cdots x \\ + \\ \vdots \\ +\\ \underbrace{ x\cdot x\cdot x\cdot \cdots{\color{red} x ' }}_{\text{also $n$ times}} \end{matrix} \,\,\right\} \text{$n$ times} \\ \\ &=\,\, \left. \begin{matrix} {\color{red} 1}\cdot x\cdot x\cdot \cdots x \\ + \\ x\cdot {\color{red} 1}\cdot x\cdot \cdots x \\ + \\ x\cdot x\cdot {\color{red} 1}\cdots x \\ + \\ \vdots \\ +\\ \underbrace{ x\cdot x\cdot x\cdot \cdots{\color{red} 1 }}_{\text{also $n$ times}} \end{matrix} \,\,\right\} \text{$n$ times} \\ \\ &=\,\, \left. \begin{matrix} x\cdot x \cdots x \\ + \\ x\cdot x \cdots x \\ + \\ x\cdot x \cdots x \\ + \\ \vdots \\ +\\ \underbrace{ x\cdot x\cdot x\cdots}_{\substack{\text{now $n-1$ times}\\\text{'cause we zonked off the $1$}}} \end{matrix} \,\, \right\} {\text{$n$ times} } \\ \\ &=\,\, \left. \begin{matrix} x^{n-1} \\ + \\ x^{n-1} \\ + \\ x^{n-1} \\ + \\ \vdots \\ +\\ x^{n-1} \\ \end{matrix} \,\,\right\} \text{$n$ times} \\ \\ &=\,\, {\large nx^{n-1} } \end{align*}\]

Written more compactly with \(\prod\)- and \(\sum\)-notation, here’s how I’d write this. 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: \[\begin{align*} &= \sum_{i=1}^{i=n} \left( x_i\right)'\prod_{\substack{k=1 \\ k\neq i}}^{k=n}x_k \\ \\ &= \sum_{i=1}^{i=n} 1\cdot \prod_{\substack{k=1 \\ k\neq i}}^{k=n}x_k \\ \\ &= \sum_{i=1}^{i=n} 1\cdot \left(\,\underbrace{x \cdot x \cdot x \cdots}_{(n-1) \text{ times}} \right) \\ \\ &= \sum_{i=1}^{i=n} \left(\,\underbrace{x \cdot x \cdot x \cdots}_{(n-1) \text{ times}} \right) \\ \\ &= \sum_{i=1}^{i=n} x^{n-1} \\ \\ &= \underbrace{ x^{n-1} + x^{n-1} + x^{n-1} + \cdots}_{n \text{ times} } \\ \\ &= {\Large nx^{n-1}} \end{align*}\]