動物方程式：多項式微分證明

2025/02/25

我發現非理工科系好像很怕這個，那我就發出來吧！

$$ \begin{align*} \frac{d}{dx} x^n & = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ \sum_{k = 0}^{n} \binom{n}{k} x^{n - k} h^k \right] - x^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ x^n + n x^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} x^{n - k} h^k \right] - x^n}{h} \\ & = \lim_{h \to 0} \frac{n x^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} x^{n - k} h^k}{h} \\ & = \lim_{h \to 0} n x^{n - 1} + \sum_{k = 2}^{n} \binom{n}{k} x^{n - k} h^{k - 1} \\ & = n x^{n - 1} \end{align*} $$

換成狗勾🐶好像就比較不恐怖了。

$$ \begin{align*} \frac{d}{d🐶} 🐶^n & = \lim_{h \to 0} \frac{(🐶 + h)^n - 🐶^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ \sum_{k = 0}^{n} \binom{n}{k} 🐶^{n - k} h^k \right] - 🐶^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ 🐶^n + n 🐶^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} 🐶^{n - k} h^k \right] - 🐶^n}{h} \\ & = \lim_{h \to 0} \frac{n 🐶^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} 🐶^{n - k} h^k}{h} \\ & = \lim_{h \to 0} n 🐶^{n - 1} + \sum_{k = 2}^{n} \binom{n}{k} 🐶^{n - k} h^{k - 1} \\ & = n 🐶^{n - 1} \end{align*} $$

來隻貓🐱就更好了！

$$ \begin{align*} \frac{d}{d🐶} 🐶^n & = \lim_{🐱 \to 0} \frac{(🐶 + 🐱)^n - 🐶^n}{🐱} \\ & = \lim_{🐱 \to 0} \frac{\left[ \sum_{k = 0}^{n} \binom{n}{k} 🐶^{n - k} 🐱^k \right] - 🐶^n}{🐱} \\ & = \lim_{🐱 \to 0} \frac{\left[ 🐶^n + n 🐶^{n - 1} 🐱 + \sum_{k = 2}^{n} \binom{n}{k} 🐶^{n - k} 🐱^k \right] - 🐶^n}{🐱} \\ & = \lim_{🐱 \to 0} \frac{n 🐶^{n - 1} 🐱 + \sum_{k = 2}^{n} \binom{n}{k} 🐶^{n - k} 🐱^k}{🐱} \\ & = \lim_{🐱 \to 0} n 🐶^{n - 1} + \sum_{k = 2}^{n} \binom{n}{k} 🐶^{n - k} 🐱^{k - 1} \\ & = n 🐶^{n - 1} \end{align*} $$