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$y = f(x) = a^x (a > 0)$について$x=0$のときの微分係数$f'(0) = \lim_{h \to 0} \frac{1}{h}(a^h - 1)$を$1$としてネイピア数を求めるやり方よりも
$\int^{x}_{1} x^{\beta}dx=\frac{1}{\beta +1}(x^{\beta + 1} - 1)$を$1$としてネイピア数求める方が個人的にはイメージしやすいなあ

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Mathtodon

A Mastodon instance named Mathtodon, where you can post toots with beautiful mathematical formulae in TeX/LaTeX style.