# Derivative of $e^x$ by series!

For this note, we must understand a few things:

• $\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ where $n!$ denotes $n$ factorial
• $n! = n(n-1)!$ because $n! = n \blue{(n-1)(n-2)(n-3)\dots 3 \cdot 2 \cdot 1} = n\blue{(n-1)!}$
• $(-1)!$ diverges. This is because if we use the Gamma Function (which is useful because $\Gamma(n+1) = n!$), we can see that $\displaystyle (-1)! = \Gamma(0) = \int_{0}^{\infty}t^{-1}e^{-t}\, dt$ which is a divergent integral

\begin{aligned} \frac{d}{dx}\left(e^x\right) = \frac{d}{dx}\left(\sum_{n=0}^{\infty} \frac{x^n}{n!} \right) = \sum_{n=0}^{\infty} \frac{d}{dx}\left( \frac{x^n}{n!} \right) &= \sum_{n=0}^{\infty} \frac{1}{n!} \cdot \blue{\frac{d}{dx}\left( x^n \right)} \\ &= \sum_{n=0}^{\infty} \frac{1}{\orange{n!}} \cdot \blue{nx^{n-1}} \\ &= \sum_{n=0}^{\infty} \frac{\sout{n}x^{n-1}}{\orange{\sout{n}(n-1)!}} \\ &= \sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!} \\ &= \frac{x^{0-1}}{(0-1)!} + \sum_{\orange{n=1}}^{\infty} \frac{x^{\orange{n-1}}}{(\orange{n-1})!} \\ &= \frac{x^{-1}}{\blue{(-1)!}} + \sum_{\orange{n=0}}^{\infty} \frac{x^\orange{n}}{\orange{n}!} \end{aligned} Since $(-1)!$ has been defined to be divergent at the start, we can see that $\displaystyle \frac{x^{-1}}{\blue{(-1)!}} = \frac{x^{-1}}{\blue{\infty}} = 0$: $\blue{\frac{x^{-1}}{(-1)!}} + \sum_{n=0}^{\infty} \frac{x^n}{n!} = \blue{0} + \green{\sum_{n=0}^{\infty} \frac{x^n}{n!}} = \green{\boxed{e^x}}$ Therefore, $\displaystyle \frac{d}{dx}(e^x) = e^x \square$ Note by James Watson
1 month, 3 weeks ago

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