Derivative of exe^x by series!

For this note, we must understand a few things:

  • ex=n=0xnn!\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} where n!n! denotes nn factorial
  • n!=n(n1)!n! = n(n-1)! because n!=n(n1)(n2)(n3)321=n(n1)!n! = n \blue{(n-1)(n-2)(n-3)\dots 3 \cdot 2 \cdot 1} = n\blue{(n-1)!}
  • (1)!(-1)! diverges. This is because if we use the Gamma Function (which is useful because Γ(n+1)=n!\Gamma(n+1) = n!), we can see that (1)!=Γ(0)=0t1etdt\displaystyle (-1)! = \Gamma(0) = \int_{0}^{\infty}t^{-1}e^{-t}\, dt which is a divergent integral

ddx(ex)=ddx(n=0xnn!)=n=0ddx(xnn!)=n=01n!ddx(xn)=n=01n!nxn1=n=0nxn1n(n1)!=n=0xn1(n1)!=x01(01)!+n=1xn1(n1)!=x1(1)!+n=0xnn!\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)! (-1)! has been defined to be divergent at the start, we can see that x1(1)!=x1=0\displaystyle \frac{x^{-1}}{\blue{(-1)!}} = \frac{x^{-1}}{\blue{\infty}} = 0: x1(1)!+n=0xnn!=0+n=0xnn!=ex\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, ddx(ex)=ex\displaystyle \frac{d}{dx}(e^x) = e^x \square

Note by James Watson
11 months ago

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