Respuesta :
[tex]y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\[/tex]
[tex]\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\[/tex]
This shows that [tex]y' = y[/tex] is true when [tex]y = e^x[/tex]
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- Note 1: A more general solution is [tex]y = Ce^x[/tex] for some constant C.
- Note 2: It might be tempting to say the general solution is [tex]y = e^x+C[/tex], but that is not the case because [tex]y = e^x+C \to y' = e^x+0 = e^x[/tex] and we can see that [tex]y' = y[/tex] would only be true for C = 0, so that is why [tex]y = e^x+C[/tex] does not work.
