<s>[INST] Why the differentiation of $e^x$ is $e^x?$ $$\frac{d}{dx} e^x=e^x$$
Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.
 [/INST] Have a look at the series representation of $e^x$ which is
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$
Taking derivative of this gives
$$\left(e^x\right)'=\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right)'=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots\right)'$$
$$=1'+x'+\left(\frac{x^2}{2}\right)'+\left(\frac{x^3}{6}\right)'+\left(\frac{x^4}{24}\right)'+\left(\frac{x^5}{120}\right)'+\dots$$
$$\implies (e^x)'=\sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\right)'$$Then, differentiating term by term gives us
$$(e^x)'=0+1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots$$
$$=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$
$$=e^x$$
$$\implies \frac{d}{dx}e^x=e^x$$
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