Question

Prove using the definition of the derivative that the derivative of e^x is actually e^x?

Answer 1

\[{e^{x+h}-e^x}\over {h}\]
\[e^{x } \frac{\left(e^h-1\right)}{h}\]
I got stuck here

Answer 2

I guess we can use L'Hopital rule

Answer 3

But when using L'Hopital then don't you eventually come to the conclusion that the derivative of e^x is e^x by saying that the derivative of e^x is e^x?

Answer 4

Using derivative to prove derivative does sound legit to me but I am not sure

Answer 5

\begin{eqnarray*}\frac{d}{dx}e^x &=& \lim_{h \to 0} \frac{e^{x + h} - e^x}{h} \\ &=& e^x \lim_{h \to 0}\frac{e^h - 1}{h} \\ &=& e^x\end{eqnarray*}
if \[\lim_{h \to 0} \frac{e^h - 1}{h} = 1.\]We will now prove that this is true.
\begin{eqnarray*}\lim_{h \to 0} \frac{e^h - 1}{h} &=& \lim_{h \to 0} \frac{\lim_{n \to \infty}\left( 1 + \frac{h}{n}\right)^n - 1}{h}\\ &=& \lim_{h \to 0} \frac{\lim_{n \to \infty} \sum_{k = 0}^n\binom{n}{k}\left(\frac{h}{n}\right)^k - 1}{h}\\ &=& \lim_{h \to 0}\frac{1 + \lim_{n \to \infty}\left( \binom{n}{1}\frac{h}{n}+\sum_{k = 2}^n\binom{n}{k}\left(\frac{h}{n}\right)^k\right) - 1}{h}\\&=&\lim_{h \to 0}\lim_{n \to \infty}\left(\binom{n}{1}\frac{1}{n}+\sum_{k = 2}^n\binom{n}{k}\left(\frac{h}{n}\right)^{k-1}\right)\\&=&\lim_{h \to 0}\left(1 + h\underbrace{\lim_{n\to\infty}\sum_{k = 2}^{n}\binom{n}{k}\left(\frac{h}{n}\right)^{k-2}}_{< \infty}\right)\\&=& 1.\end{eqnarray*}

Answer 6

Thanks a bunch this really helps.

Answer 7

serious math there