# Is the natural exponential function defined as being its own derivative?

Is $e^x$ actually defined as being the function $f$ for which $\dfrac{d}{dx}f=f$?

By which I mean not "does the identity hold", of course I know it does and that this definition is sufficient for $e$, but did Euler actually sit down and think "gee, I wonder what I can differentiate to get the same thing back"?

I guess my question is equivalent to "what was the first use of Euler's constant" or "why did Euler come up with [what we call] $e$".

• @Ian Define - or arrive at - it, yes. Nov 5, 2015 at 0:55
• I think, historically, $e$ was first defined as $e = \lim_{n \to \infty} \left( 1 + \frac 1n\right )^n$. Also, Euler didn't come up with the $e$ as a number, but rather used a letter for it, which later established as a convention. Nov 5, 2015 at 0:57
• @Kaster Oh wow, if so, that's even more interesting (to me). For some reason I assumed discovering the usefulness of $e^1$ as it were was a cool consequence. Nov 5, 2015 at 1:00
• You can do that and show that all works. Nov 5, 2015 at 1:06
• @ncmathsadist Pardon? Nov 5, 2015 at 1:07

Gregoire de Saint-Vincent and Alphonse Antonio de Sarasa around 1690 studied the question of how to compute areas under the hyperbola $xy=1$, which led to the notion of how to compute the area under the curve $y=1/x$. While the case $y=1/x^n$, $n > 1$ was simpler and solved by Cavalieri earlier, a new function had to be defined for the case $n=1$. They introduced this notion of a "hyperbolic logarithm".
Euler, about 40 years later, introduced $e$ as the constant which gave area $1$ in a letter to Goldbach.
The limit expression $\lim_{n\to\infty} (1+\frac{1}{n})^n$ was introduced by Bernoulli even earlier than this, and I'm not entirely sure when the notions were found to coincide.