Solving e^a = a^(-1) ?

Line 2: This is because the fact that two functions are equal at one point does not imply their derivatives are equal at that point. It could be, but not always. However, if two functions are equal in an entire open interval, then their derivatives will be equal in said open interval.

Bonus: $e^x = x^{-1} \iff x e^{x} - 1 = 0$ . Let $f(x) = xe^{x} - 1$ . It is a continuous function with the property that $f(0) = -1$ and $f(1) = e - 1$ so by the intermediate value theorem, at least one root exists in the interval $(0,1)$ . To find there is only one root, $f'(x) = xe^{x} + e^x = e^{x} (x + 1 )$ which is larger than 0 for $x>-1$ so in the interval where roots exist, it is strictly increasing. Thus there must be one and only one root. Applying Newton's method to this function inside this interval we find that $x \approx 0.567143$