$f''(e^{x})$ and it's definition I'm working through "The Calculus Tutoring Book" by Carol and Robert Ash (0-7803-1044-6).  In chapter 3.3, when discussing derivatives of basic functions, they show and define the $D_{x}e^{x}$ and subsequently define $e$.  Page 67 for those that have the book.
In any event, they define a "base", eventually, they land at $(1)$: $$D_{x}b^x = \lim_{\Delta x\to 0} b^x\lbrack \frac{b^{\Delta x}-1}{\Delta x} \rbrack$$  I have no problem here.  And I understand why they effectively ignore the constant $b^x$ when you factor it out $(2)$:
$$\frac{b^{\Delta x}-1}{\Delta x}$$ to $(3)$
$$D_{x}b^x=mb^x$$
And go on to state that the latter must be the slope of a line centered at $(0, 1)$
Then from the former, they land on the fact the $(4)$ $$D_{x}e^x = e^x$$
I don't see the path from $(2)$ to $(3)$ and subsequently $(4)$.
 A: What they are doing, which is fairly sloppy the way it is phrased and justified, is the following. 


*

*They assume that limits as $\Delta x\to0$ of expressions of the form $(2)$ exist, and that they are numbers $m(b)$ (depending on $b$)

*They argue "by picture" that, as $b$ moves between $1$ and $100$, the numbers $m(b)$ range from close to zero to very big

*They assume that $m(b)$ depends continually on $b$

*They assume that a continuous function satisfies the Intermediate Value Theorem

*They put together all the above to conclude that there exists a number $e$ such that $m(e)=1$. 

*It follows, by $(1)$, that $(e^x)'=e^x$. 
Comment: the notation you are using for derivatives is highly unusual, and it will be confusing for   anybody who sees is. One often writes $f'(x)$ for the derivative of the function $f$ at the point $x$. When you write $f'(e^x)$, what we all read is "the derivative of the function $f$, evaluated at the point $e^x$", which is not what you meant.
A: I don't have the book but what they are doing in general is showing that $\frac{d}{dx} \ b^x = ln(b)\cdot b^x\cdot 1$ ..........where the $1$ is the derivative of the exponent x (chain rule)
This is a general rule so for example $\frac{d}{dx}\ 5^{3x}= ln(5)\cdot 5^{3x}\cdot 3$
When applying this rule to $e^x$ we get $\frac{d}{dx} \ e^x = ln(e)\cdot e^x\cdot 1 = e^x$
