Derivative of $e^{x}$ by definition if you plug in $e^{x}$ into the definition of a derivative, you'll end up with
the limit $\lim_{h\to 0}\frac{e^{h-1}}{h}$. This limit is a ''standard limit'' and equals 1.
However if i wouldnt know this is a standard limit that equals 1, but use a taylor series instead to evaluate that limit, would that be justified? because the coefficient of a taylor series of a function involves a derivative, you're making use of something that you actually dont know yet (namely the derivative of $e^{x}$)
So the question is: is a taylor series justified to use in the definition of a derivative?
 A: If you define the exponential as $\displaystyle e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$, and differentiate term by term (you can prove this is ok...) you get
$$
(e^x)' = \left(1+x+x^2/2 + x^3/3! + \cdots\right)' = 1+x+x^2/2 +x^3/3! + \cdots) = e^x.$$
A: Here's how I approach these topics. (i) Define a function $\exp:\mathbb R \to \mathbb R$ by
$$\exp(x)=1+\sum_{i=1}^{\infty}\frac{x^n}{n!}.$$ (ii) Differentiate this defining equation to prove that $$\exp'(x)=\exp(x)$$(iii) Define $$e=\exp(1).$$ (iv) Prove $$\exp(x)=e^x \quad \forall x \in \mathbb R.$$ (v) Prove that 
$$e=\lim_{n \to \infty}(1+\frac{1}{n})^n.$$ 
A: The definition of the limit says;
$$f^\prime (x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
Given that we care considering $f(x)=e^x$ then $f(x+h)=e^{x+h}$.Substituting this into the expression;
$$f^\prime (x)=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$$
This can be simplified;
$$f^\prime (x)= e^{x}\lim_{h\to 0}\frac{(e^h-1)}{h}$$
Refer to;
How to prove that $\lim_{h\to 0} \, \frac{e^h-1}{h}$
Then;
$$f^\prime (x)= e^{x}\lim_{h\to 0}\frac{(e^h-1)}{h}=e^x \times 1$$
And thus;
$$f^\prime(x)=e^x$$
