Talk:Derivative of Exponential Function

$$\frac{dy}{dx}=\frac{e^x(e^h-1)}{h}$$

Limit h → 0, (eh --1)→ h

$$\frac{dy}{dx}=\frac{e^xh}{h}$$

'Is the second line above correct? It seems to me that as h → 0, eh→ 1 and (eh --1)→ 0'

Definition of e
According to Stewart's Calculus: Early Transcendentals, the definition of the number e is the number such that its derivative at x=0 is 1.

This is equivalent to stating $$~\lim_{h \to 0} \frac{e^h - 1}{h} = 1 $$.

Why? This is simply because

$$f(x)=e^x\rightarrow f^'(x)=\lim_{h \to 0}\frac{e^{x+h}-e^x}{h} = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$$

Defining $$f^'(0)=1$$, we get

$$ 1 = f^'(0) = \lim_{h \to 0}\frac{e^0(e^h - 1)}{h} = \lim_{h \to 0}\frac{1\cdot(e^h-1)}{h}=\lim_{h \to 0}\frac{e^h-1}{h}$$

There ya go.