Derivative of Exponential Function

Theorem:If $$y=e^x\,$$, then $$\frac{dy}{dx}=e^x$$ Proof: Let $$y=e^x\,$$

$$\frac{dy}{dx}=\lim_{h \to 0} \frac{e^{(x+h)}-e^x}{h}$$

$$\frac{dy}{dx}=\lim_{h \to 0} \frac{e^x(e^h-1)}{h}$$

By Definition: $$ \lim_{h \to 0} \frac{e^h-1}{h} = h $$

Therefore: $$\frac{dy}{dx}=\lim_{h \to 0} {e^x}\lim_(h\to 0}\frac{e^h-1}{h} = \lim_{h \to 0}e^x$$

$$\rightarrow \frac{dy}{dx}=e^x$$