Derivative of Exponential function

Let $y=f(x)=e^x$, then $\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}$

$=\underset{h\to 0}{\lim }\frac{e^x{e}^h-e^x}{h}$

$=\underset{h\to 0}{\lim }\frac{e^x(e^h-1)}{h}$

$=\underset{h\to 0}{\lim }{e}^x\frac{(e^h-1)}{h}$

$=(e^x)\underset{h\to 0}{\lim }\frac{e^h-1}{h}$

$=(e^x)(1)$

$=e^x$

In the above, we have used the fact that $\underset{h\to 0}{\lim }\frac{e^h-1}{h}=1$