Derivative of Exponential

In order to find the derivative of $e^x$, we can use the limit definition of the derivative:

$$\frac{d}{dx}{e}^x=\underset{h\to 0}{lim}\frac{e^{x+h}–e^x}{h}$$

$$\Rightarrow \frac{d}{dx}{e}^x=\underset{h\to 0}{lim}\frac{e^x(e^h–1)}{h}$$

Since $e^x$ is independent of $h$, we can factor it out of the limit:

$$\Rightarrow \frac{d}{dx}{e}^x=e^x\underset{h\to 0}{lim}\frac{e^h–1}{h}(1)$$

We use the Taylor series expansion of $e^h$:

$$e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+\dots$$

$$\Rightarrow e^h–1=h+\frac{h^2}{2!}+\frac{h^3}{3!}+\dots$$

$$\Rightarrow \frac{e^h–1}{h}=1+\frac{h}{2!}+\frac{h^2}{3!}+\dots$$

As $h\to 0$, all terms involving higher powers of $h,h^2,h^3\dots$ tend to zero

$$\Rightarrow \underset{h\to 0}{lim}\frac{e^h–1}{h}=1$$

Thus, (1) becomes:

$$\frac{d}{dx}{e}^x=e^x$$