Derivative of Exponential Function

Theorem
Let $$e^x$$ be the exponential function.

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: $$e$$ is the number such that $$ \lim_{h \to 0} \frac{e^h-1}{h} = 1 $$

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

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

Alternative proof
We use the definition of the exponential function as the inverse of the natural logarithm function.

Let $$y = \exp x$$.

Then from Derivative of an Inverse Function, we have:

$$D \exp x = \frac 1 {D \ln x} = \frac 1 {1/y} = y = \exp x$$.