Why is the derivative of e^x itself

Using the definition (f'(x)=\lim_{h\to0}$\frac{f(x+h)-f(x)}{h}$) with (f(x)=e^x) gives ($\frac{e^{x+h}-$$e^x$}{h}=$e^x$$\frac{e^{h}-1}{h}$). The term (e^x) can be pulled out because it does not depend on (h).

Why the inner limit equals 1

The inner limit is the key step:

• Write (e^h=\sum_{n=0}^{$\infty$}$\frac{$$h^n$}{n!}).
• Subtract 1 to get $e^h-1= h+\frac{h^2}{2!}+\cdots$.
• Divide by (h) and let (h\to0); all higher‑order terms vanish, leaving 1.

Deriving the derivative step‑by‑step:

1. 1Start with (\lim_{h\to0}$\frac{e^{x+h}-$$e^x$}{h}).
2. 2Factor out (e^x): ($e^x$\lim_{h\to0}$\frac{e^{h}-1}{h}$).
3. 3Evaluate the limit of ($\frac{e^{h}-1}{h}$) using the series or the definition of (e).
4. 4Result: ($e^x$\cdot1=e^x).

Comparison of derivatives for different exponential bases: