1. **Problem Statement:** Given the function $f(x) = e^x$, show that its derivative $f'(x) = e^x$. 2. **Formula Used:** The derivative of the exponential function $e^x$ with respect to $x$ is known to be $e^x$. This is a fundamental rule in calculus. 3. **Explanation:** The function $e^x$ is unique because its rate of change at any point is equal to its value at that point. 4. **Intermediate Work:** Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x e^h - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}$$ 5. **Evaluating the Limit:** The limit $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ by the definition of the number $e$. 6. **Final Result:** Therefore, $$f'(x) = e^x \times 1 = e^x$$ This shows that the derivative of $f(x) = e^x$ is indeed $f'(x) = e^x$. 7. **Graph Description:** The graph of $y = e^x$ passes through $(0,1)$ and increases exponentially as $x$ increases, consistent with the derivative being the same as the function itself.