1. **State the problem:** We want to derive the function $f(x) = a^x$ where $a$ is a positive constant and $a \neq 1$. 2. **Recall the definition of the derivative:** The derivative of a function $f(x)$ is defined as $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition to $a^x$:** $$f'(x) = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = \lim_{h \to 0} \frac{a^x a^h - a^x}{h}$$ 4. **Factor out $a^x$:** $$f'(x) = a^x \lim_{h \to 0} \frac{a^h - 1}{h}$$ 5. **Recognize the limit:** The limit $$L = \lim_{h \to 0} \frac{a^h - 1}{h}$$ is a constant that depends on $a$. This limit is the definition of the derivative of $a^x$ at $x=0$. 6. **Express $a^x$ using the natural exponential:** Recall that $$a^x = e^{x \ln a}$$ 7. **Differentiate using the chain rule:** $$\frac{d}{dx} a^x = \frac{d}{dx} e^{x \ln a} = e^{x \ln a} \cdot \ln a = a^x \ln a$$ 8. **Final answer:** $$\boxed{\frac{d}{dx} a^x = a^x \ln a}$$ This means the derivative of $a^x$ is the original function multiplied by the natural logarithm of $a$.