1. The problem is to show the derivative of a function $f(x)$ using first principles, which means using the definition of the derivative as a limit. 2. By definition, the derivative $f'(x)$ is given by $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. To demonstrate this, choose a specific function, for example, $f(x) = x^2$. 4. Substitute into the limit definition: $$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$$ 5. Expand the numerator: $$(x+h)^2 - x^2 = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2$$ 6. Rewrite the limit as: $$f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ 7. Factor out $h$ in the numerator: $$f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h}$$ 8. Cancel $h$: $$f'(x) = \lim_{h \to 0} (2x + h)$$ 9. As $h \to 0$, the limit becomes: $$f'(x) = 2x$$ 10. Thus, using first principles, we have shown that the derivative of $f(x) = x^2$ is $f'(x) = 2x$.