1. **Stating the problem:** Use first principles (definition of a derivative) to find the derivative of the function $f(x) = x^3 - 5$. 2. **Recall the definition of the derivative:** The derivative $f'(x)$ is given by $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$$ 3. **Find $f(x+h)$:** For $f(x) = x^3 - 5$, we have $$f(x+h) = (x+h)^3 - 5.$$ Expanding using the binomial theorem gives $$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3,$$ so $$f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 5.$$ 4. **Compute the difference quotient:** $$\frac{f(x+h) - f(x)}{h} = \frac{\big(x^3 + 3x^2h + 3xh^2 + h^3 - 5\big) - (x^3 - 5)}{h} = \frac{3x^2h + 3xh^2 + h^3}{h}.$$ 5. **Simplify the quotient:** $$\frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2.$$ 6. **Take the limit as $h \to 0$:** $$\lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2 + 0 + 0 = 3x^2.$$ 7. **Final answer:** $$f'(x) = 3x^2.$$ This means the derivative of $x^3 - 5$ using first principles is $3x^2$.