1. **Problem statement:** Find the derivative $f'(x)$ using the limit definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ We will do this for two functions: (a) $f(x) = \sqrt{3x - 2}$ (b) $f(x) = 4x - 7x^2$ --- 2. **Recall:** The limit definition of the derivative calculates the instantaneous rate of change of the function at any point $x$. --- ### (a) $f(x) = \sqrt{3x - 2}$ 3. Substitute into the limit definition: $$f'(x) = \lim_{h \to 0} \frac{\sqrt{3(x+h) - 2} - \sqrt{3x - 2}}{h}$$ 4. Simplify inside the square roots: $$= \lim_{h \to 0} \frac{\sqrt{3x + 3h - 2} - \sqrt{3x - 2}}{h}$$ 5. To simplify the difference of square roots, multiply numerator and denominator by the conjugate: $$= \lim_{h \to 0} \frac{\left(\sqrt{3x + 3h - 2} - \sqrt{3x - 2}\right) \cdot \left(\sqrt{3x + 3h - 2} + \sqrt{3x - 2}\right)}{h \cdot \left(\sqrt{3x + 3h - 2} + \sqrt{3x - 2}\right)}$$ 6. The numerator becomes: $$ (3x + 3h - 2) - (3x - 2) = 3h$$ 7. So the expression is: $$= \lim_{h \to 0} \frac{3h}{h \left(\sqrt{3x + 3h - 2} + \sqrt{3x - 2}\right)}$$ 8. Cancel $h$ in numerator and denominator: $$= \lim_{h \to 0} \frac{\cancel{3h}}{\cancel{h} \left(\sqrt{3x + 3h - 2} + \sqrt{3x - 2}\right)} = \lim_{h \to 0} \frac{3}{\sqrt{3x + 3h - 2} + \sqrt{3x - 2}}$$ 9. Now take the limit as $h \to 0$: $$f'(x) = \frac{3}{2 \sqrt{3x - 2}}$$ --- ### (b) $f(x) = 4x - 7x^2$ 10. Substitute into the limit definition: $$f'(x) = \lim_{h \to 0} \frac{(4(x+h) - 7(x+h)^2) - (4x - 7x^2)}{h}$$ 11. Expand terms: $$= \lim_{h \to 0} \frac{4x + 4h - 7(x^2 + 2xh + h^2) - 4x + 7x^2}{h}$$ 12. Simplify numerator: $$= \lim_{h \to 0} \frac{4h - 7x^2 - 14xh - 7h^2 + 7x^2}{h} = \lim_{h \to 0} \frac{4h - 14xh - 7h^2}{h}$$ 13. Factor $h$ out: $$= \lim_{h \to 0} \frac{h(4 - 14x - 7h)}{h}$$ 14. Cancel $h$: $$= \lim_{h \to 0} 4 - 14x - 7h$$ 15. Take the limit as $h \to 0$: $$f'(x) = 4 - 14x$$ --- **Final answers:** (a) $$f'(x) = \frac{3}{2 \sqrt{3x - 2}}$$ (b) $$f'(x) = 4 - 14x$$