1. **Differentiate from first principle** the function $f(x) = x^2 - 3x + 1$. The first principle definition of the derivative is: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 2. Substitute $f(x+h)$ and $f(x)$: $$f(x+h) = (x+h)^2 - 3(x+h) + 1 = x^2 + 2xh + h^2 - 3x - 3h + 1$$ 3. Calculate the difference quotient: $$\frac{f(x+h) - f(x)}{h} = \frac{x^2 + 2xh + h^2 - 3x - 3h + 1 - (x^2 - 3x + 1)}{h}$$ Simplify numerator: $$= \frac{2xh + h^2 - 3h}{h}$$ 4. Cancel $h$ in numerator and denominator: $$= \frac{\cancel{h}(2x + h - 3)}{\cancel{h}} = 2x + h - 3$$ 5. Take the limit as $h \to 0$: $$f'(x) = \lim_{h \to 0} (2x + h - 3) = 2x - 3$$ **Final answer:** $$f'(x) = 2x - 3$$