1. **Stating the problem:** Differentiate the function $$L_0(x) = \frac{(x - x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)} = \frac{(x - (x_0 - h))(x - (x_0 - 2h))}{(-h)(-2h)}$$ with respect to $x$. 2. **Rewrite the function:** $$L_0(x) = \frac{(x - x_0 + h)(x - x_0 + 2h)}{2h^2}$$ since $(-h)(-2h) = 2h^2$. 3. **Apply the product rule to the numerator:** Let $f(x) = x - x_0 + h$ and $g(x) = x - x_0 + 2h$. 4. **Derivatives:** $$f'(x) = 1, \quad g'(x) = 1$$ 5. **Derivative of numerator:** $$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) = 1 \cdot (x - x_0 + 2h) + (x - x_0 + h) \cdot 1 = 2x - 2x_0 + 3h$$ 6. **Derivative of $L_0(x)$:** $$L_0'(x) = \frac{2x - 2x_0 + 3h}{2h^2}$$ 7. **Final answer:** $$\boxed{L_0'(x) = \frac{2x - 2x_0 + 3h}{2h^2}}$$ This is the derivative of the given function with respect to $x$.