1. **Problem Statement:** Find the first derivative $f'(x)$ and second derivative $f''(x)$ at $x=1$ using the forward difference operator from the given data: $$\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ f(x) & 3 & 4 & 19 & 84 & 259 & 628 \end{array}$$ 2. **Formulas:** The forward difference operator $\Delta$ is defined as $\Delta f(x) = f(x+h) - f(x)$ with step size $h=1$ here. - First derivative approximation: $$f'(x) = \frac{1}{h} \left[ \Delta - \frac{\Delta^2}{2} + \frac{\Delta^3}{3} - \frac{\Delta^4}{4} + \cdots \right] f(x)$$ - Second derivative approximation: $$f''(x) = \frac{1}{h^2} \left[ \Delta^2 - 2 \frac{\Delta^3}{2} + \frac{11}{12} \Delta^4 - \cdots \right] f(x)$$ 3. **Calculate forward differences at $x=1$: (using $h=1$)** - $\Delta f(1) = f(2) - f(1) = 19 - 4 = 15$ - $\Delta^2 f(1) = \Delta(\Delta f(1)) = \Delta f(2) - \Delta f(1) = (f(3)-f(2)) - (f(2)-f(1)) = (84-19) - (19-4) = 65 - 15 = 50$ - $\Delta^3 f(1) = \Delta(\Delta^2 f(1)) = \Delta^2 f(2) - \Delta^2 f(1)$ Calculate $\Delta^2 f(2)$: $\Delta f(2) = f(3) - f(2) = 84 - 19 = 65$ $\Delta f(3) = f(4) - f(3) = 259 - 84 = 175$ $\Delta^2 f(2) = \Delta f(3) - \Delta f(2) = 175 - 65 = 110$ So, $\Delta^3 f(1) = 110 - 50 = 60$ - $\Delta^4 f(1) = \Delta^3 f(2) - \Delta^3 f(1)$ Calculate $\Delta^3 f(2)$: Calculate $\Delta^2 f(3)$: $\Delta f(3) = 175$ (from above) $\Delta f(4) = f(5) - f(4) = 628 - 259 = 369$ $\Delta^2 f(3) = \Delta f(4) - \Delta f(3) = 369 - 175 = 194$ Calculate $\Delta^3 f(2) = \Delta^2 f(3) - \Delta^2 f(2) = 194 - 110 = 84$ Therefore, $\Delta^4 f(1) = 84 - 60 = 24$ 4. **Calculate $f'(1)$ using the series:** $$f'(1) = \frac{1}{1} \left[ 15 - \frac{50}{2} + \frac{60}{3} - \frac{24}{4} \right] = 15 - 25 + 20 - 6 = 4$$ 5. **Calculate $f''(1)$ using the series:** $$f''(1) = \frac{1}{1^2} \left[ 50 - 2 \times \frac{60}{2} + \frac{11}{12} \times 24 \right] = 50 - 60 + 22 = 12$$ 6. **Verification by finite difference simpler formulas:** - First derivative approx (forward difference): $$f'(1) \approx \frac{f(2) - f(1)}{1} = 15$$ - Central difference (more accurate): $$f'(1) \approx \frac{f(2) - f(0)}{2} = \frac{19 - 3}{2} = 8$$ Our series expansion gives $4$, which is a refined estimate considering higher order terms. - Second derivative approx (central difference): $$f''(1) \approx f(0) - 2f(1) + f(2) = 3 - 8 + 19 = 14$$ Our series gives $12$, close to this estimate. **Final answers:** $$\boxed{f'(1) = 4, \quad f''(1) = 12}$$