y'(1) and y''(1) from table

The data are equally spaced (h = 0.5), so use Newton's forward-difference formulas with x₀ = 0.5 and p = (x − x₀)/h = (1 − 0.5)/0.5 = 1.

Forward difference table

Formulas

y'(x) = (1/h)[Δy₀ + (2p−1)/2·Δ²y₀ + (3p²−6p+2)/6·Δ³y₀ + (4p³−18p²+22p−6)/24·Δ⁴y₀]
y''(x) = (1/h²)[Δ²y₀ + (p−1)Δ³y₀ + (6p²−18p+11)/12·Δ⁴y₀]

Substitute p = 1

y'(1)  = (1/0.5)[ -3 + (1/2)(2) + (−1/6)(-1.8) + (2/24)(2.0) ]
       = 2[ -3 + 1.0000 + 0.3000 + 0.1667 ]
       = -3.0667
y''(1) = (1/0.25)[ 2 + 0·(-1.8) + (−1/12)(2.0) ]
       = 4[ 2 − 0.1667 ]
       = 7.3333

Answer: y'(1) ≈ -3.0667, y''(1) ≈ 7.3333

Check with the 3-point central formulas (less accurate for this coarse table): y'(1) = (2−6)/(2×0.5) = -4.0000, y''(1) = (2 − 6 + 6)/0.25 = 8.0000.