Find f'(3) from table

The x values (2, 4, 8, 12, 16) are not equally spaced, so the difference formulas cannot be used. Fit Newton's divided difference polynomial and differentiate it.

Divided difference table

(for example f[2,4] = (23 − 20)/(4 − 2) = 1.5, f[2,4,8] = (1.75 − 1.5)/(8 − 2) = 0.041667)

Interpolating polynomial

f(x) = 20 + 1.5(x−2) + 0.041667(x−2)(x−4) + (-0.010417)(x−2)(x−4)(x−8)
       + 0.001116(x−2)(x−4)(x−8)(x−12)

Differentiate

f′(x) = 1.5 + 0.041667[(x−2) + (x−4)]
        + (-0.010417)[(x−4)(x−8) + (x−2)(x−8) + (x−2)(x−4)]
        + 0.001116[(x−4)(x−8)(x−12) + (x−2)(x−8)(x−12)
                      + (x−2)(x−4)(x−12) + (x−2)(x−4)(x−8)]

At x = 3:

(x−2)+(x−4) = 1 − 1 = 0
(x−4)(x−8)+(x−2)(x−8)+(x−2)(x−4) = 5 − 5 − 1 = −1
(x−4)(x−8)(x−12)+(x−2)(x−8)(x−12)+(x−2)(x−4)(x−12)+(x−2)(x−4)(x−8)
   = (−1)(−5)(−9) + (1)(−5)(−9) + (1)(−1)(−9) + (1)(−1)(−5) = −45 + 45 + 9 + 5 = 14

f′(3) = 1.5 + 0.041667(0) + (-0.010417)(−1) + 0.001116(14)
      = 1.5 + 0.010417 + 0.015625
      = 1.5260

Answer: f′(3) ≈ 1.5260