--- title: "Derive derivatives from Newton forward" subject: "Numerical Method" chapter: 5 key: "semester_4/numerical_method:derive-derivatives-newton-forward" source: https://www.bctengineer.com/q/semester_4/numerical_method/derive-derivatives-newton-forward site: BCT Engineering --- # Derive derivatives from Newton forward Numerical Method · Chapter 5: Numerical Differentiation and Integration - Asked in: 2073 Magh, 4 marks - Video: [Newton forward difference formula to compute the derivative l Newton's Forward Formula Derivatives](https://www.youtube.com/watch?v=S5zAWGSTCOE) — SK Maths Tutorial - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/derive-derivatives-newton-forward - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5 ## Question Derive the formula to evaluate y'(x) and y''(x) from Newton's Forward Interpolation formula. ## Short answer From Newton forward formula y(x0+ph), differentiate w.r.t. x: dy/dx = (1/h)[Δy0 + (2p-1)/2 Δ²y0 + (3p²-6p+2)/6 Δ³y0 + …]; d²y/dx² = (1/h²)[Δ²y0 + (p-1)Δ³y0 + (6p²-18p+11)/12 Δ⁴y0 + …]. At p=0, y′(x0) = (1/h)(Δy0 − ½Δ²y0 + ⅓Δ³y0 − …), y″(x0) = (1/h²)(Δ²y0 − Δ³y0 + 11/12 Δ⁴y0 − …). ## Answer Newton's forward interpolation formula, with p = (x − x₀)/h so that dp/dx = 1/h, is ```text y = y₀ + pΔy₀ + p(p−1)/2!·Δ²y₀ + p(p−1)(p−2)/3!·Δ³y₀ + p(p−1)(p−2)(p−3)/4!·Δ⁴y₀ + … ``` Expanding the coefficients in powers of p: ```text y = y₀ + pΔy₀ + (p²−p)/2·Δ²y₀ + (p³−3p²+2p)/6·Δ³y₀ + (p⁴−6p³+11p²−6p)/24·Δ⁴y₀ + … ``` #### First derivative Differentiate with respect to p and use dy/dx = (dy/dp)(dp/dx) = (1/h)(dy/dp): ```text dy/dx = (1/h)[ Δy₀ + (2p−1)/2·Δ²y₀ + (3p²−6p+2)/6·Δ³y₀ + (4p³−18p²+22p−6)/24·Δ⁴y₀ + … ] ``` #### Second derivative Differentiate again: d²y/dx² = (1/h²)·d²y/dp²: ```text d²y/dx² = (1/h²)[ Δ²y₀ + (p−1)Δ³y₀ + (6p²−18p+11)/12·Δ⁴y₀ + … ] ``` #### At the tabulated point x = x₀ (p = 0) ```text y′(x₀) = (1/h)[ Δy₀ − Δ²y₀/2 + Δ³y₀/3 − Δ⁴y₀/4 + … ] y″(x₀) = (1/h²)[ Δ²y₀ − Δ³y₀ + 11Δ⁴y₀/12 − … ] ``` These formulas are used when x is near the beginning of the table; the corresponding backward formulas are used near the end. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What substitution is used in deriving derivatives from Newton's forward formula? A. p = (x - x0)·h, so dp/dx = h B. p = x/h, so dp/dx = h C. p = (x0 - x)/h, so dp/dx = -1/h D. p = (x - x0)/h, so dp/dx = 1/h Answer: D. Using p turns y into a polynomial in p, and the chain rule gives dy/dx = (1/h) dy/dp. (Key point: Variable p) 2. What is the first-derivative formula at x = x0 (p = 0)? A. y'(x0) = (1/h)[Δy0 - Δ²y0/2 + Δ³y0/3 - Δ⁴y0/4 + …] B. y'(x0) = (1/h)[Δy0 + Δ²y0/2 + Δ³y0/3 + …] C. y'(x0) = h[Δy0 - Δ²y0/2 + …] D. y'(x0) = (1/h²)[Δy0 - Δ²y0 + …] Answer: A. Setting p = 0 in the differentiated series leaves the alternating series with coefficients 1, -1/2, 1/3, -1/4. (Key point: First derivative) 3. Why does the second-derivative formula carry the factor 1/h²? A. h is squared to keep units positive B. Differentiating twice with respect to x brings the factor 1/h twice C. Second differences are divided by 2h D. Because Δ²y is proportional to h Answer: B. d²y/dx² = (1/h²) d²y/dp². (Key point: Second derivative factor) 4. Where in a table are the forward-difference derivative formulas appropriate? A. Near the end of the table B. Only at the middle point C. Near the beginning of the table D. Only for unequally spaced data Answer: C. Forward differences are most accurate near the start; backward ones near the end. 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