At the interpolation points, we have:

f′(x_i) = p′_n(x_i) + E′_n(x_i)

The term p′_n(x_i) is a finite difference formula. Several finite difference formulas can be used to evaluate derivatives of f(x), distinguished by the degree of the polynomial and the interpolation points used.

First-Order Approximations

For a polynomial of degree 1 passing through (x₀,f(x₀)) and (x₁,f(x₁)), Newton's interpolation gives:

p₁(x) = f(x₀) + f[x₀,x₁](x − x₀)
f′(x) = p′₁(x) + E′₁(x) = f[x₀,x₁] + E′₁(x)

Using the relation with h = x₁ − x₀, we get the forward difference at x₀:

f′(x₀) = (f(x₁) − f(x₀)) / h − h f″(ξ₀) / 2, ξ₀ ∈ [x₀, x₁]

Evaluating at x₁ gives the backward difference:

f′(x₁) = (f(x₁) − f(x₀)) / h + h f″(ξ₁) / 2, ξ₁ ∈ [x₀, x₁]

The same divided difference approximates the derivative at both points, but the error term differs.

Second-Order Approximations

For a polynomial of degree 2 through points (x₀,f(x₀)), (x₁,f(x₁)), (x₂,f(x₂)):

p₂(x) = f(x₀) + f[x₀,x₁](x − x₀) + f[x₀,x₁,x₂](x − x₀)(x − x₁)
p′₂(x) = f[x₀,x₁] + f[x₀,x₁,x₂](2x − (x₀ + x₁))

Evaluating at x₀, x₁, x₂ gives second-order derivative approximations:

f′(x₀) = (−f(x₂) + 4f(x₁) − 3f(x₀)) / 2h + h² f‴(ξ₀) / 3 (forward)
f′(x₁) = (f(x₂) − f(x₀)) / 2h − h² f‴(ξ₁) / 6 (centered)
f′(x₂) = (3f(x₂) − 4f(x₁) + f(x₀)) / 2h + h² f‴(ξ₂) / 3 (backward)

The error terms are derived from the general error relation. ξ₀, ξ₁, ξ₂ lie somewhere in [x₀, x₂] and are unknown.

Remark: All these finite difference formulas are of second order. Forward, backward, and centered indicate the point at which the derivative is evaluated and the points used for calculation.

Figure illustrates the possibilities: first-order differences estimate the derivative as the slope of the segment joining (x₀,f(x₀)) and (x₁,f(x₁)), while second-order differences use a degree 2 polynomial to compute forward, centered, and backward differences at x₀, x₁, and x₂ respectively.