\[ \newcommand{\d}{\text{d}} \newcommand{\dx}{\d x} \newcommand{\ddx}{\dfrac{\d}{\d x}} \]
As definition, the derivative from right/left/both sides are defined as the limit of forward/backward/central difference quotients: \[ \begin{align} & f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h} \\ & f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0)-f(x_0-h)}{h} \\ & f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0+h)-f(x_0-h)}{2h} \end{align} \] respectively, and they are equivalent if the function \(f(x)\) is derivative at \(x_0\), hence we can use the difference quotients to approximate those derivatives for particular small \(h\): \[ \begin{align} & f'(x_0) \approx \frac{f(x_0+h)-f(x_o)}{h} \\ & f'(x_0) \approx \frac{f(x_0)-f(x_0-h)}{h} \\ & f'(x_0) \approx \frac{f(x_0+h)-f(x_0-h)}{2h} \end{align} \] The truncation error of forward/backward difference is \[ R(x) = f'(x_0) - \frac{f(x_0+h)-f(x_0)}{h} = -\frac{h}{2}f''(\xi) = O(h) \]
Proof. The Taylor formula gives \[ f(x_0 + h) = f(x_0) + hf'(x_0) + \frac{h^2}{2!}f''(\xi),~~~~\xi\in[x_0, x_0+h] \] Hence \[ R(x) = f'(x_0) - \frac{f(x_0+h)-f(x_0)}{h} = -\frac{h}{2}f''(\xi) \]
and the truncation error of central difference quotient is \[ R(x) = f'(x_0) - \frac{f(x_0+h)-f(x_0-h)}{2h} = -\frac{h^2}{6}f'''(\xi) = O(h^2), ~~~~ \xi\in[x_0-h, x_0+h] \]
Proof. The Taylor formula gives \[ \begin{align} & f(x_0 + h) = f(x_0) + hf'(x_0) + \frac{h^2}{2!}f''(x_0) + \frac{h^3}{3!}f'''(\xi_1) \\ & f(x_0 - h) = f(x_0) - hf'(x_0) + \frac{h^2}{2!}f''(x_0) - \frac{h^3}{3!}f'''(\xi_1) \end{align} \] Substract and devided by \(2h\) we get \[ \frac{f(x_0+h)-f(x_0-h)}{2h} = f'(x_0) + \frac{h^2}{2\cdot3!}f'''(\xi_1) + \frac{h^2}{2\cdot3!}f'''(\xi_2) \] Hence \[ R(x) = f'(x_0) - \frac{f(x_0+h)-f(x_0-h)}{2h} = -\frac{h^2}{6}f'''(\xi) = O(h^2), ~~~~\xi\in[x_0-h, x_0+h] \]
For the given function values, construct Lagrange intropolation polynomials \[ f(x) \approx L_n(x) = \sum_{i=0}^n l_i(x)f(x_i) \] The derivative \[ f'(x) \approx L'_n(x) = \sum_{i=0}^n l'_i(x)f(x_i) \]
The remainder term
\[ R(x) = \ddx \left( \frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n(x-x_i) \right) \]
Example. Given \((x_i, f(x_i)), i=0,1,2\), Denote \(x_2-x_1 = x_1-x_0 =h\), compute \(f'(x_0), f'(x_1), f'(x_2)\)
Solution. Apply the Lagrange interpolation: \[ L_2(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0) + \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1) + \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2) \]
Hence \[ \begin{align} f'(x) \approx L_2'(x) &= \frac{f(x_0)}{2h^2}(x-x_1+x-x_2) - \frac{f(x_1)}{h^2}(x-x_0+x-x_2) \\ &~~~~+ \frac{f(x_2)}{2h^2}(x-x_0+x-x_1) \end{align} \] Substitute \(x=x_i, i=0,1,2\), we have
\[ \begin{align} & f'(x_0) = \frac{1}{2h}(-3f(x_0) + 4f(x_1) - f(x_2)) \\ & f'(x_1) = \frac{1}{2h}(-f(x_0) + f(x_2)) \\ & f'(x_2) = \frac{1}{2h}(f(x_0) + -f(x_1) + 3f(x_2)) \\ \end{align} \]