Editing Lagrange polynomial (section)

==Derivatives==
The {{mvar|d}}th derivative of a Lagrange interpolating polynomial can be written in terms of the derivatives of the basis polynomials,

:<math>L^{(d)}(x) := \sum_{j=0}^{k} y_j \ell_j^{(d)}(x).</math>

Recall (see {{slink|#Definition}} above) that each Lagrange basis polynomial is

<math display=block>\begin{aligned}
\ell_j(x) &= \prod_{\begin{smallmatrix}m = 0\\ m\neq j\end{smallmatrix}}^k \frac{x-x_m}{x_j-x_m}.
\end{aligned}</math>

The first derivative can be found using the [[product rule#Product of more than two factors|product rule]]:

:<math>\begin{align}
\ell_j'(x) &= \sum_{\begin{smallmatrix}i=0 \\ i\not=j\end{smallmatrix}}^k \Biggl[ \frac{1}{x_j-x_i}\prod_{\begin{smallmatrix}m=0 \\ m\not = (i , j)\end{smallmatrix}}^k \frac{x-x_m}{x_j-x_m} \Biggr]
\\[5mu]
&= \ell_j(x)\sum_{\begin{smallmatrix}i=0 \\i\not=j\end{smallmatrix}}^k \frac{1}{x-x_i}.
\end{align}</math>

The second derivative is

:<math>\begin{align}
\ell_j''(x)
&= \sum_{\begin{smallmatrix}i=0 \\ i\ne j\end{smallmatrix}}^{k} \frac{1}{x_j-x_i} \Biggl[ \sum_{\begin{smallmatrix}m=0 \\ m\ne(i,j)\end{smallmatrix}}^{k} \Biggl( \frac{1}{x_j-x_m}\prod_{\begin{smallmatrix}n=0 \\ n\ne(i,j,m)\end{smallmatrix}}^{k} \frac{x-x_n}{x_j-x_n} \Biggr) \Biggr]
\\[10mu]
&= \ell_j(x)
\sum_{0 \leq i < m \leq k}
\frac{2}{(x-x_i)(x - x_m)}
\\[10mu]
&= \ell_j(x)\Biggl[\Biggl(\sum_{\begin{smallmatrix}i=0 \\i\not=j\end{smallmatrix}}^k \frac{1}{x-x_i}\Biggr)^2-\sum_{\begin{smallmatrix}i=0 \\i\not=j\end{smallmatrix}}^k \frac{1}{(x-x_i)^2}\Biggr].
\end{align}</math>

The third derivative is

:<math>\begin{align}
\ell_j'''(x)
&= \ell_j(x)
\sum_{0 \leq i < m < n \leq k}
\frac{3!}{(x-x_i)(x - x_m)(x - x_n)}
\end{align}</math>

and likewise for higher derivatives.

Note that all of these formulas for derivatives are invalid at or near a node.  A method of evaluating all orders of derivatives of a Lagrange polynomial efficiently at all points of the domain, including the nodes, is converting the Lagrange polynomial to power basis form and then evaluating the derivatives.