 Solution

Interpolate through the $s+1$ values at $t_n,\dots ,t_{n+s}$, and differentiate the interpolation polynomial at its newest node. In the backward coordinate $\tau =(t-t_{n+s})/h$, the Newton form iswith the $j=0$ factor interpreted as one and $\nabla y_m=y_m-y_{m-1}$. For $j\ge 1$, the derivative of the factor at zero is $(j-1)!/j!=1/j$. Thus the backward differentiation formula isIf $w$ denotes the forward-shift variable, ${\nabla }^j{y}_{n+s}$ corresponds to $w^{s-j}(w-1{)}^j$. Consequently the requested polynomial isThe first form is a polynomial valid also at $w=0$. The second form is convenient for relating it to the truncated series of $\log w$. Differentiation of the degree-$s$ interpolant has local derivative error $O(h^s)$, giving order $s$ for smooth solutions.