We try to estimate the derivative of 
at 
using five nodes.
at using five nodes. t = [ 0.35 0.5 0.57 0.6 0.75 ]'; % nodes
f = @(x) cos(x.^2);
dfdx = @(x) -2*x.*sin(x.^2);
exact_value = dfdx(0.5)
We have to shift the nodes so that the point of estimation for the derivative is at 
.
. w = fdweights(t-0.5,1);
fd_value = w'*f(t)
We can reproduce the weights in the finite difference tables by using equally spaced nodes with 
. For example, here are two one-sided formulas.
. For example, here are two one-sided formulas.format rat
fdweights(0:2,1)
fdweights(-3:0,1)