Recall that the Lagrange form of the
polynomial interpolating the data points $(x_0, f_0), \ldots,(x_n,f_n)$
is given by
$$p(x) =\sum^n_{j=0} f_jL_{j}(x).$$
So we approximate
$$f'(x_i) \color{var(--emphColor)}{\approx} p'(x_i) = \sum^n_{j=0}f_jL'_{j}(x_i).$$
Via interpolant
Setting $d_{ij} = L'_{j}(x_i)$, we have
\begin{align}
\begin{bmatrix}
f'(x_0)\\
\vdots\\
f'(x_n)
\end{bmatrix}
\color{var(--emphColor)}{\approx}
\begin{bmatrix}
p'(x_0)\\
\vdots\\
p'(x_n)
\end{bmatrix}
=
\begin{bmatrix}
\sum_j d_{0,j}f_j\\
\vdots\\
\sum_j d_{n,j}f_j
\end{bmatrix}
=D\mathbf{f},
\end{align}
with $\mathbf{f} = [f_0\cdots f_n]^T$.
Via interpolant
A direct calculation shows that
\begin{align}
d_{ij} = L'_{n,j}(x_i)=
\begin{cases}
\frac{w_j/w_i}{x_i - x_j}, & i\not=j\\
\sum_{i\not=j} L'_n,j(x_i), & i =j.
\end{cases}
\end{align}
Try to show this!
Let $n = 2$, consider the three equally spaced nodes
$x_0,x_1$, and $x_2$, and set $h = x_{j+1} - x_j$.
The interpolating polynomial is given by
$$p(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} \color{var(--emphColor)}{f_0}
+ \frac {(x-x_0)(x-x_2)} {(x_1-x_0)(x_1-x_2)} \color{var(--emphColor)}{f_1}
+ \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\color{var(--emphColor)}{f_2},
$$
and
\begin{align}
p'(x)&= \frac{1}{2h^2}\big[(2x-x_1-x_2)\color{var(--emphColor)}{f_0}
- 2(2x - x_0 -x_2) \color{var(--emphColor)}{f_1}
+(2x- x_0- x_1)\color{var(--emphColor)}{f_2}\big].
\end{align}
Let $n = 2$, consider the three equally spaced nodes
$x_0,x_1$, and $x_2$, and set $h = x_{j+1} - x_j$.
\begin{align}
p'(x)&= \frac{1}{2h^2}\big[(2x-x_1-x_2)\color{var(--emphColor)}{f_0}
- 2(2x - x_0 -x_2) \color{var(--emphColor)}{f_1}
+(2x- x_0- x_1)\color{var(--emphColor)}{f_2}\big].
\end{align}