Given $n + 1$ data pairs $(x_0, y_0), ... , (x_n, y_n)$, define for $j = 0, 1, ..., n$ the functions:
$p_j = \prod_{ i \neq j } (x_i - x_j ) $ and let also, $\psi(x) = \prod_{i = 0}^n (x - x_i)$
$(a).$ Show that
$$ p_j = \psi'(x_j) $$
$(b).$ Show that the interpolating polynomial of degree at most $n$ is given by
$$ p_n(x) = \psi(x) \sum_{j=0}^n \frac{y}{( x - x_j ) \psi'( x_j )}$$
These chapters are about polynomial interpolation. I tried to use the fact that when talking about barycentric weights we say that
$$ w_j = \frac{1}{p_j} \quad \text{ for j } = 0,1,..n $$
and tried realting it to $L_j(x)$ which we denote as
$$L_j(x) = \prod_{i \neq j \\ i = 0} \frac{x - x_i}{x_j - x_i} $$
and I could not use the two equations to help me with the problems. Also if you could shed some light on how to even find the derivative of $\psi(x)$ I feel as if that would help a lot. It seems that it makes a polynomial, but I feel as if this is supposed to be intuitive.
What are you trying to do in your post ? You start from $p_j=\psi'(x_j)$ to arrive at the conclusion that $p_j=\psi'(x_j)$ ?!
Let's do it with a simpler function.
If you have $f(x)=(x-x_1)(x-x_2)(x-x_3)$
The derivative will be $f'(x)=(x-x_2)(x-x_3)+(x-x_1)(x-x_3)+(x-x_1)(x-x_2)$
Now you can notice that
$\require{cancel}f'(x_1)=(x_1-x_2)(x_1-x_3)+\cancel{(x_1-x_1)(x_1-x_3)}+\cancel{(x_1-x_1)(x_1-x_2)}$ because the two other factors have $(x_1-x_1)=0$ in them so they vanish.
This is the same for $\psi'(x_k)$ only the factors free of $x_k$ will stay, and there is only one which is $p_k$. In the sum, we only keep $i=k$.
$\psi'(x)=\sum\limits_{i=0}^{n}\prod\limits_{\substack{j=0 \\j\neq i}}^{n}(x-x_j)$
$\psi'(x_k)=\sum\limits_{i=0}^{n}\prod\limits_{\substack{j=0 \\j\neq i}}^{n}(x_k-x_j)=\sum\limits_{i=k}\prod\limits_{\substack{j=0 \\j\neq i}}^{n}(x_k-x_j)=\prod\limits_{\substack{j=0 \\j\neq k}}^{n}(x_k-x_j)=p_k$.
Now look at $\displaystyle{p_n(x) = \psi(x) \sum\limits_{j=0}^n \frac{y_j}{(x-x_j) \psi'(x_j)}=\sum\limits_{j=0}^n \frac{\psi(x)}{(x-x_j)}\;\frac{y_j}{p_j} }$
And notice that $\left(\frac{\psi(x)}{(x-x_j)}\right)(x_j)=p_j$ and $\left(\frac{\psi(x)}{(x-x_j)}\right)(x_i)=0$ for $i\neq j$.
Thus you have $\psi(x_j)=y_j$ for $0\le j\le n$ and since the polynom is of degree $n$ this is the interpolating Lagrange polynomial.