Other way to write Lagrange's form (with derivative) Prove that we can write polynomial $L_{n}\in\Pi_{n}$
  which is interpolating function $f(x)$
  in $n+1$
  nodes $x_{0},\,\ldots,\, x_{n}$
  in following form:
$$L_{n}(x)=\sum_{k=0}^{n}f(x_{k})\frac{p_{n+1}(x)}{(x-x_{k})\cdot p'_{n+1}(x_{k})}$$
where $p_{n+1}(x)=(x-x_{0})(x-x_{1})\cdot\ldots\cdot(x-x_{n})$
 .
I got stuck while proving this.
Assuming that:
$p_{0}=1$
$p_{n+1}(x)=\prod_{j=0}^{n}(x-x_{j})$
$p'_{n+1}(x_{k})=
 \prod_{j=0,j\not=k}^{n}(x_{k}-x_{j})$
It's easy to show, by expanding Lagrange's form:
$$L_{n}(x)=\sum_{k=0}^{n}f(x_{k})\prod_{j=0,j\not=k}^{n}\frac{x-x_{j}}{x_{k}-x_{j}}=\sum_{k=0}^{n}f(x_{k})\frac{p_{n+1}(x_{k})}{x-x_{k}}\cdot\frac{1}{p'_{n+1}(x_{k})}=\sum_{k=0}^{n}f(x_{k})\frac{p_{n+1}(x_{k})}{(x-x_{k})\cdot p'_{n+1}(x_{k})}$$
So, only problem is to show that:
$$p'_{n+1}(x_{k})=
 \prod_{j=0,j\not=k}^{n}(x_{k}-x_{j})$$
I have tried something like this: $p'_{n+1}(x_{k})=\Sigma_{i=0}^{n}\left((x_{k}-x_{i})'\cdot\prod_{j=0,j\neq i}^{n}(x_{k}-x_{j})\right)=\Sigma_{i=0}^{n}\prod_{j=0,j\neq i}^{n}(x_{k}-x_{j})$
 , but I don't see the next point - how to get rid of that sum?
 A: We have some formula to prove. We start from formula of Lagrange interpolation polynomial: 
$$\underline{\overset{n}{\underset{k=0}{\Sigma}}f(x_{k})}\cdot\frac{p_{n+1}(x)}{(x-x_{k})p'_{n+1}(x_{k})}\overset{?}{=}\underline{\overset{n}{\underset{k=0}{\Sigma}}f(x_{k})}\cdot\overset{n}{\underset{i=0;i\neq k}{\Pi}}\frac{(x-x_{i})}{(x_{k}-x_{i})}$$
We see now that we can skip underlined parts, so we have to prove only that:
$$\frac{p_{n+1}(x)}{(x-x_{k})p'_{n+1}(x_{k})}\overset{?}{=}\overset{n}{\underset{i=0;i\neq k}{\Pi}}\frac{(x-x_{i})}{(x_{k}-x_{i})}$$
Let's expand left side:
$$\frac{p_{n+1}(x)}{(x-x_{k})p'_{n+1}(x_{k})}=\frac{\Pi_{i=0}^{n}(x-x_{i})}{(x-x_{k})\cdot p'_{n+1}(x_{k})}=\ldots$$
now we pull $(x-x_k)$ from product:
$$\ldots=\frac{(x-x_{k})\Pi_{i=0;i\neq k}^{n}(x-x_{i})}{(x-x_{k})\cdot p'_{n+1}(x_{k})}=\frac{\Pi_{i=0;i\neq k}^{n}(x-x_{i})}{p'_{n+1}(x_{k})}\qquad(1)$$
Now we just have to show what is really $p'_{n+1}(x_{k})$
 :
$$p'_{n+1}(x) = \left(\overset{n}{\underset{i=0}{\Pi}}(x-x_{i})\right)'
 = \left((x-x_{k})\cdot\overset{n}{\underset{i=0;i\neq k}{\Pi}}(x-x_{i})\right)'
 = \underset{1}{\underbrace{(x-x_{k})'}}\cdot\overset{n}{\underset{i=0;i\neq k}{\Pi}}(x-x_{i})+(x-x_{k})\cdot\left(\overset{n}{\underset{i=0;i\neq k}{\Pi}}(x-x_{i})\right)'$$
Now, we calculate our derivative in  $x_{k}$
 :
$$p'_{n+1}(x_{k})=\overset{n}{\underset{i=0;i\neq k}{\Pi}}(x_{k}-x_{i})+\underset{0}{\underbrace{(x_{k}-x_{k})\ldots}}$$
back to (1):
$$\frac{\Pi_{i=0;i\neq k}^{n}(x-x_{i})}{p'_{n+1}(x_{k})}=\frac{\Pi_{i=0;i\neq k}^{n}(x-x_{i})}{\Pi_{i=0;i\neq k}^{n}(x_{k}-x_{i})}=\overset{n}{\underset{i=0;i\neq k}{\Pi}}\frac{(x-x_{i})}{(x_{k}-x_{i})}$$
End of proof.
