Differentiate $((x^{2}-1)f_{n-1}(x))^{(n)}$. 
Differentiate $((x^{2}-1)f_{n-1}(x))^{(n)}$.

Using Leibniz rule, I obtain the following:

$\frac{x^{2}-1}{2^{n}n!}f_{n-1}^{(n)}(x)+\frac{x}{2^{n-1}(n-1)!}f_{n-1}^{(n-1)}(x)+\frac{1}{2^{n}(n-2)!}f_{n-1}^{(n-2)}(x)$

I am trying to show for Legendre polynomial $\phi_{n}(x)$ of degree $n$ that $x\phi_{n-1}(x)+\frac{x^{2}-1}{n}\phi_{n-1}'(x)=\phi_{n}(x)$, and the first two terms of the derivative above give me this, but how do I show the last term is zero? 

In order to prove that
$$\frac{x^2-1}{n}P_n' = xP_n - P_{n-1}$$
a different route should be followed.
Two other formulas will be used in the derivation below:
\begin{equation}\label{1}
(n+1)P_{n+1} = (2n+1)xP_n-nP_{n-1}
\end{equation}
and
\begin{equation}\label{2}
P_{n+1}'-P_{n-1}' = (2n+1)P_n
\end{equation}
Here we go:
\begin{equation}
\begin{split}
\frac{x^2-1}{n}P_n' &= \frac{x^2-1}{n(2n+1)} \left( P_{n+1}'-P_{n-1}' \right)'\\
 &= \frac{x^2-1}{n(2n+1)} \left( P_{n+1}''-P_{n-1}'' \right)\\
 &= \frac{1}{n(2n+1)} \left[ (n+1)(n+2)P_{n+1}-2xP_{n+1}' -(n-1)nP_{n-1}+2xP_{n-1}' \right]\\
 &= \frac{1}{n(2n+1)} \left[ (n+1)(n+2)P_{n+1} -(n-1)nP_{n-1}-2x(P_{n+1}'-P_{n-1}') \right]\\
 &= \frac{1}{n(2n+1)} \left[ (n+2)[(2n+1)xP_n-nP_{n-1}] -n(n-1)P_{n-1}-2x(2n+1)P_n \right]\\
 &= \frac{1}{n(2n+1)} \left[ n(2n+1)xP_n-n(2n+1)P_{n-1} \right]\\
 &= xP_n-P_{n-1}\\
\end{split}
\end{equation}
