Is this true for all polynomials I took $3$ random polynomials with non zero roots one having even degree and two having odd degrees


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*$f(x)=\color{red}{4}x^2-(4\sqrt3+12)x+12\sqrt3$ having roots $\color{blue}{3,\sqrt3}$ and leading coefficient $\color{red}{4}$ and calculated values of $xf'(x)$$(f'(x)$ is the derivative of $f(x))$ at both roots which are $3f'(3)$ and $\sqrt3f'(\sqrt3)$ and then sum of their reciprocals $\frac1{3f'(3)}+\frac1{\sqrt3f'(\sqrt3)}=\frac{-1}{12\sqrt3}=\frac{-1}{\color{red}{4}}\left(\frac{1}{\color{blue}{3\cdot\sqrt3}}\right)$ then repeated same thing for 

*$g(x)=\color{red}{1}x^3-\frac{20}{3}x^2-12x+\frac{32}{3}$ having roots $\color{blue}{8,-2,\frac{2}{3}}$
$\frac1{8g'(8)}+\frac1{-2g'(-2)}+\frac1{\frac{2}{3}g'(\frac2 3)}=\frac{1}{\color{red}{1}}\left(\frac{1}{\color{blue}{8\cdot-2\cdot\frac2 3}}\right)$

*$h(x)=\color{red}{1}x^5+41x^4+137x^3-1601x^2-1818x+3240 $ having roots $\color{blue}{1,-2,5,-9,-36}$
$\frac1{1h'(1)}+\frac1{-2h'(-2)}+\frac1{5h'(5)}+\frac1{-9h'(-9)}+\frac1{-36h'(-36)}=\frac{1}{\color{red}{1}}\left(\frac{1}{\color{blue}{1\cdot-2 \cdot5\cdot-9\cdot-36}}\right)$
Is this true for all polynomials? Is there any known result?
 A: It is true for all polynomials with non-zero simple roots.
This follows from the barycentric form of Lagrange interpolation:
$$
L(x)=\ell (x)\sum _{j=1}^{n}{\frac {w_{j}}{x-x_{j}}}y_{j}
$$
where
$$
\ell (x)=(x-x_{1})(x-x_{1})\cdots (x-x_{n}),
\quad
w_{j}={\frac {1}{\ell '(x_{j})}}
$$
Therefore, for the constant function $1$ evaluated at $x=0$ we have
$$
1= L(0) = -\ell (0)\sum _{j=1}^{n}{\frac {1}{x_{j}\ell '(x_{j})}}
$$
For a polynomial $f$ with simple roots $x_1, \dots, x_n$, we have
$$
f(x)=a(x-x_{1})(x-x_{1})\cdots (x-x_{n})=a\ell (x)
$$
and so
$$
\sum _{j=1}^{n}{\frac {1}{x_{j}f'(x_{j})}}
=\sum _{j=1}^{n}{\frac {1}{x_{j}a\ell '(x_{j})}}
=-\frac{1}{a\ell(0)}
=-\frac{1}{(-1)^{n}ax_1 \cdots x_n}
=\frac{(-1)^{n+1}}{ax_1 \cdots x_n}
$$
This can also be written as
$$
\sum _{j=1}^{n}{\frac {1}{x_{j}f'(x_{j})}}
=-\frac{1}{f(0)}
$$
A: This results from partial fraction decomposition.
Suppose $g(x)$ has no repeated roots, and without loss of generality is monic (leading coefficient $1$). Then it factors as $\displaystyle g(x)=\prod_{r}(x-r)$ over all roots $r$ and
$$ \frac{1}{g(x)}=\sum_{r}\frac{c(r)}{x-r} $$
for some constants $c(r)$, one for each root $r$. To find the constant $c(s)$ for a specific root $s$, first multiply the equation by the factor $(x-s)$,
$$ \frac{x-s}{g(x)}=c(s)+\sum_{r\ne s} c(r)\frac{x-s}{x-r} $$
then "evaluate" i.e. take the limit $x\to s$ to obtain
$$ \frac{1}{g'(s)}=c(s). $$
Therefore we may plug $x=0$ into
$$ \frac{1}{g(x)} =\sum_{r} \frac{1}{g'(r)(x-r)} $$
and manage negative signs to get
$$ \frac{(-1)^{\deg g}}{\displaystyle \prod r} = -\sum_{r} \frac{1}{rg'(r)}. $$
When $\deg g$ is odd, all signs can go away.
