If $P(x)$ is a polynomial of degree $n > 1$ with only simple roots $a_1,\ldots,a_n$, is it true that $\frac 1{P'(a_1)} + \cdots + \frac 1{P'(a_n)} = 0$, and, if so, what is the proof? I can see this directly for $n = 2,3,4$ with some brute force for $4$.
[Math] sum of reciprocals of derivative of polynomial at its roots
calculuspolynomialsroots
Related Solutions
First case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation "greater than one" is not necessary but the statement is trivial if $r\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.
There may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.
Second case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.
Third case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.
Fourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.
Comment case: You asked, "Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?)."
The formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.
The dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.
$a_n$ is not a root, it is the leading coefficient.
Imagine the polynomial $(x-2)(x-3) = x^2 - 5x + 6$.
The leading coefficient is $1$ but the roots are $2$ and $3$.
This part of your argument is incorrect.
Then polynomial can be factored as:
$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)$
It should read this.
Then polynomial can be factored as:
$a_n(x-{r_1})(x-r_2)(x-r_3)\ldots(x-r_n)$
From this representation it gets obvious that the free term is
$a_n(-1)^nr_1 r_2 ... r_n$
But on the other hand we know it is $a_0$.
So the product of the roots must be:
$$\frac{a_0}{(-1)^na_n} = (-1)^n \cdot \frac{a_0}{a_n} $$
For calculating the sum of the roots just compare the coefficient before $x^{n-1}$.
You will get that the sum of the roots equals $$-\frac{a_{n-1}}{a_n}$$
See also: Vieta's formulas
Best Answer
The partial fraction decomposition yields
$$\frac{1}{P(x)}= \sum_{i=1}^n \frac{c_i}{x-a_i} $$
Multiplying by $x-a_j$ yields
$$\frac{x-a_j}{P(x)}= \sum_{i=1}^n \frac{c_i(x-a_j)}{x-a_i} (*)$$
from where $c_j =\frac{1}{P'(a_j)}$. [ED: $P(x)=(x-a_j)Q(x)$, sub it back up and note that $P'(a_j)=Q(a_j)$]
Hence
$$1=\sum_{i=1}^n \frac{1}{P'(a_i)} \frac{(x-a_1)\cdots(x-a_n)}{x-a_i}$$
The coefficient of $x^{n-1}$ is
$$0= \sum_{i=1}^n \frac{1}{P'(a_i)}$$