A calculus and algebra problem If $a_1$, $a_2$...,$a_n$ be the roots of $f(x)=0$ and $A_1$, $A_2$,...,$A_t$ be the roots of $f'(x)=0$.
Prove that 
$f'(a_1)×f'(a_2)...×f'(a_n)= n^n×f(A_1)×f(A_2)...×f(A_t)$
$f'$ represents first derivative of $f(x)$
I think we can prove each of these equal to some other constant.
 A: Before I begin, I'll note that you haven't quite provided complete information. What is $f$? Is it a polynomial? Is it $\newcommand{\RR}{\mathbb{R}}\RR\to \RR$ and differentiable once? Differentiable infinitely many times? Analytic? Is it $\newcommand{\CC}{\mathbb{C}}\CC\to\CC$ and holomorphic?
It turns out that this formula only works (with a slight modification) when $f$ is a polynomial, I've provided a counterexample for the other conditions below.
Case 1: $f$ is a polynomial, $f'$ is its formal derivative.
(Note that we can assume we are working over an algebraically closed field in this case)
First note that if $f$ has a repeated root, then it will also be a root of $f'$, so both sides will be 0. Also note that $f'$ should have one less root than $f$, since its degree is one less than the degree of $f$. Thus we can assume that $f=c\prod_{i=1}^n(x-a_i)$, where all the $a_i$ are distinct. Then $f' = cn\prod_{i=1}^{n-1}(x-A_i)$.
Thus 
$$cn^n\prod_{i=1}^{n-1} f(A_i) 
= n^nc^n \prod_{i,j\in[1,n-1]\times[1,n]}(A_i-a_j)$$
$$
= \prod_{j=1}^n cn(-1)^{n-1}\prod_{i=1}^{n-1}(a_j-A_i)
$$
$$
= (-1)^{n(n-1)}\prod_{j=1}^n f'(a_j)
= \prod_{j=1}^n f'(a_j).
$$
In the special case when $f$ is monic, your statement holds. Unless I've made an error in my algebra somewhere.
Note that the product $$\prod_{i,j\in [1,n-1]\times[1,n]}(A_i-a_j)$$ is related to the discriminant of $f$, or equivalently the resultant
of $f$ and $f'$, so this result could probably be expressed in those terms as well.
Case 2: $f:\CC\to\CC$ is holomorphic (or worse).
In this case, it is false.
Consider $f(z)=ze^z$, $f'(z)=ze^z+e^z=(z+1)e^z.$ $a_1=0$, $A_1=-1$. So
$f'(a_1)=1\ne 1^1\cdot f(A_1)=-1/e$.
Note that in this example $f$ and $f'$ have each have a single real zero, so this counterexample is also a counter example for $f : \RR\to\RR$ with any degree of smoothness.
