<s>[INST] Roots of the derivative as symmetric functions of the roots of the polynomial Let $p(t)=(t^2-a_1^2)\ldots(t^2-a_n^2)$ be an even polynomial with distinct real non-zero roots. Can the roots of its derivative $p'(t)$ be expressed nicely (e.g. as rational symmetric functions) in terms of the roots of $p(t)$? How? 
 [/INST] Rational symmetric functions of the roots of the derivative can be expressed as rational symmetric functions of the roots of the polynomial, because they are rational functions of the coefficients of the derivative, thus rational functions of the coefficients.
But the individual roots cannot be:
$\frac{d}{dt}(t^2-a_1^2)(t^2-a_2^2) = 4t^3 - 2 (a_1^2 + a_2^2)t$, so the nonzero roots are $\pm \sqrt{\frac{a_1^2 + a_2^2}{2} }$.
For higher degrees, we will get more complicated algebraic functions. For instance:
$\frac{d}{dt}(t^2-a_1^2)(t^2-a_2^2)(t^2-a_3^2) = 6 t^5 - 4 (a_1^2+a_2^2+a_3^2)t^3 + 2 (a_1^2a_2^2 + a_2^2a_3^2 + a_1^2a_3^2)t $ so the nonzero roots are:
$$\pm \sqrt{ \frac{a_1^2+a_2^2+a_3^2\pm \sqrt{ \left(a_1^2+a_2^2+a_3^2\right)^2 - 3\left(a_1^2a_2^2+a_1^2a_3^2+a_2^2a_3^2\right)}}{3} }$$
and it just gets worse as the degree increases.
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