Comparative statics for quintic equation I have a quintic equation in $x$ with coefficients $a_0,\ldots,a_5$ which are functions of some exogenous parameter $\alpha$:
$$a_0(\alpha)+a_1(\alpha)x+a_2(\alpha)x^2+a_3(\alpha)x^3+a_4(\alpha)x^4+a_5(\alpha)x^5=0$$
I know I cannot find an explicit solution for $x$ in general by the Abel–Ruffini Theorem. However, I am wondering whether I can say something about the change of the solution in $\alpha$. I.e., if $x^*$ is some solution to the equation, can I say something (magnitude, sign) about $\frac{\partial x^*}{\partial\alpha}$? I have not managed to apply some Implicit Function Theorem or Envelope Theorem.
In my particular case, I have $a_5$ and $a_4$ independent of $\alpha$ just in case this simplifies the answer. Many thanks!
 A: Let 
$$
p(\alpha,x)=a_0(\alpha)+a_1(\alpha)x+a_2(\alpha)x^2+a_3(\alpha)x^3+a_4(\alpha)x^4+a_5(\alpha)x^5.
$$
Suppose $\frac{\rm d}{{\rm d} x} p(\alpha,x)\Big|_{x=x^\ast}\neq 0$. Here $x^\ast$ is a fixed root of $p(\alpha,x)$ for a fixed $\alpha$. Then by implicit function theorem there is a open retangle $(a,b)\times (c,d)$ such that $\alpha\in (a,b)$, $x^\ast\in (c,d)$ and a analytic function 
$$
(a,b)\ni \alpha\mapsto x(\alpha)\in (c,d).
$$
such that
$$
p(\alpha,x(\alpha))=0, \forall \alpha\in(a,b)
$$
and 
$$
\frac{{\rm d}}{{\rm d} \alpha}x(\alpha)
=
-\frac{\frac{{\rm d}}{{\rm d} \alpha}p(\alpha,x)}{\frac{{\rm d}}{{\rm d} x}p(\alpha,x)}
=-
\frac{
a_0'(\alpha)+a_1'(\alpha)x+a_2'(\alpha)x^2+a_3'(\alpha)x^3
}
{
a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4
},
$$
for all  $(\alpha,x)=(\alpha,x(\alpha))\in (a,b)\times(c,d).$
A: Let $J \subset \Bbb R$ be some open interval, and suppose we have a family of polynomials $p_\alpha(x) \in \Bbb C[x]$, the coefficients of which depend sufficiently differentiably (I think $C^1$ works) on some real parameter $\alpha \in J$, thusly
$p_\alpha(x) = \sum_0^n p_i(\alpha)x^i. \tag 1$
Let $x^\ast_0 \in \Bbb C$ be a zero of $p_\alpha(x)$ for some $\alpha = \alpha_0 \in J$; that is,
$p_{\alpha_0}(x_0^\ast) = \sum_0^n p_i(\alpha_0)(x_0^\ast)^i = 0. \tag 2$
If
$\dfrac{\partial p_{\alpha_0}(x_0^\ast)}{\partial x} \ne 0, \tag 3$
then it follows from the implicit function theorem that there exists a real $\epsilon > 0$ such that, for $\alpha \in I = (\alpha_0 - \epsilon, \alpha_0 + \epsilon) \subset J$, there is a differentiable function $x^\ast(\alpha):  I \to \Bbb C$ satisfying
$p_\alpha(x^\ast(\alpha)) = 0, \tag 4$
i.e.,
$\sum_0^n p_i(\alpha)(x^\ast(\alpha))^i = 0 \tag 5$
with $x^\ast(\alpha_0) = x_0^\ast$.
Having the existence of such an $x^\ast(\alpha)$ in hand, we may in fact proceed to find an expression for $\dot x^\ast(\alpha)$, $\alpha \in I$, where the dot $\dot {}$ denotes differentiation with respect to $\alpha$, 
$\dot x^\ast (\alpha) = \dfrac{dx^\ast(\alpha)}{d\alpha} \tag 6$
and so forth; if we apply the chain rule to (4)-(5) we find
$\dfrac{d\sum_0^n  p_i(\alpha)(x^\ast(\alpha))^i}{d\alpha} = \dfrac{\partial}{\partial \alpha} (\sum_0^n  p_i(\alpha)(x^\ast(\alpha))^i) + \dfrac{\partial}{\partial x}((\sum_0^n  p_i(\alpha)(x^\ast(\alpha))^i)\dot x(\alpha) = 0, \tag 7$
with
$\dfrac{\partial}{\partial \alpha} (\sum_0^n  p_i(\alpha)(x^\ast(\alpha))^i) = \sum_0^n  \dot p_i(\alpha)(x^\ast(\alpha))^i, \tag 8$
and
$\dfrac{\partial}{\partial x}((\sum_0^n  p_i(\alpha)(x^\ast(\alpha))^i) = (\sum_1^n  i p_i(\alpha)(x^\ast(\alpha))^{i - 1}; \tag 9$
inserting these into (7) we have
$ \sum_0^n  \dot p_i(\alpha)(x^\ast(\alpha))^i + \dot x^\ast(\alpha)\sum_1^n  i p_i(\alpha)(x^\ast(\alpha))^{i - 1} = 0; \tag{10}$
if we now assume that (3) holds, i.e., 
$\sum_1^n  i p_i(\alpha)(x^\ast(\alpha))^{i - 1} \ne 0 \tag{11}$
for $\alpha \in I$, we may solve (9) for $\dot x^\ast(\alpha)$:
$\dot x^\ast(\alpha) = -\dfrac{ \sum_0^n  \dot p_i(\alpha)(x^\ast(\alpha))^i}{\sum_1^n  i p_i(\alpha)(x^\ast(\alpha))^{i - 1}}, \tag{12}$
which describes the way the zero $x^\ast(\alpha)$ moves as we vary $\alpha \in I$.  Indeed, (12) may be thought of as an ordinary differential equation for $x^\ast(\alpha)$, with the initial condition $x^\ast(\alpha_0) = x^\ast_0$.  Furthermore, there is nothing exceptional about fixing $p_\alpha(x^\ast(\alpha)) = 0$; we could just as well start with any $x(\alpha_0)$ and the same equation(s) track the path of $x(\alpha)$ such that
$p_\alpha(x(\alpha)) = p_0, \in \Bbb C, \; p_0 \; \text{fixed}. \tag{13}$
It is worth noting, I think, that the condition (11) is precisely the algebraic criterion for $x^\ast(\alpha)$ to be a zero of $p_\alpha(x)$ of multiplicity one.  Perhaps similar formulas exist when $x^\ast(\alpha)$ is of greater multiplicity; these may involve taking higher derivatives of the various expressions given above.
It is easy to specialize the above formulas to the case $n = 5$, the quintic case; with (1) taking the form
$p_\alpha(x) = \sum_0^5 p_i(\alpha)x^i, \tag{14}$
we see that $\dot x^\ast(\alpha)$ as given by (12) becomes
$\dot x^\ast(\alpha) = -\dfrac{ \sum_0^5  \dot p_i(\alpha)(x^\ast(\alpha))^i}{\sum_1^5  i p_i(\alpha)(x^\ast(\alpha))^{i - 1}}$
$= -\dfrac{\dot p_5(\alpha)(x^\ast(\alpha))^5 + \dot p_4(\alpha)(x^\ast(\alpha))^4 + \dot p_3(\alpha)(x^\ast(\alpha))^3 + \dot p_2(\alpha)(x^\ast(\alpha))^2 + \dot p_1(\alpha)x^\ast(\alpha) + \dot p_0(\alpha)}{5p_5(\alpha)(x^\ast(\alpha))^4 + 4p_4(\alpha)(x^\ast(\alpha))^3 + 3p_3(\alpha)(x^\ast(\alpha))^2 + 2p_2(\alpha)x^\ast(\alpha) + p_1(\alpha)} \tag{15}$
and if we further assume $p_5$ and $p_4$ do not depend on $\alpha$, this expression simplifies to
$\dot x^\ast(\alpha) = -\dfrac{ \sum_0^5  \dot p_i(\alpha)(x^\ast(\alpha))^i}{\sum_1^5  i p_i(\alpha)(x^\ast(\alpha))^{i - 1}}$
$= -\dfrac{\dot p_3(\alpha)(x^\ast(\alpha))^3 + \dot p_2(\alpha)(x^\ast(\alpha))^2 + \dot p_1(\alpha)x^\ast(\alpha) + \dot p_0(\alpha)}{5p_5(\alpha)(x^\ast(\alpha))^4 + 4p_4(\alpha)(x^\ast(\alpha))^3 + 3p_3(\alpha)(x^\ast(\alpha))^2 + 2p_2(\alpha)x^\ast(\alpha) + p_1(\alpha)}. \tag{16}$
