Differentiability of simple zero of a polynomial curve Suppose $f(x, t) = x^n + a_{n-1}(t) x^{n-1} + \dots + a_0(t) \in \mathbb R[x]$ where each $a_j(t) \in C^{\infty}(\mathbb R)$. If $x_0 \in \mathbb C$ is a simple zero of $f(x, t_0)$, I want to know if there is a $C^{\infty}$ function $\eta: I \to \mathbb C$ such that $\eta(t_0) = x_0$ and $\eta(t)$ is a zero for $f(x, t)$ for every $t \in I$ and $I$ is some interval with $t_0 \in I$. 
Clearly, a zero should be a function over the coefficients and thus a function over $t$. By chain rule, we can formally differentiate with respect to $t$ and get $f(x(t))' = f'(x(t)) x'(t)$ here $x$ can be equivalently viewed as $\eta$. At $t_0$, $f'(x(t_0)) = f'(x_0) \neq 0$. I want to say by inverse function theorem, $\eta$ is defined for some $I$. But I got suspicious over the first step. I am not convinced why I can do that? 

This is essentially the implicit function theorem in this context.
The condition that $x_0$ is a simple root of $f(x,t_0)$ is the statement that $(\partial f/\partial x)\neq 0$ at $(x_0,t_0)$.
As in the proof of the implicit function theorem, you can start with $\eta_0(t)=x_0$ and use successive iterates of the Newton-Raphson method to provide better and better approximations of the desired $\eta(t)$ as
$$
    \eta_{k+1}(t) = \eta_k(t) - \frac{f(\eta_k(t),t)}{(\partial f/\partial t)(x_0,t_0)}
$$
Obviously, this will only work sufficiently close to $t_0$.
