Let $P(x) = a_{n}x^n + a_{n - 1}x^{n - 1} + \cdots + a_{0}$ be a polynomial with real coefficients. I have to prove that the simple roots of $P(x)$ depend differentiably on the coefficients of $P(x)$ by using the implicit function theorem. My question is the following. What does it mean that the roots depend differentiably on the coefficients?
The solution of the problem, written by a professor, is the following.
Let $F \colon \mathbb{R}^{n + 2} \longrightarrow \mathbb{R}$ given by $F\left(\alpha_n, \dots, \alpha_0, x\right) = \alpha_{n}x^n + \cdots + \alpha_0$, and $\overline{x}$ be a simple root of $P(x)$. Then $F\left(a_n, \dots, a_0, \overline{x}\right) = 0$. Since $F$ is of class $\mathcal{C}^\infty$ and $\left[D_{x}F\left(a_n, \dots, a_0, \overline{x}\right)\right] = \left[P'\left(\overline{x}\right)\right] \neq [0]$ because $\overline{x}$ is a simple root of $P(x)$, by the implicit function theorem, there exist $V \subseteq \mathbb{R}^{n + 1}$ open neighbourhood of $\left(a_n, \dots, a_0\right)$ and $T \colon V \longrightarrow \mathbb{R}$ such that $F\left(\alpha_n, \dots, \alpha_0, T\left(\alpha_n, \dots, \alpha_0\right)\right) = 0$. Hence $T\left(\alpha_n, \dots, \alpha_0\right)$ is a root of $\alpha_{n}x^n + \cdots + \alpha_0$.
Just for your intuition, it may be useful to think about "depending continuously" first. So what would it mean if the simple roots of a polynomial depend continuously on the coefficients?
The picture is this: If you have some root $\overline{x}$ of your polynomial $P\colon \mathbb{R} \to \mathbb{R}$, that is some number satisfying $P(\overline{x}) = 0$, then if you would perturb the coefficients slightly you still would have a root $\hat{x}$ which is close to the original root $\overline{x}$. "A small enough change in the input causes only a small change in the output."
Let's look at a simple example: $P(x) = x^2 - a$ for some nonzero real number $a \in \mathbb{R}$. You will have no trouble believing that the two distinct simple roots of $P$ are given by $\pm \sqrt{a}$, so clearly there is a dependence of the roots on the coefficient $a$. It's also very clear what happens when we change $a$ a little bit, since the roots are just $\pm \sqrt{a}$ and the squareroot is continuous, the roots will also only change a little bit. But even better! Away from the origin, the squareroot is actually differentiable! So the roots of $P$ do not merely depend continuously on $a$ but actually they depend differentiably on $a$.
The rest of your post (the proof) shows that this is true in more generality than just in this simple example. Maybe the only thing to add to the proof you have posted is that the implicit function theorem gives you differentiability of the function $T$, which gives you your result of differentiable dependence.