Showing that $(1+|x|^2e^{|x|^2})^{-1}$ is polinomially bounded. how proves that, for any $\alpha=(\alpha_1,\cdots, \alpha_n)$ multi-index, and $x\in\mathbb{R}^n$, and enough large $|x|$, 
$D^{\alpha} ((1+|x|^2e^{|x|^2})^{-1})$ is polynomially bounded? (i.e. exists $N=N(\alpha), C=C(\alpha)$ constants with $|D^{\alpha}((1+|x|^2e^{|x|^2})^{-1})|\leq C[1+|x|^2]^{N})$)
I have that: Let $f(t)=(1+te^{t})^{-1}$ then $f(|x|^2)=(1+|x|^2e^{|x|^2})^{-1}$
Now, 
\begin{align}D^{\alpha} ((1+|x|^2e^{|x|^2})^{-1})&={\partial_{x_1}^{\alpha_1}}\cdots {\partial_{x_n}^{\alpha_{n}}} f(|x|^2)\\
&={\partial_{x_n}^{\alpha_n}}\cdots {\partial_{x_1}^{\alpha_{1}}} f(|x|^2)\\
&={\partial_{x_n}^{\alpha_n}}\cdots \underbrace{\partial_{x_1}\cdots \partial_{x_1}}_{\alpha_1 \text{times}} f(|x|^2)\\
&={\partial_{x_n}^{\alpha_n}}\cdots \underbrace{\partial_{x_1}\cdots \partial_{x_1}}_{\alpha_1-1\text{ times}} f'(|x|^2)2x_1\\
&={\partial_{x_n}^{\alpha_n}}\cdots \underbrace{\partial_{x_1}\cdots \partial_{x_1}}_{\alpha_1-2\text{ times}} (f''(|x|^2)2^2x_1^{2}+f'(|x|^2)2)\\
\vdots
\end{align}
and the calculations become very complicated ...
Is there a book or text where there are similar exercises? to see how is the procedure to calculate many derivatives.
 A: I apologize, not so easy!
Instead of actually looking at induction, let us try to create a lemma which will show that $D^{\alpha}$ looks like a function with a certain form. Then, we will show that every function of such a form is polynomially bounded.
So we will create this lemma, after observing the forms of the first derivatives of $f$. We will show that lemma by induction, though.
First, let $L[\{d_i\}]$, where $d_i$ are variables/constants, denote some linear combination of these variables. For example, $L[e^t,t]$ could be $2e^t - 5t$ or $77e^t + 12 t$, but not $5te^t$, for example. 

Claim : For each $\alpha$, we have:
  $$
D^{\alpha}f(x) = \sum_{k=0}^{2^{|\alpha|}}q_k(x_1,...,x_n)\frac{L[\{t^ie^{jt} : 0 \leq i,j < k\}]}{(1+te^t)^{k}}
$$

where $p_i,q_i$ are polynomials of total degree at most $|\alpha|$, and $t = |x|^2$. (That is, the numerators are fixed linear combinations of the given parameters). Note that the sum is from $k=0$ to $2^{|\alpha|}$.

To prove, we differentiate and check. Let's go through the details.
The base case should be $\alpha = 0$, let's go with $\alpha = 1$. Here, the derivative as we aleady saw is $2x_i\frac{(e^t+te^t)}{(1+te^t)^2}$, which fits with $q_0,q_1 \equiv 0$,$q_2 = 2x_i$ and $L = te^t + e^t$.
Let's now assume that this is true for $|\alpha| = n$ and try to prove it for $|\alpha| = n+1$.
WLOG, let us assume we are differentiating w.r.t $x_1$. For the polynomials $q_k$, we denote the derivative with respect to the first variable by $q_k'$.
For this, let's differentiate each term of the form $q_k(x_1,...,x_n)\frac{L[t^ie^{jt} : 0 \leq i,j < k]}{(1+te^t)^k}$.
For this, let us use the product rule, with parts $q_k$ and the fraction. Now, what is the derivative of the fractional part?
Note that $(1+te^t)^k = L[\{t^ie^{it} : 0 \leq i \leq k\}]$ by the binomial theorem. Now, the derivative of this is also some $L[\{t^ie^{it} : 0\leq i \leq k\}]$. Something similar applies for the numerator. Recall the derivative of $\frac uv$ is $\frac{vu'-uv'}{v^2}$. With the $u,v$ given here in this fraction here, I leave you to verify that $vu'-v'u \in L[\{t^ie^{jt} : 0 \leq i,j < 2k\}]$.
With that, the derivative is :
$$
q_k'(x_1,...,x_n)\frac{L[t^ie^{tj} : 0 \leq i,j < k]}{(1+te^t)^k} + 2x_1q_k(x_1,...,x_n)\frac{L[{t^ie^{tj} : 0 \leq i,j < 2k}]}{(1+te^t)^{2k}}
$$
where $q_k'$ and $2x_1q_{k}$ are polynomials of degree at most $|\alpha| + 1$. Furthermore, $2k \leq 2^{|\alpha| + 1}$, and now things are clear : after combining "like terms" as per the summation of the theorem, we get the result.

With this in mind, we have this lemma.

For every $k$, the function $f_k(x) = \frac{L[\{t^ie^{tj} : 0 \leq i,j < k\}]}{(1+te^t)^k}$ for $t = |x|^2$, is bounded on $\mathbb R^n$.

The reason is simple : divide top and bottom of $f_k$ by $t^ke^{tk}$ to get :
$$
\frac{L[\{t^{i}e^{tj} : i,j < 0\}]}{\left(1+\frac{1}{te^t}\right)^k}
$$
as $t \to \infty$, this function has limit $0$ since the top has limit zero and the bottom has limit $1$. Therefore, it is bounded for large $|x| >R$. For small $|x|$, you have that it is a continuous function for $|x| \leq R$ which is compact so is bounded. Now you can conclude.

Now what? Put all the $q_k$s together, along with the bounding constants, and you have a polynomial which is sitting there bounding all of them. Things are all of degree atmost $\alpha$. Maybe taking say $N= \alpha + 1$ then does the job. 
A: We must prove that
 $|D^{\alpha}(1+|x|^2e^{c|x|^2})^{-1}|\leq C(1+|x|^2)^{N}$ equivalent to
$|\partial_{x_n}^{\alpha_n}\cdots \partial_{x_1}^{\alpha_1}f(|x|^2)|\leq C(1+|x|^2)^N$ where $f(t)=(1+te^{ct})^{-1}$.
Afirmation. For any $\alpha_1$ no negative, $$\partial_{x_1}^{\alpha_1}f(|x|^2)=\sum_{i=0}^{\alpha_1} C_i f^{(i)}(|x|^2)x_1^{N_i}$$ with $C_i,N_i$ constants. To see this, notice that:
For $\alpha_1=0$, $f(|x|^2)=C_0f^{(0)}(|x|^2)x_1^{N_0}$ with $C_0=1$ and $N_0=0$ holds.
For $\alpha_1=1,\ f'(|x|^2)2x_1=C_0f^{(0)}(|x|^2)x_1^{N_0}+C_1f'(|x|^2)x_1^{N_1}$ with $C_0=0,\ C_1=2,\ N_1=1$ holds.
For $\alpha_1=2,\ f''(|x|^2)2^2x_1^2+f'(|x|^2)2=C_0f^{(0)}(|x|^2)x_1^{N_0}+C_1f'(|x|^2)x_1^{N_1}+C_2f''(|x|^2)x_1^{N_2}$ con $C_0=0,\ C_1=2, N_1=0,\ C_2=2^2,\ N_2=2$  holds
Now that we have a formula for the first $\alpha_1 $-derivatives from our function we must continue with $\partial_{x_2}^{\alpha_2} \sum_{i=0}^{\alpha_1} C_i f^{(i)}(|x|^2)x_1^{N_i}$. Similarly, $$\partial_{x_2}^{\alpha_2} \sum_{i=0}^{\alpha_1} C_i f^{(i)}(|x|^2)x_1^{N_i}=\sum_{i,j=0}^{\alpha_1,\alpha_2}C_iD_jf^{(i+j)}(|x|^2)x_i^{N_i}x_j^{N_j}$$.
Analogously
, $$\partial_{x_n}^{\alpha_n}\cdots \partial_{x_1}^{\alpha_1}f(|x|^2)=\sum_{i_1,\ldots, i_n=0}^{\alpha_1 \ldots, \alpha_n} C_{i_1}^{1}\cdots C_{i_n}^{n}f^{(i_1+\cdots +i_n)}(|x|^2)x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}$$
further, 
$((1+te^{t})^{-1})^{(k)}=\sum_{i=1}^{k}C_i(1+te^{t})^{-1-i}{(te^{t})^{(1)}}^{P_{i1}}\cdots {(te^{t})^{(n)}}^{P_{in}}$ with $\sum_{j=1}^{n}P_{ij}=i$ and $C_i$ constants.
Therefore: (with $a(|x|^2)=|x|^2e^{|x|^2})$
\begin{align*}|\partial_{x_n}^{\alpha_n}\cdots \partial_{x_1}^{\alpha_1}f(|x|^2)|&=| \sum_{i_1,\ldots, i_n=0}^{\alpha_1 \ldots, \alpha_n } C_{i_1}^{1}\cdots C_{i_n}^{n}\sum_{i=1}^{i_1+\cdots+i_n} C_i(1+a(|x|^2)^{-1-i}(a'(|x|^2)^{P_{i1}}\cdots (a^n(|x|^2)^{P_{in}}x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}|\\
&= C(1+a(|x|^2)^{-1-i}|(a'(|x|^2)^{P_{i1}}|\cdots |(a^n(|x|^2)^{P_{in}}||x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}|\\
 &\leq C(1+a(|x|^2)^{-1-i}|(1+a(|x|^2))^{P_{i1}}|\cdots |(1+a(|x|^2)^{P_{in}}||x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}|  \\
  &=C|(1+a(|x|^2)^{-1-i}|(1+a(|x|^2))^{\sum_{j=1}^{n}P_{ij}}||x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}| \\
    &=C|(1+a(|x|^2)^{-1-i}|(1+a(|x|^2))^{i}||x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}| \\
  &=C1+a(|x|^2)^{-1}|x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}|\\
 &=C(1+a(|x|^2))^{-1}|x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}|\\
 &=C(1+|x|^2e^{c|x|^2})^{-1}|x_1^{N_{i_1}}\cdots x_n^{N_{i_n}}|\\
 &\leq \tilde{C}(1+|x|^2)^{N}
\end{align*}
for some $\tilde{C}$ and $N$ constants.
I have a doubt:
$(te^{t})^{(k)}=e^{t}(t+k) $
It is correct the folowwing inequality?
$e^{t}(t+k)\leq C(k)te^{ct}$ for some $C(k)$ constant
