Is there an easy way to see that all derivatives are bounded? 
Show that all derivatives of $f:\mathbb{R}\to\mathbb{R}$ given by $$f(x):=\frac{1}{\sqrt{x^2+1}+1}$$ are bounded.

It's easy to see that all derivatives are continuous. So the only potential problem is that a derivative might blow up at $\infty$. Is there an easy way to see that this does not happen? I guess deriving an explicit form of the n-th derivative is not the way to do it (I'd doubt there is an easy closed form).
 A: It is clear by composition that $f$ is smooth. So we need only prove that the derivatives are bounded on, say $A=\{x\;;|x|\geq 2\}$. Since $f$ is even we can restrict to $I=[2,+\infty)$.
This form of $f$ is not tractable. Observe that for $x\neq 0$, 
$$
f(x)=\frac{\sqrt{x^2+1}-1}{(\sqrt{x^2+1}+1)(\sqrt{x^2+1}-1)}=\frac{\sqrt{x^2+1}-1}{x^2}=\frac{\sqrt{x^2+1}}{x^2}-\frac{1}{x^2}.
$$
It is trivial to check that the derivatives of $\frac{1}{x^2}$ are all bounded on $I$, so we can focus on 
$$
g(x)=\frac{\sqrt{x^2+1}}{x^2}=\frac{1}{x}\sqrt{1+\frac{1}{x^2}}.
$$
Now we can use the power series representation of $(1+u)^{1/2}$ at $0$, whose radius of convergence is $1$, to get 
$$
g(x)=\frac{1}{x}\sum_{n\geq 0}\binom{1/2}{n}\left(\frac{1}{x^2}\right)^n=\sum_{n\geq 0}\binom{1/2}{n}x^{-2n-1} \quad \forall x\geq 2
$$
with normal convergence as $0\leq x^{-1}\leq \frac{1}{2}<1$. Term-by-term differentiation turns out well, as these yield after $k$ differentiations
$$
\frac{1}{x^{k+1}}\sum_{n\geq 0}\binom{1/2}{n}(-2n-1)(-2n-2)\cdots(-2n-k-1)(x^{-2})^n.
$$
Of course $\frac{1}{x^{k+1}}$ is bounded on $I$, and the remaining power series evaluated at $x^{-2}$ still has radius $1$. So the latter converges normally on $I$ for each $k$. Hence a simple induction justifies simultaneously term-by-term differentiation and boundedness on $I$.
A: If you know some complex analysis, you can just say something like
The function is analytic and bounded in the strip $|\Im z|\le 1/2$. Apply the Cauchy estimate in each circle of radius $1/2$ centered on the real line and conclude.
If you don't, but still strongly prefer to avoid any talks about "the general form of the expression for the $n$-th derivative", you can just say that the function $[0,\infty)\ni y\to \frac{1}{1+y}$ is bounded with all derivatives (explicit formula), so, by the chain rule, it will suffice to show that all derivatives of $F(x)=\sqrt{x^2+1}$ are bounded. However, $F(x)^2=x^2+1$, $F'(x)$ is bounded by an explicit formula, and, by the product rule for the $n$-th derivative with $n\ge 2$, we have
$$
2F^{(n)}F+\sum_{k=1}^{n-1}{n\choose k}F^{(k)}F^{(n-k)}=\begin{cases}2,&n=2\\0,&n>2\end{cases}
$$
from where the conclusion follows by induction using the inequality $F\ge 1$.
A: Hint: You don't need an explicit form for the derivatives, you only need their leading behaviour for large $x$. You could do this by considering what gives the largest contribution.
Warning - $\sin x^2$ is bounded but its derivative is not. You'll have to be slightly careful in what you throw away.

Let $u=\sqrt{1+x^2}\sim x$. Then $u'=x/u\sim 1$. As a result you can check that $u^{(n+1)}\sim x^{-n}$ since all derivatives are just polynomials in $x$ over powers of $u$ and there is no difference in order between differentiating with respect to $x$ or $u$.
Similarly you can deduce that any derivative of $1/(1+u)$ has the form of a sum of fractions which in turn have the form of a product of derivatives of $u$ in the numerator and a power of the denominator below. Then the numerator is at most order 1 and the denominator decays.

An alternative method is to write
$$f'(x)=-f(x)^2\times x/u$$
Then one can proceed by induction.
