Diffeomorphisms of $\mathbb{R}^n$ I recently realized that I don't know any non-linear diffeomorphisms of the plane (or $\mathbb{R}^n$ in general) except for linear ones, so I want to ask rather broad questions hoping to be pointed to the appropriate literature.
1) Are there simple ways of constructing autodiffeomorphisms of $\mathbb{R}^n$ that can be expressed in closed form? UPD: Ok, there's e.g. $(x, y) \mapsto (x, y + f(x))$, where $f: \mathbb{R} \to \mathbb{R}$ is smooth, so I call this one back - kind of, if you know some exciting and unusual family, feel free to share :)
2) Is every autodiffeomorphism of $\mathbb{R}^n$ isotopic to a linear one? Obviously, every one is homotopic to any other due to $\mathbb{R}^n$ being contractible, but since $\mathrm{GL}(\mathbb{R}^n)$ is not connected, $\mathbb{R}^n$ is a k-space), and homotopies behave nicely under differentials at a point, even some linear autodiffeomorphisms are not isotopic, if I'm not mistaken.
 A: The answer to your question (2) is yes. 
The proof goes like this.  Let $f$ be a diffeomorphism.  We find an isotopy from $f$ to a diffeomorphism $g$ with $g(0)=0$.  The isotopy is
$$(x,t) \longmapsto f(x)-tf(0) $$
when $t=0$ this is $f$, when $t=1$ you get $f(x)-f(0)$. 
Next, given a diffeo $g$ with $g(0)=0$ we isotope $g$ to a linear diffeomorphism.  The map is this:
$$(x,t) \longmapsto \frac{g((1-t)x)}{1-t}$$
for $t \in [0,1)$ and at $t=1$ we have
$$(x,1) \longmapsto Dg_0(x)$$
You can check this map is continuous provided $g$ is $C^1$. 
Regarding your 1st question, one of the most common techniques is to integrate vector fields.  
A: Consider $\mathbb{R}^2$ first. Let $f$ be a smooth function on $\mathbb{R}^2$. If we consider a matrix
$$ \begin{bmatrix}
\cos(f(x,y)) &  -\sin(f(x,y))\\
 \sin(f(x,y)) &  \cos(f(x,y))
\end{bmatrix}$$ This is non linear map and it gives a diffeomorphism on $\mathbb{R}^2$. Using these $2\times2$ blocks we can build diffeomorphisms on $\mathbb{R}^n$ 
just like we build rotations from the $2\times2$ blocks. We can use hyperbolic cos and sine also.
$$ \begin{bmatrix}
\cosh(f(x,y)) &  \sinh(f(x,y))\\
 \sinh(f(x,y)) &  \cosh(f(x,y))
\end{bmatrix}$$
Another set of non linear diffeomorphisms are given by,
$$\begin{bmatrix} 1 & f(x,y)\\0 & 1\end{bmatrix}$$ we can choose any non zero constants on the diagonals instead of $1$ and can have lower triangular matrices also. Product of all these types also gives non linear diffeomorphisms.
