Form of $SU(N)$ gauge transformations in $SU(N)$ Yang-Mills theory For $SU(N)$ Yang-Mills theory, instantons correspond to finite action solutions $A_\mu(x)$ of the Euclidean equation of motion. The requirement of finite action demands that $A_\mu(x)$ is a pure gauge at the boundary of $\mathbb{R}^4$ given by $$A_\mu(x)=-\frac{i}{g}(\partial_\mu U)U^{-1}$$ where $U\in SU(N)$. 
For $SU(2)$, therefore, we have, $$U(x)=\exp[i\theta_a(x)T^a]$$ where $T^a=\sigma^a/2$ are the generators of $SU(2)$ in the fundamental representation. However, $U$ is taken to be $$U=\frac{x_4+i\boldsymbol{\sigma}\cdot\textbf{x}}{\tau}$$ where $\tau^2=x_4^2+\textbf{x}\cdot\textbf{x}$ while studying instantons of class $n=1$. 
Is the latter expression of $U$ a special case of the former expression? In that case, how is the latter expression derived from the former?
 A: This has nothing to do with instantons or quantum field theory, it is just an elementary fact about $2\times 2$ matrices:
The Pauli matrices $\sigma^i$ together with the identity $\mathbf{1}_2$ form a basis of the vector space of 2-by-2 matrices. Therefore, $U(x)$, as a 2-by-2 matrix-valued function, may be written as
$$ U(x) = \zeta(x)\mathbf{1}_2 + \omega_i(x)\sigma^i$$
and your expression for $U(x)$ follows from the choices $\zeta(x) = x_4/r,\omega_i(x) = x_i/r$ for $r = \sqrt{x^2}$.
If you want to see how you need to choose the $\theta_a(x)$ in the exponential for $U(x)$ you wrote, simply use the standard relation
$$ \exp(\mathrm{i}\alpha\vec n\cdot\vec\sigma) = \mathbf{1}_2\cos(\alpha) + \mathrm{i}(\vec n\cdot\vec \sigma)\sin(\alpha)$$
and compare coefficients to get the $\theta_a = \alpha n_a$.
A: Of course, it is. The only condition one imposes on $U$ is that it is unitary. You can easily check that the latter matrix is.
\begin{equation}U^{\dagger}=\frac{x_4-\mathrm{i} \vec{\sigma} \cdot \vec{x}}{\sqrt{x_4^2+\vec{x}^2}};
\end{equation}
\begin{equation}U^{\dagger} \cdot U=\frac{x_4-\mathrm{i} \vec{\sigma} \cdot \vec{x}}{\sqrt{x_4^2+\vec{x}^2}} \cdot \frac{x_4+\mathrm{i} \vec{\sigma} \cdot \vec{x}}{\sqrt{x_4^2+\vec{x}^2}}=\frac{x_4^2+\vec{x}^2}{x_4^2+\vec{x}^2}=1.
\end{equation}
