I need to compute second distributional derivative of the function
$$ g(x) = cos|x-2|, $$
but I'm not sure about my solution.
\begin{align} \left<g'', \varphi \right> = \left<g, \varphi''\right> &= \int^{\infty}_{-\infty} \underbrace{cos|x-2|}_u \cdot \underbrace{\varphi''(x)}_{v'} dx \\ &= \Big| \begin{array}{@{}cc@{}} u=cos|2-x| & u'=sin(2-x) \\ v = \varphi'(x) & v'= \varphi''(x) \end{array} \Big|\\ &=\underbrace{\left[ cos|x-2| \cdot \varphi'(x) \right]^\infty_{-\infty}}_0 - \int^\infty_{-\infty} \underbrace{sin(2-x)}_u \cdot \underbrace{\varphi'(x)}_{v'} dx\\ &= \Big| \begin{array}{@{}cc@{}} u=sin(2-x) & u'=- cos(2-x) \\ v = \varphi(x) & v'= \varphi'(x) \end{array} \Big|\\ &= - \underbrace{\left[ sin(2-x) \cdot \varphi(x) \right]^\infty_{-\infty}}_0 + \int^\infty_{-\infty} - cos(2-x) \cdot \varphi(x) dx \end{align}
So, is $- cos(2-x)$ the correct distributional derivative?
Your computations are exact but you can bypass them.
In fact, there is a general result that says that, till its $n$th derivative (including it), the distributional derivatives and the ordinary derivatives of a $C^n$ function coincide.
In this case where in fact we deal with a $C^{\infty}$ function ($\cos(|x-2|)$ is identical with $\cos(x-2)$), there will be no problem at any order of derivation.