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 P: 3 OK I found the answer. To get rid of the derivatives of Dirac deltas, I used the identity $f(x) \delta ' (x-y) = f(y) \delta ' (x-y) - f'(x) \delta (x-y)$ which follows from the convolution between delta and $(f \times \varphi)'$ ($\varphi$ is a test function). This also produces the remaining wanted term.