Understanding a proof related to mollifiers

Question

I am reading the proof of the following statement

If $D^{\alpha}f$ exists weakly then $D^{\alpha}(f*\eta^{\epsilon}) = D^{\alpha}f* \eta^{\epsilon}$

https://www.math.ucdavis.edu/~hunter/pdes/pde_notes.pdf (page 55, bottom)

We have;

$ D^{\alpha}\eta^{\epsilon}*f = \int D^{\alpha}_x\eta^{\epsilon}(x-y)f(y)$. How do I get the next step that

$\int D^{\alpha}_x\eta^{\epsilon}(x-y)f(y) = (-1)^{|\alpha|}\int D^{\alpha}_y\eta^{\epsilon}(x-y)f(y)$

Answer 1

Its just chain rule, I'll state a 1D version: $$ \frac{d}{dx}[f(x-y)] =f'(x-y) = -\frac{d}{dy}[f(x-y)]$$ since $$\frac{d}{dy}[f(x-y)] = f'(x-y) \cdot \underbrace{\frac{d}{dy}(x-y)}_{=-1}$$