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Apr 23 at 9:09 vote accept Nyquist-er
Apr 22 at 23:09 comment added AHT @Nyquist-er, yes, that's correct. In general, the convolution and differentiator operator commute, i.e., $D[f*g] = D[f]*g = f*D[g]$ if $f$ and $g$ are absolutely integrable and at least one of them has an absolutely integrable (L1) weak derivative. In signal processing, these conditions are typically hold. Perhap, you may want to say that in the second last equality, the time-invariance property is used.
Apr 22 at 21:49 history answered AHT CC BY-SA 4.0