= Shift derivative of a step response
{title2=$-\partial_b[H(t-b)K(t-b)]=H(t-b)K'(t-b)$}
For a step-response kernel with $K(0)=0$, differentiating the start time produces no delta term from the response itself, because its coefficient is zero. The negative shift derivative is the impulse response. More generally an additional $\delta(t-b)K(0)$ term appears if the kernel has a nonzero initial value. The identity holds distributionally as well as for the ordinary smooth pieces.
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