= Solution
Use the <shift derivative of a step response>, writing $y_3(t,b)=H(s)K(s)$ with $s=t-b$. Differentiate with respect to the forcing location:
$$
-\frac{\partial y_3}{\partial b}=\delta(s)K(s)+H(s)K'(s).
$$
The possible delta term vanishes because $K(0)=0$, while $K'(s)=g(s)$. Therefore, as a distribution and also as the ordinary piecewise response,
$$
\boxed{-\frac{\partial y_3(t,b)}{\partial b}=H(t-b)g(t-b)=y_2(t,b)}.
$$
Moving the start of a unit step differentiates its forcing into a negative impulse. The zero initial conditions and <linearity> ensure that the corresponding responses obey the same identity, including the zero-frequency cases.
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