= Solution
For the <parameter sensitivity of autorepressed expression>, unregulated <Gene expression>[production] has $Y_{\rm st}=\beta/\alpha$, so its logarithmic <Sensitivity analysis>[sensitivity] is one. In the strong Hill approximation, $Y_{\rm st}=(\beta K^n/\alpha)^{1/(n+1)}$, with $K,\alpha,n$ held fixed. Hence
$$
\boxed{S(Y_{\rm st},\beta)=1\quad\text{without feedback},\qquad
S(Y_{\rm st},\beta)=\frac1{n+1}\quad\text{with strong Hill repression}.}
$$
The ideal logic level $Y_{\rm st}\simeq K$ instead gives $S\simeq0$, because changing <Gene expression>[production] primarily changes the switching duty fraction, not the threshold. Finite Hill repression has a residual <Gene expression>[production] <Sensitivity analysis>[sensitivity].
The exact Hill <Steady state>[steady-state] relation makes the comparison continuous. Write $s=(Y_{\rm st}/K)^n$; then $\beta=\alpha Y_{\rm st}(1+s)$ and $d\beta/dY_{\rm st}=\alpha[1+(n+1)s]$. Thus
$$
\boxed{S(Y_{\rm st},\beta)=\frac{1+s}{1+(n+1)s}.}
$$
It tends to one for weak repression and to $1/(n+1)$ for strong repression. The calculation concerns a local relative parameter change, not the response to an arbitrary large <Gene expression>[production] perturbation.
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