We again assume that the degradation can be modelled as exponential decay with characteristic time $\tau_\text{X}$. We start with protein formation with an inhibiting transcription factor R. Then:

$$\begin{aligned} \text{formation rate of the protein X} &= \frac{V_{\max}\cdot K_m}{K_m+[\text{R}]}\\ \text{degradation rate of the protein X} &= \frac{[\text{X}]}{\tau_\text{X}} \end{aligned}$$

The differential equation for the concentration of the protein X is:

$$\begin{aligned}\frac{\dd[\text{X}]}{\dd t}&= \text{formation rate}- \text{degradation rate}\\ &= \frac{V_{\max}\cdot K_m}{K_m+[\text{R}]}-\frac{[\text{X}]}{\tau_\text{X}}\end{aligned}$$ This again is a differential equation of limited exponential growth. We first determine the steady state solution $[\text{X}_{ss}$ by setting $\frac{\dd[\text{X}]}{\dd t}$ equal to zero: $$[\text{X}_{ss}]=\frac{V_{\max}\cdot K_m}{K_m+[\text{R}]}\cdot \tau_\text{X}$$ The differential equation can be rewritten as $$\frac{\dd[\text{X}]}{\dd t}=\frac{1}{\tau_X}\left([\text{X}_{ss}-[\text{X}]\right)$$ Then the time course of the concentration is $$[\text{X}]=[\text{X}_{ss}]\cdot\left(1-e^{-\frac{t}{\tau_\text{X}}}\right)$$