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\end{equation}  \begin{equation}  k(t)\equiv(\frac{d\frac{1}{(S(t)^{n-1)}}}{dt})  \end{equation} When it comes to mixed second order reactions, the rate law does not allow a survival function to be obtained for the process because the rate law only allows the survival function of one reactant to be looked at. The rate law of a mixed second order reaction is  \begin{equation}  \int\frac{dx}{([A_0]-x)([B_0]-x)}=k(t)   \end{equation}  Taking the time derivative of the left hand side gives   \begin{equation}  \int\frac{\frac{dx}{dt}}{{([A_0]-x)([B_0]-x)}}dt=k(t)  \end{equation}  And from that definition of k(t), we get the inequality.