We have the following equations:
$$V_T=I_BR_T+V_{BE}+(I_B+I_C) R_E \tag1$$
$$=>I_B=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E} \tag2$$
$$I_C=\beta I_B+(1+\beta)I_{CO} \tag3$$
Substituting $I_B$ from 2nd eq to 3rd eq yields:
$$I_C=\beta\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E}+(1+ \beta)I_{CO} \tag4$$
$$I_C(1+\frac{\beta R_E}{R_T+R_E})=\frac{\beta(V_T-V_{BE})}{R_T+R_E}+(1+\beta)I_{CO} \tag {4a}$$
Also given that:
$$S=\frac{\partial I_C}{\partial I_{CO}}=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}} \tag5$$
$$S'=\frac{\partial I_C}{\partial V_{BE}}=-\frac{\beta}{(1+\beta)R_E+R_T} \tag6$$
How can I differentiate $I_C$ w.r.t $\beta$ to obtain
$$S''=\frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)+\beta SR_E}{R_T+R_E}+SI_{CO} \tag7 ]$$
provided $I_{CO}$ and $V_{BE}$ are constant?
Rewrite $$I_C(1+\frac{\beta R_E}{R_T+R_E})=\frac{\beta(V_T-V_{BE})}{R_T+R_E}+(1+\beta)I_{CO} \tag {4a}$$ as $$I_C(1+a{\beta})=b \beta+(1+\beta)I_{CO}\implies I_C=\frac{b \beta+(1+\beta)I_{CO} }{1+a{\beta} } $$ and go on.