transistors - Stability factors of a voltage divider bias circuit

I am trying to derive the stability factors of a voltage divider bias circuit. I could derive S,S', however, I'm facing some problem with the derivation of S''. Please help me in this regard.

How to find \$\frac{\partial I_C}{\partial \beta}\$?

The equations given in my book are these:

For a voltage divider circuit:

Kirchhoff's voltage law at the base circuit gives:

$$V_T=I_BR_T+V_{BE}+(I_B+I_C) R_E \tag1$$ $$I_B=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E} \tag2$$ The collector current \$ I_C \$ is given by: $$I_C=\beta I_B+(1+\beta)I_{CO} \tag3$$

Substituting \$ I_B \$ from 2nd eq to 3rd eq yields:

$$I_C=\beta\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E}+(1+ \beta)I_{CO} \tag4$$

$$ I_C(1+\frac{\beta R_E}{R_T+R_E})=\frac{\beta(V_T-V_{BE})}{R_T+R_E}+(1+\beta)I_{CO} \tag {4a}$$

I understand and verified that the above equations came as a result of manipulating equations derived from Kirchhoff's voltage law and other current equations.

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In my book, they haven't derived the following equations(5,6,7), but have jotted down the equations as given below:

Then the stability factor can thus be expressed as: $$S=\frac{\partial I_C}{\partial I_{CO}}=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}} \tag5$$

$$S'=\frac{\partial I_C}{\partial V_{BE}}=-\frac{\beta}{(1+\beta)R_E+R_T} \tag6$$

$$S''=\frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)-\beta SR_E}{R_T+R_E}+SI_{CO} \tag7 ]$$

I could derive equation eq 5 and 6 however I am facing the following issues with equation 7:

1) I am getting $$ \frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)-\beta SR_E}{R_T+R_E}-SI_{CO} $$ instead of $$\frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)+\beta SR_E}{R_T+R_E}+SI_{CO} $$

2) Why do we need to devive equation (7) ,in which stability factor \$S''\$ is in terms of other stability factors. Why wouldn't it be sufficient to say that \$ S'' \$ is given by eq 9a? or $$\frac{\partial I_C}{\partial \beta}=\frac{V_T-V_{BE}-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{I_{CO}}{(1+\frac{\beta R_E}{R_T+R_E})}$$

3) How does eq 7 or 9a account for the fact that the Voltage divider circuit provides better stability against \$ \beta \$.?

4)If anyone has come across these equations, eq 5,6 and 7 in any standard book, then please mention the name of the book.

DERIVATION(MY ATTEMPT):

Partial differentiating eq 4a with respect to \$ \beta \$:

$$\frac{\partial I_C}{\partial \beta}(1+\frac{\beta R_E}{R_T+R_E})+I_C\frac{R_E}{R_T+R_E}=\frac{V_T-V_{BE}}{R_T+R_E}+I_{CO} \tag 8$$ $$\frac{\partial I_C}{\partial \beta}(1+\frac{\beta R_E}{R_T+R_E})=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E}+I_{CO} \tag 9$$

$$\frac{\partial I_C}{\partial \beta}=\frac{V_T-V_{BE}-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{I_{CO}}{(1+\frac{\beta R_E}{R_T+R_E})}\frac{1+\beta}{1+\beta} \tag {9a}$$

From eq 4a:

$$ I_C=\frac{\beta(V_T-V_{BE}}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}\frac{1+\beta}{1+\beta}+\frac{(1+\beta)I_{CO}}{(1+\frac{\beta R_E}{R_T+R_E})} \tag {10}$$

$$I_C=\frac{\beta(V_T-V_{BE})S}{(1+\beta)(R_T+R_E}+SI_{CO} \tag{11}$$ $$=> (V_T-V_{BE})=\frac{(1+\beta)(R_T+R_E)}{\beta S}(I_C-SI_{CO}) \tag{12}$$

Substituting eq 12 in eq 10

$$\frac{\partial I_C}{\partial\beta}=\frac{\frac{(1+\beta)(R_T+R_E)}{\beta S}(I_C-SI_{CO})-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{S}{1+\beta} I_{CO} \tag{13}$$

$$ \frac{\partial I_C}{\partial \beta}=\frac{\frac{(1+\beta)(R_T+R_E)}{\beta S}I_C-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{S}{1+\beta}I_{CO}-\frac{\frac{(1+\beta)(R_T+R_E)}{\beta S}SI_{CO}}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})} \tag{14}$$

$$=\frac{1+\beta}{1+\beta}\frac{(1+\beta)(R_T+R_E)I_C-\beta SR_EI_C}{\beta S(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{S}{1+\beta}I_{CO}-\frac{S}{\beta}I_{CO} \tag{15}$$

$$=\frac{SI_C[(1+\beta)(R_T+R_E)-\beta SR_E}{\beta(1+\beta)S(R_T+R_E)}+SI_{CO}[\frac{\beta-1-\beta}{\beta(1+\beta)}] \tag{16}$$

$$\frac{\partial I_C}{\partial \beta}=\frac{I_C[(1+\beta)(R_T+R_E)-\beta SR_E}{\beta(1+\beta)(R_T+R_E)}-\frac{SI_{CO}}{\beta(1+\beta)} \tag{17}$$

$$\frac{\partial I_C}{\partial \beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)-\beta SR_E}{R_T+R_E}-SI_{CO} ]\tag{18}$$

The sign of the last term should be +ve. Where might have I gone wrong?