Z_{out} = - dV_{out} / dI_{out}
The minus sign in the derivative comes from the fact the output impedance has the effect of decreasing V_{out}. The output current I_{E,} which is related to the base current.
V_{in} = I_{B}R_{S} + V_{BE} +V_{E} [V_{out } = V_{E}]
V_{out} = V_{E} = V_{in} – I_{B}R_{S} – V_{BE} [I_{out} = I_{E} and I_{B} = I_{E}/(β + 1)]
V_{out} = V_{in} – {I_{E}/(β + 1)}R_{S} – V_{BE} = V_{in} – {I_{out}/( β + 1)}R_{S }– V_{BE}
= - I_{out}{R_{S}/(β + 1)} + (V_{in} – V_{BE})
So, Z_{out} = - dV_{out}/dI_{out} = -d/dI_{out} [ - I_{out}{R_{S}/(β + 1)} + (V_{in} – V_{BE}) ]
Z_{out }= R_{S}/(β + 1)
Thus we obtain the result that the impedance of the source, as viewed by the load, is reduced by the factor ~ 1/ β.
Z_{out} = {R_{S}/( β + 1) ≈ R_{S}/β |