Resistance of a semiconductor diode
For the ideal resistor, the voltage across is proportional to the current through and thus, their ratio is the constant \$R\$:

$$\frac{v_R}{i_R} = R $$

For the ideal (semiconductor) diode, we have

$$i_D = I_S(e^{\frac{v_D}{nV_T}}-1)$$

Inverting yields

$$v_D = nV_T\ln (1 + \frac{i_D}{I_S}) $$

thus, the diode voltage is not proportional to the diode current, i.e., the ratio of the voltage and current is not a constant.

$$\frac{v_D}{i_D} = \frac{nV_T}{i_D}\ln (1 + \frac{i_D}{I_S}) \ne R$$

Now, the small-signal or dynamic resistance is just

$$\frac{dv_D}{di_D} = \frac{nV_T}{I_S + i_D} \approx \frac{nV_T}{i_D} $$


  how is it different from the normal resistance


As shown above, the diode static resistance (ratio of the diode voltage and current) differs from and is, in fact, larger than the diode dynamic resistance by the factor of \$\ln (1 + \frac{i_D}{I_S})\$

$$\frac{v_D}{i_D} =  \frac{dv_D}{di_D} \ln (1 + \frac{i_D}{I_S})$$ 

which is to say that, in typical operating ranges, the diode dynamic resistance is much smaller than then diode static resistance.


  Does the power dissipation relation, \$P=I^2r\$ hold in case of dynamic
  resistances?


The instantaneous power associated with the diode is

$$p_D = v_D i_D = nV_Ti_D\ln (1 + \frac{i_D}{I_S}) \ne i_D^2\frac{nV_T}{i_D} = nV_Ti_D $$

Since the power associated with a circuit element is always the product of the voltage across and current through, one would not use the dynamic resistance but, rather the static resistance.