The retarded potentials are
V( r , t)=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho\left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime}, \quad A ( r , t)=\frac{\mu_{0}}{4 \pi} \int \frac{ J \left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime} .
We want a source localized at the origin, so we expand in powers of r ^{\prime}, keeping terms up to first order:
ᴫ ^{2}=\left( r – r ^{\prime}\right) \cdot\left( r – r ^{\prime}\right)=r^{2}-2 r \cdot r ^{\prime}+\left(r^{\prime}\right)^{2} ,
ᴫ \approx r\left(1-\frac{ r \cdot r ^{\prime}}{r^{2}}\right) ,
\frac{1}{ᴫ} \approx \frac{1}{r}\left(1+\frac{ r \cdot r ^{\prime}}{r^{2}}\right) ,
t-\frac{ᴫ}{c} \approx t-\frac{r}{c}+\frac{ r \cdot r ^{\prime}}{r c}=t_{0}+\frac{ r \cdot r ^{\prime}}{r c} ,
where t_{0} \equiv t-r / c is the retarded time for a source at the origin. Thus (Taylor expanding about t_{0})
\rho\left( r ^{\prime}, t_{r}\right)=\rho\left( r ^{\prime}, t-\frac{ᴫ}{c}\right) \approx \rho\left( r ^{\prime}, t_{0}\right)+\frac{ r \cdot r ^{\prime}}{r c} \dot{\rho}\left( r ^{\prime}, t_{0}\right) ,
\frac{\rho\left( r ^{\prime}, t_{r}\right)}{ᴫ}=\frac{1}{r}\left(1+\frac{ r \cdot r ^{\prime}}{r^{2}}\right)\left[\rho\left( r ^{\prime}, t_{0}\right)+\frac{ r \cdot r ^{\prime}}{r c} \dot{\rho}\left( r ^{\prime}, t_{0}\right)\right] \approx \frac{1}{r} \rho\left( r ^{\prime}, t_{0}\right)+\frac{ r \cdot r ^{\prime}}{r^{3}} \rho\left( r ^{\prime}, t_{0}\right)+\frac{ r \cdot r ^{\prime}}{r^{2} c} \dot{\rho}\left( r ^{\prime}, t_{0}\right) ,
and hence
V( r , t)=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{1}{r} \int \rho\left( r ^{\prime}, t_{0}\right) d \tau^{\prime}+\frac{ r }{r^{3}} \cdot \int r ^{\prime} \rho\left( r ^{\prime}, t_{0}\right) d \tau^{\prime}+\frac{ r }{r^{2} c} \cdot \int r ^{\prime} \dot{\rho}\left( r ^{\prime}, t_{0}\right) d \tau^{\prime}\right] .
The first integral is the total charge of the dipole, which is zero; the second integral is the dipole moment, and the third is the time derivative of the dipole moment:
V( r , t)=\frac{1}{4 \pi \epsilon_{0}} \frac{\hat{ r }}{r^{2}} \cdot\left[ p \left(t_{0}\right)+\frac{r}{c} \dot{ p }\left(t_{0}\right)\right] .
By the same reasoning, the vector potential is
A ( r , t)=\frac{\mu_{0}}{4 \pi}\left[\frac{1}{r} \int J \left( r ^{\prime}, t_{0}\right) d \tau^{\prime}+\frac{1}{r^{3}} \int\left( r \cdot r ^{\prime}\right) J \left( r ^{\prime}, t_{0}\right) d \tau^{\prime}+\frac{1}{r^{2} c} \int\left( r \cdot r ^{\prime}\right) \dot{ J }\left( r ^{\prime}, t_{0}\right) d \tau^{\prime}\right].
From Eq. 5.31 1 \int J \left( r ^{\prime}, t\right) d \tau^{\prime}=\dot{ p }(t), \text { and } J =\rho v ^{\prime}=\rho\left(d r ^{\prime} / d t\right) \text { is already first order in } r^{\prime} , so the second and third integrals are second order, and we drop them.
\int_{ \nu } J d \tau=d p / d t (5.31)
A ( r , t)=\frac{\mu_{0}}{4 \pi}\left[\frac{\dot{ p }\left(t_{0}\right)}{r}\right] .
In calculating the fields, remember that the dipole moments themselves depend on r, through their argument \left(t_{0}=t-r / c\right). For a function of t_{0} alone,
\nabla f\left(t_{0}\right)=\frac{d f}{d t_{0}} \nabla t_{0}=\dot{f}\left(-\frac{1}{c} \nabla r\right)=-\frac{1}{c} \dot{f} \hat{ r } ,
so you simply replace \nabla \text { by }-\frac{\hat{ r }}{c} \frac{d}{d t} . \text { For a structure of the form }\left[f\left(t_{0}\right)+\frac{r}{c} \dot{f}\left(t_{0}\right)\right]
\nabla \left[f\left(t_{0}\right)+\frac{r}{c} \dot{f}\left(t_{0}\right)\right]=-\frac{\hat{ r }}{c} \dot{f}+\frac{r}{c}\left(-\frac{\hat{ r }}{c} \ddot{f}\right)+\frac{\dot{f}}{c} \hat{ r }=-\frac{ r }{c^{2}} \ddot{f} .
Using the product rule for ∇(A · B):
\nabla V=\frac{1}{4 \pi \epsilon_{0}}\left\{\frac{\hat{ r }}{r^{2}} \times\left( \nabla \times\left[ p +\frac{r}{c} \dot{ p }\right]\right)+\left[ p +\frac{r}{c} \dot{ p }\right] \times\right. \left.\left( \nabla \times \frac{\hat{ r }}{r^{2}}\right)+\left(\frac{\hat{ r }}{r^{2}} \cdot \nabla \right)\left[ p +\frac{r}{c} \dot{ p }\right]+\left(\left[ p +\frac{r}{c} \dot{ p }\right] \cdot \nabla \right) \frac{\hat{ r }}{r^{2}}\right\}
=\frac{1}{4 \pi \epsilon_{0}}\left\{\frac{\hat{ r }}{r^{2}} \times\left(-\frac{ r }{c^{2}} \times \ddot{ p }\right)+ 0 +\left[\frac{\hat{ r }}{r^{2}} \cdot\left(-\frac{ r }{c^{2}}\right)\right] \ddot{ p }+\frac{1}{r^{3}}\left(\left[ p +\frac{r}{c} \dot{ p }\right]-3 \hat{ r }\left(\hat{ r } \cdot\left[ p +\frac{r}{c} \dot{ p }\right]\right)\right)\right\}
=\frac{1}{4 \pi \epsilon_{0}}\left\{-\frac{\hat{ r }(\hat{ r } \cdot \ddot{ p })}{r c^{2}}+\frac{[ p +(r / c) \dot{ p }]-3 \hat{ r }(\hat{ r } \cdot[ p +(r / c) \dot{ p }])}{r^{3}}\right\} .
The others are easier:
\nabla \times A =\frac{\mu_{0}}{4 \pi}\left[\frac{1}{r}( \nabla \times \dot{ p })-\dot{ p } \times\left( \nabla \frac{1}{r}\right)\right]=\frac{\mu_{0}}{4 \pi}\left[\frac{1}{r}\left(-\frac{\hat{ r }}{c} \times \ddot{ p }\right)+\left(\dot{ p } \times \frac{\hat{ r }}{r^{2}}\right)\right]=-\frac{\mu_{0}}{4 \pi}\left\{\frac{\hat{ r } \times[(\dot{ p }+(r / c) \ddot{ p }]}{r^{2}}\right\} ,
\frac{\partial A }{\partial t}=\frac{\mu_{0}}{4 \pi}\left(\frac{\ddot{ p }}{r}\right) .
The fields are therefore
E ( r , t)=- \nabla V-\frac{\partial A }{\partial t}=-\frac{\mu_{0}}{4 \pi}\left\{\frac{\ddot{ p }-\hat{ r }(\hat{ r } \cdot \ddot{ p })}{r}+c^{2} \frac{[ p +(r / c) \dot{ p }]-3 \hat{ r }(\hat{ r } \cdot[ p +(r / c) \dot{ p }])}{r^{3}}\right\} ,
B ( r , t)= \nabla \times A =-\frac{\mu_{0}}{4 \pi}\left\{\frac{\hat{ r } \times[\dot{ p }+(r / c) \ddot{ p }]}{r^{2}}\right\}
(with all dipole moments evaluated at the retarded time t_{0}=t-r / c).