I Derivative of the retarded vector potential

[Salmone]

In a problem of an oscillating electric dipole, under appropriate conditions, one can find, for the potential vector calculated at the point $\vec{r}$, the expression $\vec{A}=\hat{k}\frac{\mu_0I_0d}{4\pi}\frac{cos(\omega(t-r/c))}{r}$ where: $\hat{k}$ is the direction of the $z-axis$ where the dipole is oscillating, $I_0$ is the current ($I(t)=I_0cos(\omega t)$), $d$ is the distance between the charges of the dipole and $r$ is the distance between the origin of the system and the point where I want to calculate the potential vector. Let $\vec{p}=\hat{k}qd=\frac{\hat{k}dI_0}{\omega}sin(\omega t)$ be the dipole moment, it is possible to rewrite the potential vector as $\vec{A}=\frac{\mu_0}{4\pi}\frac{\vec{\dot p(t-r/c)}}{r}$ where $\vec{\dot p}$ is the derivative with respect to time.
If we use Lorentz Gauge we have $-\epsilon_0\mu_0\frac{\partial V}{\partial t}=\vec{\nabla} \cdot \vec{A}$, since $\vec{A}$ is directed along the $z-axis$ we have $div \vec{A}=\frac{\partial A}{\partial z}$ so $\frac{\partial V}{\partial t}=-\frac{1}{\epsilon_0 \mu_0}\frac{\partial A}{\partial z}$.

The question: how do you do this last derivative?

The result is: $\frac{\partial V}{\partial t}=\frac{1}{4\pi\epsilon_0} \left( \frac{- \ddot p(t-r/c)}{cr}+\frac{\dot p(t-r/c)}{r^2} \right)\frac{z}{r}$

 

[jasonRF]

When you took calculus, did you learn the chain rule?