Are Vector Differential Identities in Electromagnetic Theory Consistent?

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Vector differential identities!

In chapter 20 of "Foundations of Electromagnetic theory" by Reitz,Milford and Christy,there is calculation which seems to make use of:[itex]\vec{\nabla}\times\dot{\vec{p}}=\Large{\frac{\vec{r}}{r}\times\frac{ \partial \dot{\vec{p}}}{\partial r}}[/itex] where [itex]\dot{\vec{p}}=\large{\frac{d}{d \tau}} \int_V \vec{r}'\rho(\vec{r}',t-\frac{r}{c})dv' \ (\tau=t-\frac{r}{c})[/itex].But I can't prove it and worse is that it seems to be inconsistent with the formula for curl in spherical coordinates.
There is also another identity mentioned in the problems of chapter 1 which seems as strange:
[itex]\vec{\nabla}\cdot\vec{F}(r)=\large{\frac{\vec{r}}{r}\cdot\frac{d\vec{F}}{dr}}[/itex]

Is there any suggestion?
Thanks
 
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Your second identity doesn't seem strange if there is no theta or phi component to F:
[tex] \operatorname{div}\, \mathbf F <br /> = \nabla\cdot\mathbf F <br /> = \frac1{r^2} \frac{\partial}{\partial r}(r^2 F_r) + \frac1{r\sin\theta} \frac{\partial}{\partial \theta} (\sin\theta\, F_\theta) + \frac1{r\sin\theta} \frac{\partial F_\phi}{\partial \phi}.[/tex]

then it would reduce to:

[tex] \operatorname{div}\, \mathbf F <br /> = \nabla\cdot\mathbf F <br /> = \frac1{r^2} \frac{\partial}{\partial r}(r^2 F_r)[/tex]

and then if it was a very large r value you'd be left with your identity.
 
Both are simple chain rule consequences, i'll
illustrate with the divergence since it's quicker
[tex]\nabla \cdot {\bf F}(r) = \sum_i \frac{\partial F_i(r)}{\partial x^i}<br /> = \sum_i \frac{\partial F_i(r)}{\partial r} \frac{\partial r}{\partial x^i}[/tex]
now [itex]\frac{\partial r}{\partial x^i}=\frac{x^i}{r}[/itex], so
[tex] \nabla \cdot {\bf F}(r) <br /> = \sum_i \frac{\partial F_i(r)}{\partial r} \frac{x^i}{r} = \frac{{\bf r}}{r} \cdot \frac{\partial {\bf F}(r)}{\partial r}.[/tex]

As an aside to get the result from the spherical formula you have to keep all three terms.
even though [itex]{\bf F}[/itex] only depends on [itex]r[/itex], when you break it into spherical
components, for example [itex]F_\theta = {\widehat \theta} \cdot {\bf F}[/itex], depends on
[itex]r, \theta[/itex] and [itex]\phi[/itex].