Derivative of Dirac Delta Function [closed]

I need to show that

$$\partial_{\mu}\delta(R^2(\tau)) ~=~ -\frac{R_{\mu}}{R \cdot u} \frac{\partial}{\partial \tau}\delta(R^2(\tau)),$$

where

$$R_{\mu} ~=~ x_{\mu} - r_{\mu}(t),$$

and

$$u_{\mu} ~=~ \frac{d}{d\tau}r_{\mu}.$$

I'm really just not at all sure how to approach this problem.

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Hi Mary, and welcome to Physics Stack Exchange! This is a site for conceptual questions about physics, not general homework help. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. In particular, what step of the process of taking these derivatives gives you trouble? See our FAQ and homework policy for more information. – David Zaslavsky May 11 '12 at 17:34

closed as too localized by David Zaslavsky♦May 11 '12 at 17:33

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.

Let $${\bf R} ~:=~ {\bf x} -{\bf r}(t), \qquad {\bf u}~:=~ \frac{d{\bf r}}{dt}, \qquad {\bf \nabla} ~:=~ \frac{\partial}{\partial {\bf x}}.$$
$$\left(({\bf R} \cdot {\bf u}) {\bf \nabla} +{\bf R}\frac{d}{dt}\right) \delta(R^2) ~=~\left(({\bf R} \cdot {\bf u}) 2{\bf R} +{\bf R}(2{\bf R}\cdot\frac{d{\bf R}}{dt})\right) \delta(R^2)$$ $$~=~2\left(({\bf R} \cdot {\bf u}) {\bf R} - {\bf R}({\bf R}\cdot {\bf u})\right) \delta(R^2)~=~0.$$