# About the curl of B using Biot-Savart Law

1. Dec 12, 2009

### yanyan_leung

I am reading the Griffiths about finding the curl of B using Biot-Savart Law. I do not understanding the step between equation (5.52) and (5.53) which finding the x components of the following:
$$(\boldsymbol{J}\cdot\nabla^{\prime})\dfrac{\hat{\xi}}{\xi^{2}}$$
where
$$\mathbf{\mathbf{\xi}}=\mathbf{r}-\mathbf{r}^{\prime}$$
I don't know why it said using product rule 5 in his book, can get the following result:
$$\left(\boldsymbol{J}\cdot\nabla^{\prime}\right)\dfrac{x-x^{\prime}}{\xi^{3}}=\nabla^{\prime}\cdot\left[\dfrac{(x-x^{\prime})}{\xi^{3}}\mathbf{J}\right]-\left(\dfrac{x-x^{\prime}}{\xi^{3}}\right)\left(\nabla^{\prime}\cdot\mathbf{J}\right)$$
Is there anyone know how to deduce this result?

2. Dec 12, 2009

### tiny-tim

Welcome to PF!

Hi yanyan_leung ! Welcome to PF!
That's ∑i (Ji∂/∂xi) (f) = ∑i ∂/∂xi (fJi) - f ∑i ∂/∂xi (Ji) …

now just use the product rule on the middle term.