Lienard-Wiechert Potential derivation, chain rule

[Pnin]

Homework Statement:

I am reading a book on my own

Relevant Equations:

chain rule

I want to follow the Lienard-Wiechert potential derivation in Robert Wald's E-M book, page 179. I do not understand $$dX(t_\text{ret})/dt$$ on the right side. I assume the chain rule is applied, but I can't see how.

$$ \frac{\partial[x'^i - X^i(t - |\mathbf x - \mathbf x'|/c)]}{\partial x'^j} = \delta^i{}_j - \frac{x^j - x'^j}{c|\mathbf x - \mathbf x'|} \frac{dX^i}{dt}(t_\text{ret}) $$

$$ t_\text{ret} = t - \frac 1c |\mathbf x - \mathbf x'| $$


∂[x′i−Xi(t−|x−x′|/c)]∂x′j=δij−xj−x′jc|x−x′|dXidt(tret)

tret=t−1c|x−x′|

 

[hutchphd]

Homework Statement:: I am reading a book on my own
Relevant Equations:: chain rule

I want to follow the Lienard-Wiechert potential derivation in Robert Wald's E-M book, page 179. I do not understand $$dX(t_\text{ret})/dt$$ on the right side. I assume the chain rule is applied, but I can't see how.

$$ \frac{\partial[x'^i - X^i(t - |\mathbf x - \mathbf x'|/c)]}{\partial x'^j} = \delta^i{}_j - \frac{x^j - x'^j}{c|\mathbf x - \mathbf x'|} \frac{dX^i}{dt}(t_\text{ret}) $$

$$ t_\text{ret} = t - \frac 1c |\mathbf x - \mathbf x'| $$


∂[x′i−Xi(t−|x−x′|/c)]∂x′j=δij−xj−x′jc|x−x′|dXidt(tret)

tret=t−1c|x−x′|

Did you read and understand footnote 19?