Given $$ g(\tau_e, t, \vec x, \vec \zeta) \equiv \tau_e - t + \frac{1}{c_0} |\vec x - \vec y(\vec \zeta, \tau_e)| = 0 $$

The textbook "Aeroacoustics" said applying for chain rule, will have $$ \left(\frac{\partial g}{\partial x_j}\right)_{\tau_e=const} + \left(\frac{\partial g}{\partial \tau_e}\right)_{\vec x=const}\frac{\partial \tau_e}{\partial x_j} =0$$

How did it come?

  • $\begingroup$ You can get parentheses (and other paired delimiters) to adjust to the size of their content by preceding them with \left and \right. $\endgroup$
    – joriki
    Commented Jul 14, 2016 at 19:41
  • $\begingroup$ What's being held constant in $\frac{\partial\tau_e}{\partial x_j}$? $\endgroup$
    – joriki
    Commented Jul 14, 2016 at 19:42
  • $\begingroup$ Also, what do you mean by $\left(\frac{\partial g}{\partial x_j}\right)_{\tau_e=\text{const}}$ when $g$ depends on further variables beyond $\vec x$ and $\tau_e$? $\endgroup$
    – joriki
    Commented Jul 14, 2016 at 19:43
  • $\begingroup$ @joriki, thanks for the comments! For your second, question, it was not written in the textbook...I copied exactly what has been said in the textbook to here. $\endgroup$ Commented Jul 14, 2016 at 19:52

1 Answer 1


I guess it was because

assuming $$\vec y = \text{const}$$ $$ \vec \zeta = \text{const}$$

so consider $\tau_e$ as a function of $x_j$, the result is the got.

  • 1
    $\begingroup$ I think it's best to add this to the question, instead of answering. $\endgroup$
    – rubik
    Commented Jul 14, 2016 at 19:53

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