Question on a part of the Euler-Lagrange-Equation.

Do the parentheses after the differentiation operator

$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial f'}\right)$

have any syntactic meaning or is it just optional, style, etc.?

I have seen the E-L-Eq. both with those parens and without.


  • $\begingroup$ Usually they mean same. $\endgroup$
    – zkutch
    Jul 16, 2022 at 8:49
  • $\begingroup$ I like to use parentheses to emphasize the application of an operator to its argument (specifically, prior to evaluation). So $\partial_i(f)$ differs from $f_i$ in that the operator has already been applied to $f$ in the $f_i$ expression, even though $\partial_i(f) = f_i$. An easy example being $f(x,y) = xy$ with $\partial_x(f) = \partial_x(xy) = y = f_x$. Stylistic i.m.o. $\endgroup$
    – Kevin
    Jul 16, 2022 at 10:18
  • $\begingroup$ @von spatz: Please feel free to roll back any edit I have made. $\endgroup$
    – Narasimham
    Jul 16, 2022 at 10:28

1 Answer 1


When mixed derivatives are taken (independent variable $x$ instead of $t$), the order of differentiation does not matter in.. in wholly partial or wholly full mode.

But such interchange leads to possible incorrect results.

To avoid possible confusion and incorrect result the correct order i.e., inside to outside is specifically and necessarily indicated by usage of parentheses.

E.g., for $ L=y \;y'$

$$\dfrac{d}{dx}\left(\dfrac{\partial L}{\partial y'}\right)= \dfrac{dy}{dx}$$


$$ \dfrac{\partial}{\partial y'}\left(\dfrac{dL}{dx}\right)= \dfrac{\partial}{\partial y'} (yy^{''}+y^{'2})= 2\dfrac{dy}{dx}.$$


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