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In Tong's QFT notes at the bottom of page 14, it is claimed that if a change $x\mapsto x-\epsilon$ is made, the Lagrangian changes in the following way:

$$\mathcal L(x)\rightarrow \mathcal L(x)+\epsilon^\nu \partial_\nu\mathcal L(x). \tag{1.40}$$

This of course comes from the Taylor expansion, what isn't clear to me is what $\mathcal L$ should be an explicit function of $x$? $\mathcal L$ is usually define to be $\mathcal L(\phi(x),\partial_\mu\phi(x),...)$, it's not clear to me why this can be straightforwardly Taylor expanded.

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You can use the chain rule to handle any implicit dependence on $x$,

\begin{equation} \partial_\mu \mathcal{L}\left(\phi(x), \partial_\nu \phi(x)\right) = \frac{\partial \mathcal{L}}{\partial \phi} \partial_\mu \phi + \frac{\partial \mathcal{L}}{\partial \partial_\nu \phi} \partial_\mu \partial_\nu \phi. \end{equation}

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  • $\begingroup$ So if we have a function which implicitly depends on a variable, say $f(g(x))$, the Taylor series of $f$ in $x$ is done in the exact same way? $\endgroup$
    – Charlie
    Commented May 30, 2022 at 20:29
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    $\begingroup$ @Charlie Yes. Since $$ g(x + \epsilon) = g(x) + g'(x) \epsilon $$ Then $$ f(g(x+\epsilon)) = f(g(x) + g'(x) \epsilon) = f(g(x)) + f'(g(x)) g'(x) \epsilon $$ The second term on the RHS above is the chain rule (multiplied by $\epsilon$). $\endgroup$
    – Andrew
    Commented May 30, 2022 at 20:36

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