Help with taking derivative of Lagrangian scalar model of graviton Quick question. Given Lagrangian density
$$\mathcal{L} = -\frac12 h \Box h + \frac13 \lambda h^3 + Jh ,\tag{3.69}$$
where the scalar $h$ represents the gravitational potential, and given the Euler-Lagrange equation 
$$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}h)} - \frac{\partial \mathcal{L}}{\partial h} = 0 \tag{1}$$
we have the equations of motion (according to Schwartz's QFT eq. 3.70) 
$$\Box h -\lambda h^2 - J = 0. \tag{3.70}$$
I get $$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}h)} = 0 \tag{2}$$ 
and 
$$\frac{\partial \mathcal{L}}{\partial h} = -\frac12 \Box h + \lambda h^2 + J\tag{3}$$
so where did I go wrong with the factor of $-\frac12$?
 A: You have to be careful with the way you've defined your Lagrangian. You're treating the second derivative of $h$, $\Box h$, as if it were independent of both $h$ and $\partial_\mu h$. If you integrate by parts then you get a new Lagrangian that is much cleaner:
$$
\mathcal L'=\frac{1}{2}\partial_\mu h\partial^\mu h+\frac{1}{3}\lambda h^3+Jh
$$
we are, of course, assuming that $h\partial_\mu h$ vanishes on the boundary of spacetime, so that the new Lagrangian gives the same equation of motion.
Applying the Euler-Lagrange equation to this new Lagrangian will give you the correct result. Cheers!
A: *

*The correct field-theoretic Euler-Lagrange (EL) equation reads in general
$$ 0~\approx~\frac{\delta S}{\delta h}
~=~\frac{\partial {\cal L}}{\partial h} 
-\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}h)} + \sum_{\mu\leq \nu} \frac{d}{dx^{\mu}} \frac{d}{dx^{\nu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\partial_{\nu}h)} - \ldots,\tag{A} $$
where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher-derivative terms.
If you include second-derivative terms, you will get agreement with Schwartz' eom (3.70).

*Alternatively, you can start by removing higher-derivative terms in the action via integration by parts. Then the EL equation (A) will not contain higher-derivative terms. That is the strategy of Damian Sowinski's answer.
A: Use functional derivation and notations of  Schwartz's book
$$
\square \equiv \partial _{\mu }^2\equiv \partial _{\mu }\partial ^{\mu }
$$
so
$$
h \square h=\partial _{\mu }^2h
$$
