Vrftsjtryk

I have been trying to derive the Einstein equation from the Einstein-Hilbert action
$$ S[g_{mu nu}] = frac{1}{16 pi} int_M text{d}^4x sqrt{-g}R $$
The standard derivation states that the variation $delta S =0$ when we vary the metric components. In this derivation, we use the fact that $delta (sqrt{-g}R)= delta sqrt{-g} R + sqrt{-g}delta R$. To me this seems quite obvious but I thought I would try and prove this.

My understanding is that if we have an action
$$ S[q] = int text{d}t L(q,dot{q},t)$$
we vary it as
$$ delta S[q] = int text{d}t delta L(q,dot{q},t)$$
so
$$ delta L= frac{partial L}{partial q} delta q + frac{partial L}{partial dot{q}} delta dot{q} \
= left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q + frac{text{d}}{text{d}t} left( frac{partial L}{partial dot{q}} delta q right)$$
As we integrate over $delta L$, we ignore the last term as this produces a boundary term which we can take to vanish, therefore I say

$$ delta L = left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q $$

Okay, now let's say my Lagrangian is a product: $L(q,dot{q},t) = f(q,dot{q},t)g(q,dot{q},t)$. Plugging this into the above formula for the variation, I have

$$ frac{partial L}{partial q} = frac{partial f}{partial q} g + frac{partial g }{partial q} f $$

$$ frac{partial L}{partial dot{q}} = frac{partial f}{partial dot{q}}g + frac{partial g}{partial dot{q}} f$$

$$ frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} = left( frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right) g + left( frac{text{d}}{text{d}t}frac{partial g}{partial dot{q}} right) f + frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t} + frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}$$

so the variation of this Lagrangian is

$$ delta L = left( frac{partial f}{partial q} - frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right)g delta q + left( frac{partial g}{partial q} - frac{text{d}}{text{d}t} frac{partial g}{partial dot{q}} right)f delta q - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}delta q$$

or

$$ delta L = g delta f + f delta g - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial g}{partial dot{q}} frac{text{d}f}{text{d}t}delta q $$

Now I can't seem to get rid of those horrible extra terms - I can't see how they would produce a boundary term when integrated. Maybe my understanding was incorrect and variations do not obey the product rule? Many standard resources suggest that varying the Einstein-Hilbert action obeys the product rule... is this just an exception?

My question:

How can I show that variations obey $delta (fg) = f delta g + g delta f$

It is true that
$$ delta L = f delta g + g delta f,$$
but the expressions for variations are:
begin{align} delta f&=frac{partial f}{partial q} delta q+ frac{partial f}{partial dot q}delta dot q = left[ frac{partial f}{partial q} - frac{d}{dt}left(frac{partial f}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q} delta qright)\ delta g&=frac{partial g}{partial q} delta q+ frac{partial g}{partial dot q}delta dot q = left[ frac{partial g}{partial q} - frac{d}{dt}left(frac{partial g}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial g}{partial dot q} delta qright)\delta L&=frac{partial L}{partial q} delta q+ frac{partial L}{partial dot q}delta dot q = left[ frac{partial f}{partial q}g+frac{partial g}{partial q}f - frac{d}{dt}left(frac{partial f}{partial dot q}g + frac{partial g}{partial dot q}f right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q}g delta q + frac{partial g}{partial dot q}f delta qright)end{align}
In each case, the variation is equal to the Euler-Lagrange equation times $delta q$, plus a total derivative. The total derivatives are important!