One-forms in differentiable manifolds and differentials in calculus Suppose that we have this metric and want to find null paths:
$$ds^2=-dt^2+dx^2$$
We can easily treat $dt$ and $dx$ "like" differentials in calculus and obtain for $ds=0$
$$dx=\pm dt \to x=\pm t$$
Now switch to the more abstract and rigorous one-forms in differentiable manifolds. 
Here $$\mathrm{d}t (v)$$ is a one-form that takes a tangent vector from $T_p$ and returns a real number, $\mathrm {d}t(v) \in \mathbb {R}$.
The tangent vector to a curve $x^{\mu}(\lambda)$ in the basis $\partial_\mu$ is 
$$v=\frac {dx^\mu}{d\lambda}\partial_\mu$$
Now apply the one-form to this vector
$$\mathrm {d}t(\frac {dx^\mu}{d\lambda}\partial_\mu)=\frac{dx^\mu}{d\lambda}\mathrm {d}t(\partial_\mu)$$
$$ =\frac{dx^\mu}{d\lambda}\frac {\partial t}{\partial x^\mu}$$
$$=\frac {dt}{d\lambda}$$
Now the above metric, in terms of one-forms read 
$$0=-\mathrm {d}t^2(v,v)+\mathrm {d}x^2(v,v)=-\mathrm {d}t(v)\mathrm {d}t(v)+\mathrm{d}x(v)\mathrm {d}x(v)$$
$$=-(\frac {dt}{d\lambda})^2+(\frac {dx}{d\lambda})^2$$
If we use the chain rule $$\frac {dx}{dt}=\frac {dx}{d\lambda}\frac {d\lambda}{dt}$$
We eventually obtain $$dx=\pm dt \to x=\pm t$$
The above is from Carroll's page 77. He reminds us that we should stick to the second more formal one-forms, that is not using differentials as in calculus, because in the first shortcut we have "sloppily" did not make the distinction between $\mathrm {d}t^2(v)$ and $dt^2$.
Now my question is that can someone please provide an example that unlike the above example, treating the one-form in differentiable manifolds as a differential in calculus will indeed produce incorrect results or conclusions. 
 A: I think that I have found a counterexample where you cannot overlook the difference between a one-form and a differential.
Consider a classical field theory.
When applying the least action I see that a term is considered total derivative.
We say that
$$\int \partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x= \int d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)= (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)$$
And then because the variation at the spatial infinity vanishes this terms is equal to zero.
I do not get the calculation from $$\partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$ to
$$d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial t}dt+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x}dx+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial y}dy+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial z}dz$$
$$\neq \frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$
Can you expand this to fill the gap for me.
Also, why we require "spatial infinity" here, isn't it also true that $$\delta \phi$$ in the $$\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi$$ vanishes at any two endpoints of the events path, but why we require infinity here?
