Confusing Total Derivative and Partial Derivative in Classical Field Theory - Noether Theorem I'm really confused about total derivatives and partial derivatives.
My multivariable calculus book (Guidorizzi vol 2 Um Curso de Calculo) says that if I have a function like $f(a(u,v),b(u,v))$ then the following is true:
$$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}\tag{1}$$
But studying classical field theory, precisely in Noether's theorem for fields, I came across that
$$\frac{d\mathcal{L}}{dx^{\mu}}=\frac{\partial \mathcal{L}}{\partial\phi_{\alpha}}\frac{\partial\phi_{\alpha}}{\partial x^{\mu}}+\frac{\partial \mathcal{L}}{\partial \phi_{\alpha,\mu}}\frac{\partial \phi_{\alpha,\mu}}{\partial x^{\mu}}+\frac{\partial \mathcal{L}}{\partial x^{\mu}}\tag{2}$$
with $\mathcal{L}=\mathcal{L}(\phi_{\alpha}(x^{\mu}),\phi_{\alpha,\mu}(x^{\mu}),x^{\mu})$
Note: $\alpha$ fields $\phi$ for $x^\mu$ arguments ($\mu=0,1,2,3$) in lagrangian density $\mathcal{L}$
This is called total space-time derivative (also someone can give me a reference than explain and define formally ?, for bibliography reference purpose please!)
The problem is, I can understand why equation (2) is true and make sense. After all, we are varying in all $\mathcal{L}$ arguments. It seems like the (1) equation doesn't make sense if equation (2) is true. I'm thinking that equation (1) is a notation abuse and in fact should be a total derivative 
$$\frac{\text d f}{\text d u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}$$
Am I right?
I have a third question: If $f$ depends explicitly on $u$ then
$f(a(u,v),b(u,v),u)$  and equation (1) doesn't make sense:
$$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}+\frac{\partial f}{\partial u}$$
All of my reasoning makes me believe that in fact that is 
$$\frac{\text d f}{\text d u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}+\frac{\partial f}{\partial u}$$
Is my calculus book wrong along with several other references I had studied?
 A: In math a total derivative is simply the full derivative of a (possibly multivariate) function, and can be represented by a matrix at each point. In math, the notation $\partial f/\partial x$ vs $df/dx$ is simply that in the first case $f$ may depend on more variables, where in the second or is emphasized to only depend on a single argument. There is no difference in how the chain rule applies.
There is a difference in notation between math an physics: in $f(x)$, $f(r)$, $f(p)$ and $f(t)$ means the same function $f$ evaluated in $x$, $r$, $p$ or $t$, and it is clear what is meant by $f(4)$ when 4 is in the domain. In other words, there is a functional relationship between the arguments and the function value, and this is independent of the names given to the arguments.
In physics, $f(x)$ often means the value of some quantity in the point $x$, $f(r)$ the value of the same function or quantity after a change to polar coordinates, $f(p)$ the value of the Fourier transform of the original quantity in the point $p$, and $f(t)$ the composition of $f$ with some other function that depends on time. This is some kind of a conceptual relationship between an underlying physical quantity, and some other physical quantities, between which there does exist a functional relationship but that is not explicit in the notation.
When dealing with a Lagrangian density $\mathcal L(\phi_\alpha, \phi_{\alpha,\mu}, x^\mu)$, in math there is no such thing as $\frac{d\mathcal L}{dx^\mu}$, because $\mathcal L$ is not a function of a single variable $x^\mu$. $\frac{\partial\mathcal L}{\partial x^\mu}$ does of course exist, and is the same as in physics. A mathematically correct notation for the former would be
$$\frac{\partial}{\partial x^\mu}\left(\mathcal L\circ\Phi\right)(x^\mu)$$
where $\Phi(x^\mu) = \left(\phi_{\alpha}\left(x^{\mu}\right), \phi_{\alpha, \mu}\left(x^{\mu}\right), x^{\mu}\right)$.
A: Neither of your books are incorrect. For the calculus book you have two independent variables $u$ and $v$. Therefore, you need to use the partial derivative $\partial$.
In your physics book all functions depend on only one independent variable$^*$ $x^\mu$. Therefore, you use the total derivative $\text d$. 
So, in your first case $\text df/\text du$ doesn't make sense, unless $v$ also depends on $u$.

For your "third" case, to avoid confusion it would be best to replace the $u$ dependence with some function say $c(u)$. Then you have $f(a(u,v),b(u,v),c(u))$. Then your partial derivative $\partial f/\partial u$ will have a term
$$\frac{\partial f}{\partial c}\frac{\text dc}{\text d u}$$
Therefore we end up with
$$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}+\frac{\partial f}{\partial c}\frac{\text dc}{\text du}$$
i.e. the term you have confusion about really should just include any $u$ dependence that has not been considered as a part of $a$ or $b$ already.
And this brings up a point that is worth mentioning. The break up of $f$ into $a$, $b$, and $c$ is completely subjective. In reality we just have $f(u,v)$. But sometimes it's more useful to specify composite functions to do our math. Note that this exactly what we do in calculus 1 for single-variable functions. For example, to find the derivative of $f(x)=(x+1)^2$, we first define a new function $g(x)=x+1$. Then we rewrite $f$ as $f(g(x))=\left[g(x)\right]^2$. Then we use the chain rule to find $\text df/\text dx$.

$^*$ I know this isn't actually a single variable, but in this case it is being treated as one. If you wanted to break it up into each part, then each part would involve a partial derivative.
