Partial Derivative vs Total Derivative: Function depending Implicitly and Explicitly on Variable I have a quick question that I hope someone can shed some light on for me. Its to do with partial derivatives and total derivatives (In this case I was working on Lagrangian Formalism but its really a general question). Im a theoretical physics student so although I do study a good deal of mathematics, we dont really go into the same level of detail that say a mathematician would (Lets just say some serious abuse of notation does occur more than one would like)
So lets assume I have some function $L = L(q_i(t),q_i'(t))$.
If I take the partial derivative of $L(q_i(t),q_i'(t))$ then, through the chain rule, I get
$$\frac {\partial L}{\partial t} = \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}.$$
Sound so far.
Lets assume now that the function $L \to L(t,q_i(t),q_i'(t))$ (i.e an explicit time dependence is now introduced in addition to the implicit time dependence.) 
 If I now take the total derivative of said function, then I get 
$$\frac {dL}{dt} = \frac{\partial L}{\partial t} + \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}.$$
But how would I now compute $\frac {\partial L}{\partial t}$? If I just apply the same logic and the chain rule, I would say that 
$$\frac {\partial L}{\partial t} = \frac{\partial L}{\partial t} + \frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}.$$
However, here comes my issue with the whole thing. If the above is true, then that would mean that 
$$\frac {dL}{dt} = \frac {\partial L}{\partial t}$$
and also that
$$\frac {\partial L}{\partial q_i} \frac {\partial q_i}{\partial t} + \frac {\partial L}{\partial q_i'} \frac {\partial q_i'}{\partial t}=0.$$
Is it just that for this particular scenario the total and partial derivatives are equal, or am I missing something here? I've heard that the Leibniz notation breaks down somewhat here and can lead to confusion (something to do with the fact that the $\frac{\partial L}{\partial t}$'s on either side of the equation mean two totally different things); is there something that Im doing wrong?
Thanks very much for any help you can give!
 A: Your problem comes from an abuse of notation (that is very common in this field), namely that different things are denoted with the same letters. You have a function $L$ of $2n+1$ variables, namely
$$(t,q_1,\ldots, q_n,q_1',\ldots, q_n')\mapsto L(t,q_1,\ldots, q_n,q_1',\ldots, q_n')\ .\tag{1}$$
This function is given a priori, even before you know that $t$ should be time, and that, later on, the $q_i$ and $q_i'$ will be functions of $t$, whereby $q_i'(t)={d\over dt}q_i(t)$. This function $L$ has partial derivatives
$${\partial L\over\partial t},\quad {\partial L\over\partial q_i},\quad {\partial L\over\partial q_i'}\ ,$$
all of them  functions of the same variables given in $(1)$.
Now you start doing physics, with the already sketched interpretation of $t$. In order to do this properly we denote the "running time" with $\tau$. We then have the functions
$$\tau\mapsto t(\tau)=\tau, \quad \tau\mapsto q_i(\tau),\quad \tau\mapsto q_i'(\tau):={d\over d\tau}q_i(\tau)\ .$$
Plugging these functions into $L$ we obtain a new function
$$\tau\mapsto {\tt L}(\tau):=L\bigl(\tau,q_1(\tau),\ldots, q_n(\tau),q_1'(\tau),\ldots, q_n'(\tau)\bigr)\ .$$
This function ${\tt L}$ of one variable $\tau$ has a derivative
$${\tt L}'(\tau)={\partial L\over\partial t}\cdot1+\sum_i{\partial L\over\partial q_i}q_i'(\tau)+\sum_i{\partial L\over\partial q_i'}q_i''(\tau)\ .$$
A: I think the main point is to always ask "how do I currently look at the function". You can always break a function to different blocks and look at it differently. 
It's true that since $L = L(q_i(t),q_i'(t))$ only really depends on $t$, you can't do partial derivative on $t$, only total derivative. But you could take a partial derivative with regard to $q_i$ or $q'_i$. So the total derivative with regards to $t$ would look:
$$\frac {dL}{dt} = \frac {\partial L}{\partial q_i} \frac {dq_i}{dt} + \frac {\partial L}{\partial q_i'} \frac {d q_i'}{d t}.$$
And $\frac {\partial L}{\partial q_i}$ exist - e.g. if $L(t^2, sin(t)) = t^2sin(t)$ then $\frac{\partial L}{\partial q_i} = \frac{\partial L}{\partial t^2} = sin(t)$.
This way of mixing total and partial derivatives is used for example in the Method of Characteristics to solve the (Partial Differential) transport equation. 
Now, you could decide that $q_i = t$, then you would get $L = L(t,q_i'(t))$. Now you can take the partial derivative w.r.t. $t$. It would be equal to whatever the function is, keeping $q_i'(t)$ as a constant. 
e.g. if $L(t, sin(t)) = t^2sin(t)$ then $\frac{\partial L}{\partial t} = 2tsin(t)$. I.e. you keep the $sin(t)$ as a constant.
SO - your mistake is twofold: 


*

*as mentioned, in the first case you computed the total derivative, not the partial (which doesn't exist w.r.t. $t$); and 

*in the 2nd case, the partial derivative (w.r.t. $t$) is NOT equal to the total derivative (w.r.t. $t$). i.e.
$$\mathbf {\frac {\partial L}{\partial t} \neq \frac{dL}{dt}} = \frac{\partial L}{\partial t} + \frac {\partial L}{\partial q_i} \frac {d q_i}{d t} + \frac {\partial L}{\partial q_i'} \frac {d q_i'}{d t}.$$
$\frac {\partial L}{\partial t}$ in the 2nd case would mean you keep $q_i, q_i'$ as constants, and there's no further derivation for it unless you have the actual function $L$. 
So, going back to my 1st point - the total derivative of $t^2 sin(t)$ will always be the same, but you may decide to break it to partial derivatives of $t$ and $sin(t)$, or $t^2$ and $sin(t)$ or something else. And sometimes these partial derivatives actually have some meanings.
