# Confusion about partial and total derivatives.

I stumbled upon this: And there is something that specifically bothers me. Say $$L(q(t),\dot{q}(t),t)$$

Then I claim it makes no sense that $\frac{\partial L}{\partial t}$ only differentiates the explicit dependencies of $L$ on $t$ without also differentiating the contributions of $q(t)$ and $\dot{q}(t)$.

Why do I claim this? Well suppose we have a function such that $$f(x(r,\theta),y(r,\theta))$$

Then the operator $\frac{\partial}{\partial r}$ does act on the functions and not only on "explicit dependences".

What is going on here?

It might look as if I'm an advanced student because of the image I've posted, but I really am not, so don't be misguided (if that were to be the case).

Thanks.

• The notation for partial derivatives is awful. Your doubt is 200% reasonable. EDIT: One way to go around this is by adopting Rudin's PMA notation for partial derivative: $D_1f=\frac{\partial f}{\partial x}$, for example. EDIT2: In fact, I think this notation is one of the worst of mathematics. – Aloizio Macedo Sep 24 '15 at 7:30

Suppose the variables were named $(a,b,c)$ to begin with (instead of $(q,\dot q,t)$), so that you simply have a function $L(a,b,c)$. Then there's nothing strange about computing the partial derivative $\frac{\partial L}{\partial c}(a,b,c)$, right? And there is also nothing strange about inserting values $$a=q(t), \quad b=\dot q(t), \quad c=t$$ after you have computed this derivative: $\frac{\partial L}{\partial c}(q(t),\dot q(t),t)$. And this is exactly what is meant by the notation $$\frac{\partial L}{\partial t}(q(t),\dot q(t),t).$$
For comparison, if instead you first insert these values into $L(a,b,c)$, so that you get a function of $t$ alone, and then take the (ordinary) derivative of this function (with respect to $t$ of course, since it's now the only variable in sight), then you get the "total derivative" which is denoted by $$\frac{dL}{dt}(q(t),\dot q(t),t).$$
• Sorry to comment a thread far from now! But I really wonder what is $\frac{dL}{dq}$ in your example – FaDA Aug 17 at 0:16
• @FaDA: I don't understand your question. $dL/dq$ doesn't make much sense. – Hans Lundmark Aug 17 at 9:58
• Thanks for your reply! It doesn't have too much physical meaning, but when performing integration over a space of (q,p,t) it is useful when doing integration by parts. Here is my question math.stackexchange.com/questions/3325649/… In this thread which gives two links, the second link is the case where $dL/dq$ is useful in the integration by parts like I mentioned – FaDA Aug 17 at 10:57
• @FaDA: I still don't understand. You would have to clarify what you mean by $dL/dq$ to begin with. – Hans Lundmark Aug 17 at 13:45