Variations wrt. Time Derivatives

Here is the cheat sheet:

• On one hand, in a partial differentiation the variable $x,dot{x},ddot{x},ldots$ are independent. E.g. the partial derivatives $frac{partial dot{x}}{partial x}=0$ and $frac{partial x}{partial dot{x}}=0$ are zero.

• On the other hand, in a functional differentiation the variable $x,dot{x},ddot{x},ldots$ are dependent. E.g. the functional derivative $frac{delta dot{x}(t)}{delta x(t^{prime})}=delta^{prime}(t!-!t^{prime})$ and $frac{delta ddot{x}(t)}{delta x(t^{prime})}=delta^{primeprime}(t!-!t^{prime})$ are derivatives of the Dirac delta distribution, while $frac{delta x(t)}{delta dot{x}(t^{prime})}$ or $frac{delta ddot{x}(t)}{delta dot{x}(t^{prime})}$ are ill-defined/meaningless.

This is explained further in e.g. this and this related Phys.SE posts.

This is a fairly common misconception. The action functional $S$ eats a function $q$ and spits out the following number:

$$S[q] := int_{t_1}^{t_2} Lbig(q(t),dot q(t), tbig) dt$$

where the Lagrangian $L: mathbb R^3 rightarrow mathbb R$ is just a function of three variables. One might have, for example,

$$L(a,b,c) = frac{1}{2} mb^2 - frac{1}{2} momega^2 a^2$$

where $m$ and $omega$ are constants. In that case, the action would be

$$S[q] := int_{t_1}^{t_2} Lbig(q(t),dot q(t),tbig) = int_{t_1}^{t_2} left(frac{1}{2}m dot q^2(t) - frac{1}{2}momega^2 q^2(t) right) dt$$

---

Given some $(a,b,c)$, we can linearize the Lagrangian to find its value at a nearby point $(a+delta a,b+delta b,c+delta c)$:

$$L(a+delta a,b+delta b,c+delta c) = L(a,b,c) + left[ big(partial_1Lbig)cdot delta a+ big(partial_2Lbig)cdot delta b +big(partial_3Lbig)cdot delta cright]$$

where $partial_nL$ is the derivative of $L$ with respect to its $n^{th}$ slot. Therefore, if we add a small $eta$ to $q$, we get

$$S[q+eta]=int_{t_1}^{t_2} Lbig(q(t)+eta(t),dot q(t) + dot eta(t),tbig) dt $$ $$simeq int_{t_1}^{t_2} Lbig(q(t),dot q(t),tbig) dt + int_{t_1}^{t_2} big[(partial_1 L) eta(t) + (partial_2 L) dot eta(t) big] dt$$

Integration by parts then gives

$$S[q+eta]-S[q]simeq int_{t_1}^{t_2} big[(partial_1 L) - frac{d}{dt}(partial_2 L)big]eta dt$$

where we've used the fact that $eta(t_1)=eta(t_2)=0$. If we demand that this vanish for arbitrary (differentiable) $eta$, we must have that

$$frac{d}{dt}left[bigg(partial_2 Lbigg)big(q(t),dot q(t),tbig)right] = bigg(partial_1 Lbigg)big(q(t),dot q(t),tbig)$$

---

Notation being what it is, it is standard to say something like $$L(x,v,t) = frac{1}{2}m v^2 + frac{1}{2}momega^2x^2$$ $$frac{partial L}{partial x} = momega^2 x$$ $$frac{partial L}{partial v} = mv$$

But this is a bit misleading. What we've written as $frac{partial L}{partial x}$ is really the derivative of $L$ with respect to its first slot, evaluated at the point $(x,v,t)$. The same is true when we write $L = L(q,dot q,t)$; the fact that $q$ and $dot q$ are related to each other by differentiation is irrelvant, because $q(t)$ and $dot q(t)$ are the values (not functions!) we plug in to the first and second slots after we take the partial derivative of $L$.

---

I have an equation $L=x+dot x+ddot x$. What is the partial derivative of this with respect to $dot x$?

$L$ is a function, not a functional; it doesn't know how to take derivatives. It is a map which eats numerical values and spits out a numerical value. The only way to have a Lagrangian like that is to define $L(a,b,c)=a+b+c$, and then plug $x$ into the first slot, $dot x$ into the second slot, and $ddot x$ into the third slot$^dagger$. If you do this, then $big(partial_2 Lbig)(x,dot x,ddot x)$, which we we would usually write as $frac{partial L}{partial dot x}$ in an abuse of notation, would be equal to $1$.

---

$^dagger$ Note that an action functional which involves second derivatives is mathematically problematic under most circumstances.

I am going to guess that you are working in the context of a Lagrangian, trying to derive the equation of motion. The Lagrangian for one-dimensional motion along a single direction $x$ is a function $L(dot{x},x)$. (It can also be a function of $t$, but that’s not directly relevant for this discussion.) To derive the Euler-Lagrange equations, you need to take partial derivatives of $L$ with respect to both $x$ and $dot{x}$, and this is done by considering $x$ and $dot{x}$ to be completely independent variables.

Instead of $L(dot{x},x)$, you can think of the Lagrangian as a function of two completely separate variables $L(y,x)$. Then the partial derivate with respect to $dot{x}$ is essentially defined to be $$frac{partial L(dot{x},x)}{partialdot{x}}equivleft.frac{partial L(y,x)}{partial y}right|_{y=dot{x}},$$ taking the derivative of $L$ with respect to its first argument and evaluating it at that argument equal to $dot{x}$.

When dealing with more complicated theories (involving not point particles but fields), defined in terms of an action $S=int dt,L$, the relationship between $partial/partial x$ and $partial/partialdot{x}$ can be made more apparent, but at the at the level you are talking about, what I have described is probably the best way to think about things. This means, by the way, that the partial derivative $partial L/partialdot{x}$ of the $L=x+dot{x}+ddot{x}$ in the question is just 1—as are $partial L/partial x$ and $partial L/partialddot{x}$