What is partial derivatives? Just differentiate the explicit variables is enough?

Question

From Euler-Lagrange equation, $$L=L(q(t),\dot q(t),t)$$$$\cfrac{dL}{dt}=\cfrac{\partial L}{\partial q}\cfrac{dq}{dt}+\cfrac{\partial L}{\partial \dot q}\cfrac{d\dot q}{dt}+\cfrac{\partial L}{\partial t}$$ So, from pure mathematics perspective, it is just saying that for a function $$f=f(x,y,t)=x+y$$$$x=x(t)=2t$$$$y=y(t)=3t^2$$$$\cfrac{\partial f}{\partial t}=0\quad!!!$$ If this is really true, how about that $$f=f(x,y)=x+y$$$$x=x(t)=2t$$$$y=y(t)=3t^2$$$$\cfrac{\partial f}{\partial t}=?$$ Also how about that, $$f=f(x,y)=x+y$$$$x=x(s)=2s$$$$y=y(t)=3t^2$$$$\cfrac{\partial f}{\partial t}=?$$

Answer 1

The concept you are looking for is the multivariable chain rule.

In your first example, the symbol $t$ takes on two distinct meanings: one as a independent variable, and another as an identity function of that independent variable. This is an abuse of notation, but a very common one. It's conceptually helpful to re-write the example as $f=f(x,y,z)$ where $x(t)=2t$, $y(t)=3t^2$, and $z(t)=t$. From here applying the chain rule is straight-forward, but since this is the most notationally complex of the examples you've given, I'll discuss it last.

In your second example, you have a function $f: \Bbb R^2 \to \Bbb R$, $f=f(\vec {x})$, where $\vec x: \Bbb R \to \Bbb R^2$ is itself a function of $t$. By the chain rule, $$ \frac{\partial f}{\partial t} = \nabla f \cdot \frac{\partial}{\partial t}\vec x = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} = (1)(2)+ (1)(6t) = 3+6t $$

Likewise, in the third case, we have $f: \Bbb R^2 \to \Bbb R$, $f=f(\vec {x})$ where $\vec x: \Bbb R^2 \to \Bbb R^2$ is a function of both $s$ and $t$. By the same principal, $$ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} = (1)(0) + (1)(6t) = 6t. $$

Returning to the first example, we have a function $f: \Bbb R^3 \to \Bbb R$, $f=f(\vec x)$, where $\vec x: \Bbb R \to \Bbb R^3$ is given by $$ \vec x(t) = \langle x(t), \,y(t), \,z(t)\rangle = \langle 2t, \,3t^2, \, t \rangle. $$ Applying the chain rule to this case gives us $$ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial t} = (1)(2) + (1)(6t) + (0)(1) = 3 + 6t. $$

Answer 2

You seem to be unclear on the semantics of a partial derivative.

The notation $\frac{\partial f(x,y,t)}{\partial x}$ means to

• temporarily pretend $x$, $y$, and $z$ are independent variables,
• differentiate $f$ with respect to the independent variable $x$, then
• stop pretending that $x$, $y$, and $z$ are independent variables.

We use partial derivatives to find the rates of change of the outputs of the formal function $f$ with respect to each of its inputs independently. This can be described using any of the phrases

• Differentiate $f$ with respect to $x$ holding all other variables constant. (The last phrase means that even if $x$, $y$, and/or $z$ are not independent variables, and this lack of independence means that you can not independently vary them, you must perform the calculation as if you could.)
• Differentiate $f$ with respect to its formal argument, $x$. (Here, we treat $f$ as a machine that takes three arguments and converts them to output. The clause "$\partial f(x,y,z)$" asserts that the first argument is labelled "$x$".)
• Differentiate $f$ with respect to its first slot. ("Slot" means an input to the function. With "$f(x,y,z)$", we announce that $f$ has three slots.)

Notice that this makes all of the following meaningless:

• $\frac{\partial f}{\partial x}$: Without a prior declaration of the labels of the formal arguments of $f$, there is no way to known which argument of $f$ is to be varied. So if we had previously announced "$f(a,b,c)$", this is undefined, and if we had previously announced "$f(x,y,z)$", this is defined. But the notation must be augmented by a prior announcement to resolve this difference.
• $\frac{\partial f(x,y,z)}{\partial t}$: $t$ is not a formal argument to $f$.
• $\frac{\partial f(x,x,x)}{\partial x}$: Which formal argument is varying?
• $\frac{\partial f(x(t),y(t),z(t))}{\partial t}$: $t$ is not a formal argument of $f$.

Note that one may declare $f(x,y,t)$ and then compute $\frac{\partial f(a,b,c)}{\partial a}$. This is the partial derivative of $f$ with respect to its first formal argument. That we really are talking about positions of formal arguments rather than modelled quantities is sometimes indicated by this notation $$ f_1 = \partial_1 f = \frac{\partial f(a,b,c)}{\partial a} \text{,} $$ all three of which are notation for the partial derivative of $f$ with respect to its first formal argument.

Turning to your examples:

• Your first example has "$f(x,y,t) = x + y$" declares that the formal arguments of $f$ are labelled $x$, $y$, and $t$. The rate of change of $f(x,y,t) = x+y$ with respect to $t$, the third formal argument, is zero: $\frac{\partial f}{\partial t} = 0$.
• Your second example has "$f(x,y) = x+y$". Since $f$ is not declared to have a formal parameter $t$, the expression $\frac{\partial f}{\partial t}$ is undefined; it does not have a value.
• Your third example has the same deficiency as the second: $t$ is not a formal argument of $f$, so $f$ does not have a partial derivative with respect to $t$.

Finally, it can be useful to write in words what your first, total derivative, expression means:

• $\cfrac{dL}{dt}$ is the total derivative of $L(q(t),\dot q(t),t)$ with respect to the independent variable $t$. It has three contributions:
• $\cfrac{\partial L}{\partial q}\cfrac{dq}{dt}$: the rate of change of $L$ with respect to variation in its first formal argument, times the rate of change of the input to that argument with respect to $t$ (This is an example of the chain rule, where nested rates of change are multiplied.),
• $\cfrac{\partial L}{\partial \dot q}\cfrac{d\dot q}{dt}$: the rate of change of $L$ with respect to variation in its second formal argument, times the rate of change of the input to that argument with respect to $t$, and
• $\cfrac{\partial L}{\partial t}$: the rate of change of $L$ with respect to variation in its third formal argument, which is already bound to the variable $t$.