Total derivative of explicit variable and not implicit

Question

Say you have the following function, with the following information:

$F = mk\dot x^2 + k^2gx$

1. $m, k, g$ are some constants
2. $x$ under the hood is the function of $t$ (we could say implicitly depends)
3. $\dot x$ is the derivative of $x$ with respect to $t$

What would be the $\frac{dF}{dx}$ (total derivative) of this? Can we do that, and if not, why?

Answer 1

1. If you ignore that $x$ and $\dot{x}$ are functions of $t$ and just treat them as variables, then the equation defines a function of two variables ($x$ and $\dot{x}$).

2. If you take account of the fact that $x$ and $\dot{x}$ depend on $t$, then the equation defines a function of one variable ($t$).

Only in case 1 does it make sense to differentiate with respect to $x$, but then it is only a partial derivative (as $f$ is a function of two variables). Let us call the function in (1) $f$ and the function in (2) $h$.

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Here $f:\mathbb{R}^2\to\mathbb{R}$ is given by

$$f(a,b)=k^2ga+mkb^2,$$

while $h:\mathbb{R}\to\mathbb{R}$ is given by

$$h(t)=f(x(t),\dot{x}(t))$$

i.e. $h$ is the composition $f\circ (x,\dot{x})$.

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We can calculate the following derivatives of $f$:

$$f_1(a,b)=k^2g\qquad \text{and} \qquad f_2(a,b)=2mkb. $$

We can calculate the derivative of $h$ using the chain rule

$$h'(t)=f_1(x(t),\dot{x}(t))\cdot\dot{x}(t)+f_2(x(t),\dot{x}(t))\cdot\ddot{x}(t)=k^2g\cdot\dot{x}(t)+2mk\dot{x}(t)\cdot\ddot{x}(t)$$

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In the case where $F=f$, then $\partial F/\partial x$ corresponds to $f_1$ (note the use of $\partial$ rather than $d$ because $F$ is a function of two variables in this case).