The following screen scrapes are from my notes on special relativity. I was motivated by Feynman's treatment of the four-dimensional gradient, as well as John Archibald Wheeler's approach to special relativity as an exact(sic) analog to plane trigonometry. Everything is written out in excruciating detail in order to study the structure.
RCC means rectangular Cartesian coordinates. The symbol $c$ has nothing directly to do with the speed of light, which I write as $\mathbb{c}$.
My question pertains to the block tagged $\dagger_{3}$. In particular, how to honestly justify replacing $dy/dx\mapsto{m}=\tan\left(\theta\right).$ That is, can I simply use the identity stated previously, or am I in fact deriving that identity through the implicit differentiation shown? To put it yet another way: is it pedantically correct that I had no business writing $m=\tan (\theta )=\frac{s}{c}=\frac{d\mathit{y}}{dx}=-\frac{d\mathit{x}}{dy}$ before I proved it (and gave it meaning) in $\dagger_{3}?$
All of the following is informal and reflects incomplete thoughts.
Background
I will confess, my question is not well stated. I will work on finding a better way of posing it.
For now I will share what led me to ask the question. Rather than following Feynman's derivation (which I think is pretty slick), I wanted to show that it can be done using the multivariable chain rule.
For simplicity, we are suppressing the $y$ and $z$ dimensions.
The objective is the transformation law for the gradient of a scalar field $\varphi$ going between inertial frames. We express the components of the Lorentz transformation using hyperbolic trig function, as well as in terms of Minkowski coordinates. The $\bar{\mathcal{S}}$ system moves in the positive $x$ direction relative to the $\mathcal{S}$ system with signed speed $\beta.$ Denote by $x_{\bar{o}}$ the $\mathcal{S}$ coordinate of the $\bar{\mathcal{S}}$ origin. Using these we write the Lorentz transformation matrix taking coordinates in $\mathcal{S}$ to coordinates in $\bar{\mathcal{S}}.$ My use of $\sigma=\sinh\theta$ is non-standard (but it should be standard).
\begin{align*} \beta= & \tanh\theta=\frac{dx_{\bar{o}}}{dt}\\ \gamma= & \cosh\theta=\frac{1}{\sqrt{1-\beta^{2}}}\\ \sigma= & \sinh\theta=\gamma\beta\\ \left\{ \Lambda^{\bar{\alpha}}{}_{\alpha}\right\} = & \left[\begin{array}{cc} \gamma & -\gamma\beta\\ -\gamma\beta & \gamma \end{array}\right]=\left[\begin{array}{cc} \gamma & -\sigma\\ -\sigma & \gamma \end{array}\right]\\ = & \left[\begin{array}{cc} \frac{\partial\bar{t}}{\partial t} & \frac{\partial\bar{t}}{\partial x}\\ \frac{\partial\bar{x}}{\partial t} & \frac{\partial\bar{x}}{\partial x} \end{array}\right] \end{align*}
The form form of the matrix with $-\gamma\beta$ is familiar to anybody who has studied SR. The form with $-\sigma$ obviously follows. The form of using partial derivatives is simply a statement of what the matrix does. The use of partial derivatives makes sense because $\bar{t}$ and $\bar{x}$ are dependent on both $t$ and $x.$
We first consider a displacement along the world-line of a mass-point in the un-barred $\mathcal{S}$ inertial system. That is, $\Delta t=\Delta\tau$ and $\Delta x=0.$ Feynman writes $\Delta\varphi$ in terms of both coordinate systems using the multivariable chain rule. Then he uses his version of the ``$\gamma\beta$'' matrix to express the $\bar{\mathcal{S}}$ displacement components in terms of $\mathcal{S}$ components. He then divides through by $\Delta\tau$ to get the transformation law for the time component of the gradient. Finally, he promulgates the corresponding law for the space component.
\begin{align*} \Delta\varphi= & \frac{\partial\varphi}{\partial\tau}\Delta\tau\\ = & \frac{\partial\varphi}{\partial\bar{t}}\Delta\bar{t}+\frac{\partial\varphi}{\partial\bar{x}}\Delta\bar{x}\\ = & \frac{\partial\varphi}{\partial\bar{t}}\gamma\Delta\tau-\frac{\partial\varphi}{\partial\bar{x}}\gamma\beta\Delta\tau\\ \frac{\partial\varphi}{\partial t}= & \gamma\left(\frac{\partial\varphi}{\partial\bar{t}}-\frac{\partial\varphi}{\partial\bar{x}}\beta\right)\\ \frac{\partial\varphi}{\partial x}= & \gamma\left(\frac{\partial\varphi}{\partial\bar{x}}-\frac{\partial\varphi}{\partial\bar{t}}\beta\right) \end{align*}
Note that Feynman doesn't bother mentioning that $\Delta t$ is a proper time interval, but it seems enlightening to do so. Notice that the transformation coefficients taking $\left\{ \frac{\partial\varphi}{\partial\bar{t}},\frac{\partial\varphi}{\partial\bar{x}}\right\} $ to $\left\{ \frac{\partial\varphi}{\partial t},\frac{\partial\varphi}{\partial x}\right\} $ are the same as those taking $\left\{ t,x\right\} $ to $\left\{ \bar{t},\bar{x}\right\} ,$ which was the main point of Feynman's derivation. Now lets examine these formulae a bit closer.
We will omit the $\bar{o}$ from $x_{\bar{o}},$ leaving it implied. Using $\gamma=\frac{\partial\bar{t}}{\partial t}$ from the above matrix equation and $\beta=\frac{dx}{dt}$ we rewrite our transformation as
\begin{align*} \frac{\partial\varphi}{\partial t}= & \frac{\partial\bar{t}}{\partial t}\left(\frac{\partial\varphi}{\partial\bar{t}}-\frac{\partial\varphi}{\partial\bar{x}}\frac{dx}{dt}\right)\\ \frac{\partial\varphi}{\partial x}= & \frac{\partial\bar{x}}{\partial x}\left(\frac{\partial\varphi}{\partial\bar{x}}-\frac{\partial\varphi}{\partial\bar{t}}\frac{dx}{dt}\right) \end{align*}
This almost looks like a chain rule, but not exactly. We know from the principle of relativity that $-\beta=\frac{d\bar{x}}{d\bar{t}},$ where $\bar{x}=\bar{x}_{o}$ is the $\bar{S}$ space coordinate of the $\mathcal{S}$ system. So we try that substitution
\begin{align*} \frac{\partial\varphi}{\partial t}= & \frac{\partial\bar{t}}{\partial t}\left(\frac{\partial\varphi}{\partial\bar{t}}+\frac{\partial\varphi}{\partial\bar{x}}\frac{d\bar{x}}{d\bar{t}}\right)=\frac{\partial\varphi}{\partial\bar{t}}\frac{\partial\bar{t}}{\partial t}+\frac{\partial\varphi}{\partial\bar{x}}\frac{d\bar{x}}{d\bar{t}}\frac{\partial\bar{t}}{\partial t}\\ \frac{\partial\varphi}{\partial x}= & \frac{\partial\bar{x}}{\partial x}\left(\frac{\partial\varphi}{\partial\bar{x}}+\frac{\partial\varphi}{\partial\bar{t}}\frac{d\bar{x}}{d\bar{t}}\right)=\frac{\partial\varphi}{\partial\bar{x}}\frac{\partial\bar{x}}{\partial x}+\frac{\partial\varphi}{\partial\bar{t}}\frac{d\bar{x}}{d\bar{t}}\frac{\partial\bar{x}}{\partial x} \end{align*}
The first equation looks close. If we can show $\frac{d\bar{x}}{d\bar{t}}\frac{\partial\bar{t}}{\partial t}=\frac{\partial\bar{t}}{\partial t}\left(?\right)$ we'll have it. In the second equation we need $\frac{d\bar{x}}{d\bar{t}}\frac{\partial\bar{x}}{\partial x}=\frac{\partial\bar{t}}{\partial x}\left(?\right),$ which appears a bit more formidable.
The fact that $\beta=\frac{dx}{dt}=-\frac{d\bar{x}}{d\bar{t}}$ appear as full derivatives, whereas all the other coordinate derivatives appear as partials, also had me puzzled. And finally, I wanted to explain how $\beta=\frac{dt}{dx}=-\frac{d\bar{t}}{d\bar{x}}$ (note the placement of $t$ and $x$) arises from directly differentiating the transformation mapping, where the transformation is defined by the requirement that spacetime intervals are Minkowski invariant.
This is the point at which I decided to examine the analogous problem using standard trigonometry.