Suppose $$t' = t$$ $$x' = x - vt$$
I need to prove that
$$\frac{\partial{}}{\partial{x}} = \frac{\partial{}}{\partial{x'}}$$ $$\frac{\partial{}}{\partial{t}} = \frac{\partial{}}{\partial{t'}} - v \frac{\partial{}}{\partial{x'}}$$
Now, for the first part, I was thinking about doing something along the lines of,
$$\frac{\partial{}}{\partial{x}} = \frac{\partial{}}{\partial{x'}} \frac{\partial{x'}}{\partial{x}} = \frac{\partial{}}{\partial{x'}} \cdot \frac{\partial{(x -vt)}}{\partial{x}} = \frac{\partial{}}{\partial{x'}} \cdot 1 = \frac{\partial{}}{\partial{x'}}$$
a) Is this even valid? I sort of feel like I'm cheating here, which brings me to,
b) What about $\frac{\partial{}}{\partial{t}}$ then? If I do the same thing as I did for (a), I'll get $\frac{\partial{}}{\partial{t}} = \frac{\partial{}}{\partial{t'}}$ which, in fact, is what's making me suspicious about (a).
One way is to relate the partials via the coordinate-free definition of derivative. Given a function $u$, fix $z \in \mathbb{R}^2$ and consider $\nabla u (z)$ as a linear map from $\mathbb{R}^2$ to $\mathbb{R}$.
From here, when I give coordinate vectors, they are in the $x,t$ coordinates. The $x$ partial derivative at $z$ is $\nabla u (z)$ evaluated at $(1,0)$, the $t$ at $(0,1)$ and the $t$ partial is this linear map evaluated at at $(0,1)$. The $x'$ one is the linear map evaluated at the element of $\mathbb{R}^2$ with $x',t'$ coordinates $x'=1$ and $t'=0$; i.e. $(1,0)$. The $t'$ one is when $x'=0$ and $t'=1$; i.e. $(-v,1)$. Using linearity of the map gives the answer.