Suppose you have a function $f(x',t')$ with $x'=g(x,t)$ and $t'=h(t)$. In other words, you have $f= f(x'(x,t),t'(t))$. Now, suppose that $h(t) = t$, meaning $t'=t$
Can we then say $ \frac{\partial{f}}{\partial{t'}} = \frac{\partial{f}}{\partial{t}} $ ?
Intuitively, it's tempting to say yes. However, it seems to be a source of error in the problems I am working on, and I can't quite understand why that argument wouldn't hold. My present understanding has it that change won't propagate towards the independent variable $t$, so we while we would have $ \frac{dt'}{dt} = 1$, we would also have $ \frac{dt}{dt'} = 0 $ This might be totally inaccurate, though.
So my question is this; Am I correct in saying that $ \frac{\partial{f}}{\partial{t'}} \neq \frac{\partial{f}}{\partial{t}} $ ? Why or why not? Feel free to direct me towards already existing explanations on this topic!
By applying the ordinary partial differentiation rules we have $$ \begin{split} \frac{\partial f}{\partial t}&=\frac{\partial f}{\partial x'}\frac{\partial x'}{\partial t} + \frac{\partial f}{\partial t'}\frac{\mathrm{d} t'}{\mathrm{d} t}\\ &=\frac{\partial f}{\partial x'}\frac{\partial g}{\partial t} + \frac{\partial f}{\partial t'}\frac{\mathrm{d} h}{\mathrm{d} t} \end{split}, $$ and if $t'=h(t)=t\iff \frac{\mathrm{d} t'}{\mathrm{d} t}=1$, then $$ \frac{\partial f}{\partial t}=\frac{\partial f}{\partial x'}\frac{\partial g}{\partial t} + \frac{\partial f}{\partial t'}\:. $$ Thus we have that $$ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial t'}\:\:\text{ if and only if }\:\: \frac{\partial g}{\partial t}=0\iff g(x,t)=g(x) $$ i.e. the two partial derivatives are equal if and only if $t'=t$ and $g(x,t)=g(x)$, so you're right in saying that, in the general case, $$ \frac{\partial f}{\partial t} \neq \frac{\partial f}{\partial t'} $$ Note that in solving the posed problem, I considered first the general case and then analyzed the particular one: it is a methodology I always advice to use since important details can be overlooked unless you have a strong training.