Now, for the calculus part that you seem not to get either:
$$f=f(t,x(\theta))$$
Therefore:
$$\frac{df}{d \theta}=\frac{\partial f}{\partial t} \frac{dt}{d \theta}+\frac{\partial f}{\partial x} \frac{dx}{d \theta}=\frac{\partial f}{\partial t} 0+\frac{\partial f}{\partial x} \frac{dx}{d \theta}$$
Contrary to what you might think, $$\frac{dt}{d \theta}=0$$
Tach said:
Nonsense, [$$dt/d\theta = 0$$] implies that t is not a function of $$\theta$$
In the example given by Tach we have something like:
$$y = f(t,x(\theta))$$
where I have introduced y as a new defendent variable. It is confusing to use f both as the name of a function and as the name of a variable.
Tach claims that if $$\frac{dt}{d\theta}=0$$ then $$t$$ is independent of $$\theta$$.
Here's a concrete example. Take:
$$y = f(t, x(\theta)) = t - x = t - 2\theta$$
where I have defined $$f(a,b)=a-b$$ and $$x(\theta)=2\theta$$.
We then have:
$$t = y + 2\theta$$
and
$$\frac{dt}{d\theta} = \frac{\partial y}{\partial \theta} + 2 = -2 + 2 = 0$$
Tach's claim seems to make sense to me.
However, by defining the variable y as above, we have assumed that t and $$\theta$$ are independent from the start, so it's not so surprising to get this result.
I'm not sure whether it is relevant, but it's also perhaps worth considering what happens if we look at a
particular value for y. As an example, consider y=3.
We then have
$$3=t-2\theta$$
which is easily arranged to give:
$$t = 3 + 2\theta$$
In other words, the function
$$3=f(t,x(\theta))$$
implicitly makes t dependent on $$\theta$$.
Obviously, in this case $$\frac{dt}{d\theta} \ne 0$$.
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I don't know if any of this helps.