I am having trouble understanding Feynman's explanation of four gradient.
In section 25-3 of Vol. 2 of the Feynman lectures, he explains why the four gradient is not $(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}) $ by showing that derivatives for $\phi$ become (we let c = 1):
\begin{align} \frac{\partial \phi}{\partial t} &= \gamma \left(\frac{\partial \phi}{\partial t'} - v \frac{\partial \phi}{\partial x'}\right)\tag{1}\\ \frac{\partial \phi}{\partial x} &= \gamma \left(\frac{\partial \phi}{\partial x'} - v \frac{\partial \phi}{\partial t'}\right)\tag{2} \end{align}
Then he states that since the correct Lorentz transformations are:
\begin{align} t &= \gamma \left(t' + v x'\right)\tag{3}\\ x &= \gamma \left(x' + v t'\right)\tag{4} \end{align}
And concluded that since the signs don't match, (1) and (2) must be wrong. A full explanation can be found here: http://www.feynmanlectures.caltech.edu/II_25.html
I did not understand Feynman's logic, so I decided to check the correct formulas for the four gradient using (3) and (4).
I looked at the partial derivative of $\phi$ with respect to $t$. The correct four gradient states that it should be:
$$\frac{\partial \phi}{\partial t} = \gamma \left(\frac{\partial \phi}{\partial t'} + v \frac{\partial \phi}{\partial x'}\right)\tag{*}$$
We differentiate (3), (4):
\begin{align} \frac{\partial t}{\partial t'} &= \gamma \left(1 + v \frac{\partial x\prime}{\partial t'}\right)\tag{5}\\ \frac{\partial t}{\partial x'} &= \gamma \left(\frac{\partial t'}{\partial x'} + v\right)\tag{6} \end{align}
For simplicity, let $u\equiv \partial x'/\partial t'$. We put (5), (6) into (*) to get:
\begin{align} \frac{\partial \phi}{\partial t} &= \gamma \left(\frac{\partial \phi}{\partial t} \frac{\partial t}{\partial t'} + v \frac{\partial \phi}{\partial t} \frac{\partial t}{\partial x'}\right) \\ &= \gamma \left(\frac{\partial \phi}{\partial t}\right) \left( \frac{\partial t}{\partial t'} + v \frac{\partial t}{\partial x'}\right) \end{align}
\begin{align}1 &= \gamma \left(\gamma(1 + vu) + v \gamma\left(\frac{1}{u} + v\right)\right) \\ 1 - v^2 &= 1 + vu + \frac{v}{u} + v^2\end{align}
And this last equality is clearly false. So where did I make the mistake? And how can I correctly prove (*)?
Also, another question: when doing implicit differentiation on the x, t values, we can see that:
$$\frac{\partial t}{\partial t'} = \gamma \left(1 + v \frac{\partial x'}{\partial t'}\right),$$ but, $$\frac{\partial t'}{\partial t} = \gamma \left(1 - v \frac{\partial x}{\partial t}\right).$$ So then, $$1 = \frac{\partial t}{\partial t'} \frac{\partial t'}{\partial t}$$ is false. How can this be so!?