# Covariant derivative-Differential

I was trying to prove that the derivative-four vector are covariant. This can be proved only if you consider the time and space derivatives to be

$\dfrac{\partial}{\partial t^\prime}=\dfrac{\partial}{\partial t}\dfrac{\partial t}{\partial t^\prime} + \dfrac{\partial}{\partial x_1} \dfrac{\partial x_1}{\partial t^\prime}$

$\dfrac{\partial}{\partial x_1^\prime}=\dfrac{\partial}{\partial t}\dfrac{\partial t}{\partial x_1^\prime} + \dfrac{\partial}{\partial x_1} \dfrac{\partial x_1}{\partial x_1^\prime}$

I was trying to understand but I'm afraid I can't.. if oyou for instance add the terms in the second part of the equations you get

$\dfrac{\partial}{\partial t}\dfrac{\partial t}{\partial t^\prime} + \dfrac{\partial}{\partial x_1} \dfrac{\partial x_1}{\partial t^\prime}=\dfrac{\partial}{\partial t^\prime} + \dfrac{\partial}{\partial t^\prime} =2 \dfrac{\partial}{\partial t^\prime}\neq \dfrac{\partial}{\partial t^\prime}$

How is that possible?

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Would math.stackexchange.com be a better home for this question? Basically the question (v2) is treating the chain rule in an incorrect manner. – Qmechanic May 30 '13 at 9:00
This is just the chain rule for partial derivatives. It is not true that $\partial/\partial t' = \partial t/\partial t' \partial/\partial t$. Just try it on some function that depends on both $x,t$ to see it break. – Michael Brown May 30 '13 at 9:02
And it is written incorrectly. You should have the operators on the right, $\partial t/\partial t'\partial/\partial t\ldots$. Otherwise you will do unnecessary differentiations of $\partial t/\partial t'$ etc. – Peter Kravchuk May 30 '13 at 12:17

As noticed by Michael Brown, your last equation is totally wrong.

For instance take : $$t' = t - x_1$$ $$x'_1 = t + x_1$$, so that $$t = \frac{(t' + x'_1)}{2}$$ So, you have $$\frac{\partial t}{\partial t'} = \frac{1}{2}$$

Consider the function $f(x_1, t) = (t - x_1) = t'$

Then $$\frac{\partial t}{\partial t'} \frac{\partial}{\partial t} f(x_1, t) = \frac{1}{2} \frac{\partial}{\partial t} f(x_1, t) = \frac{1}{2}$$

But $$\frac{\partial}{\partial t'} f(x_1, t) = \frac{\partial}{\partial t'}t' = 1$$

So $$\frac{\partial t}{\partial t'} \frac{\partial}{\partial t} \neq \frac{\partial}{\partial t'}$$

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