Partial differentation of two variables

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We have a function: ##\phi'(t',x')##.

We want to find: ##\frac{\partial\phi'}{\partial x}##.

I know that the answer is: ##\frac{\partial\phi'}{\partial x} = (\frac{\partial\phi'}{\partial t'} \cdot \frac{\partial t'}{\partial x}) + (\frac{\partial\phi'}{\partial x'} \cdot \frac{\partial x'}{\partial x})##.

I do not know how to achieve this answer. It appears to be some sort of chain rule question, but if you could do through every single step that would be great.
 
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I have actually solved this myself, but thought I would add the answer for anyone else that I wondering how to do it.

Consider the ##\phi'(t')## part first.

##\frac{\partial}{\partial x}\phi'(t')##

We need to have the denominator of the differentiation fraction to be the same function as the function being differentiated. Hence as we are differentiating a function ##(t')##, we need the denominator of the differentiation fraction to be ##\partial t'##.

We introduce ##\frac{\partial t'}{\partial t'}## which is a valid operation as this is just multiplying by 1, which makes no difference to the equation, such that we now have: ##\frac{\partial t'}{\partial t'} \cdot \frac{\partial \phi'}{\partial x} (t')##. I have also moved the ##\phi'## to be the numerator of the second fraction, but this is just a matter of notation.

As it is a multiplication we can swap the denominators around, such that we have: ##\frac{\partial t'}{\partial x} \cdot \frac{\partial \phi'}{\partial t'} (t')##.

This is can be re-arranged to match the answer I wrote earlier, although this is a redundant step, and the ##(t')## does not need to be written anymore either.

##\frac{\partial \phi'}{\partial t'} (t') \cdot \frac{\partial t'}{\partial x}##.

Repeat this process for the ##\phi'(x')## part and then add the two together!
 
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