# Partial differentation of two variables [solved]

1. Nov 21, 2015

### epsilon

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.

2. Nov 21, 2015

### epsilon

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!