Partial differentation of two variables

[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.

 

[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!