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I am trying to prove that the above is true when performing the change of variable shown. Here is my attempt:

What I am not quite understanding is why they choose to isolate the partial derivative of ##z## on the right side (as opposed to the left) that I have in my last line. This ultimately amounts to switching the sign of ## \frac{1}{c} \frac {\partial}{\partial t} ## but what differentiates the ## \frac {\partial}{\partial z} ## on the left and right sides? Why is it correct to switch terms over and isolate the partial derivative of ##z## that I have in my solution's right side? It seems like they just add ## \frac{1}{c} \frac {\partial}{\partial t} ## but this has slightly confused me as to why ## \frac {\partial}{\partial z} ## appears invariant upon exchange of ## \pm \frac{1}{c} \frac {\partial}{\partial t} ##. Any explanations for what I am certainly missing would be greatly appreciated.

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# I Change of variable - partial derivative

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