Why can we substitute any arbitrary variable in for this expression of the CDF?

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I'm trying to follow this example of a variable transformation, which defines the PDF and CDF as follows:

pdf and cdf definitions PROBLEM: The above definition is used in solving this problem: remark problem variable transformation

Question: Now, where I get confused is why we are allowed to differentiate $F_X(x_1)$ to get $f_Y(y)$? I would have thought for sure that by the definition we would need to differentiate $F_Y(y)$ in order to get $f_Y(y)$. Why are we allowed to do that?

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Yes, you should differentiate $F_Y(y)$, and just before that statement they showed $F_Y(y) = F_X(g^{-1}(y))$.