STUCK

So basically, since my last devlog, the main feature of the new derivative calculator that I had proudly announced is now coming to bite me. The simplification is so tough. Let me explain you all the turn of events. (Simplification is still not complete)
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basically, my initial idea of simplification was just removing all the temporary zeros and ones that were generated by the derivative itself. Then I thought of increasing the initial scope to support some other simplifications like (‘’, ‘2’, (’’, ‘2’, ‘x’)) should be solved to (‘’, ‘4’, ‘x’). I managed to achieve this simplification too but, I found out there are a lot more simplification cases that I need to handle after we get the derivative. for example: x - 3x, which should be simplified to -2x. For that I had to build an approach which would take the coefficient and base then operate on the coefficients
for the same bases according to the operator. Therefore, for x the coeff is 1 and then for 3x the coeff is 3 but that is incorrect. In the case of x-3x, the coefficients should be 1 and -3 so that when i add them i get -2 as the final coefficient for x. To do that, I had to switch all binary subtraction expression and turn them into additive unary negative for op = ‘+’ so that x - 3x turns into x + (-3x). Now, all this actually worked but the main problem was that now the flattened list looks like ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ [‘x’, (’’, ‘-1’, (‘’, ‘3’, ‘x’))] which means I cannot reliably extract the coefficients now. I first have
to handle the recursive case of flattening the node again, turning (’’, ‘-1’, (‘’, ‘3’, ‘x’)) to the flattened version
[‘-1’, ‘3’, ‘x’] and then i would have to run it back through combine where it would be formed as calc = -3 and symbols = ‘x’ and then rebuild it as the node (’’, ‘-3’, ‘x’). Only after that can i actually extract the coefficients and compare the bases of the two expressions which would be 1 and -3 and calculate them according to the op => ‘+’ to finally give -2 as calc and ‘x’ as symbols and then rebuild that again into (‘’, ‘-2’, ‘x’) to solve it.
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Now this was just one case for one operator but i need to handle both ‘+’ and ’’ as the both have associativity and there are other cases for other ops too like x^2/x should just simplify to x. I have already added so much code that I myself feel like the approach is very wrong and the more I build the more edge cases I am adding because right now the amount of if-else handling for specific cases is crazy so I really think that I need to reconsider my approach for simplification. The derivative calculator is taking too much time considering it is an intermediate step but I think it would be worth it because this simplification step is something that I have never done before so I would be learning something anyway.