The Freshman's Product Rule (f⋅g)' = f'⋅g' is true only if f/f' + g/g' = 1pic.twitter.com/6eHFtV7BwI
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Counter-example: f = (x-1)² and g = x² at x = 1. We have (f⋅g)' = f'⋅g', but f/f' + g/g' = 1/2. https://sagecell.sagemath.org/?z=eJxFjEEKwkAMRfc5RegmSatMx6XQqxSUIWlBrBSRuUBX3tKTmCmMbpL_Et5_XVamTAK5xwEjqE_OxyjjCcxz9g2pXNOsyiqQrII56I-0LTw5anChQwvJAB7rfH9iw_p5byaEg_-pZDo3B7wuy429hHMv5eW5TSYFpaoalPY62-3496aqRfkCJ882PQ==&lang=sage&interacts=eJyLjgUAARUAuQ== … It's because it's the other way around: it's true IF, but not only if.
The function [f*g]' is not equal to f'g'.. more over D[(x-1)^2] at x=1 is zero and so you'd be dividing by zero. You have to evaluate both functions at the same x.
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It is, at x = 1. And I am evaluating at the same x. You don't divide by zero because the pole is cancelled out. But anyway, my point is that the logic is reversed. You're using "p only if q" (p => q), when it should be "p if q" (p <= q).
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I'm not sure about that, the proof goes both ways. Suppose f and g are non-constant functions. It follows thatpic.twitter.com/MtPxH6xVh0
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