The Freshman's Product Rule (f⋅g)' = f'⋅g' is true only if f/f' + g/g' = 1pic.twitter.com/6eHFtV7BwI
PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …
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The Freshman's Product Rule (f⋅g)' = f'⋅g' is true only if f/f' + g/g' = 1pic.twitter.com/6eHFtV7BwI
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.
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).
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
multiply by f'g' in the first line, and divide by f'g' in the second
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