No Freshmen were harmed in the making of this tweet.
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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
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We can construct a function satisfying the freshman's product rule from f(x), by solving the differential equation for g(x). Try it out yourself!
pic.twitter.com/GcWrV04O5g
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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.
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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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What if f and g are both constant functions, e.g.?
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Then from the first equality we see that we will have 0=0, so it will "work".
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Don't drink and derive, kids!
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f'g+g'f=f'g' -> f'(g-g')+g'f=0 -> (g-g')/g' + f/f'=0 -> g/g'+f/f'=1
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Reminds me to (an "almost" similar case):https://twitter.com/SamuelGWalters/status/1136833777190154240?s=19 …
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This is astounding, at least to me. Does this mean f(x)/f'(x) + g(x)/g'(x) =1?
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