So how many times have you tried doing (fg)' and instead of doing fg' + f'g you directly did f'g'? Im just wondering if this situation ever really exists? For what functions is (fg)' = f'g' Thanks.
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Originally Posted by federernadal So how many times have you tried doing (fg)' and instead of doing fg' + f'g you directly did f'g'? Im just wondering if this situation ever really exists? For what functions is (fg)' = f'g' Thanks. Well let's see... If we let we have There you go, any function that has the property of will result in
This is really impressive!! Did you do that from scratch or have you been asked this before? Thanks!
Originally Posted by federernadal This is really impressive!! Did you do that from scratch or have you been asked this before? Thanks! From scratch - it's a method called "Separation of Variables", basically it means getting each side to be only in terms of one variable.
Yup I'm in Diff Eq class now. I had no trouble following your method, but it would be difficult for me to come up with it from scratch. You are a professor at which university?
Originally Posted by federernadal Yup I'm in Diff Eq class now. I had no trouble following your method, but it would be difficult for me to come up with it from scratch. You are a professor at which university? I do teach at a university (RMIT University in Melbourne, Australia) but I'm far from being a Professor. I'm still working on my Masters :P
Great stuff. Not seen that done before. Looks really neat if we simplify bracket to f'/(f'-f)
Originally Posted by biffboy Great stuff. Not seen that done before. Looks really neat if we simplify bracket to f'/(f'-f) I prefer to only have one derivative in it, which is why I wrote it as I did
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