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Topic:
Konyberg  for you!
Replies:
1
Last Post:
Nov 10, 2017 5:57 PM




Re: Konyberg  for you!
Posted:
Nov 10, 2017 5:57 PM


So how do you use your new calculoose to derive:
f(x) = ln(x), then f'(x) = ?
Am Freitag, 10. November 2017 23:36:03 UTC+1 schrieb John Gabriel: > On Wednesday, 8 November 2017 18:43:28 UTC5, John Gabriel wrote: > > "It confuses me that you can set m+n equal to zero after you cancel out m+n in the ratio. Surely this is wrong?" > > > > Well, consider first that I proved the sum of all the terms in m and n must be zero using 8th grade algebra (see http://thenewcalculus.weebly.com) but also note that the slope of the tangent line does NOT depend on m or n so given the derivative as an expression in x and Q(x,m,n), it follows Q(x,m,n) must be zero, that is, m=n=0. > > Read my new article on this topic: > > https://drive.google.com/open?id=1DuqrOjc5EIXoHpj6SnzUhFy4RPAAKVcE



