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

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Posts: 5,511
Registered: 9/25/16
Re: Konyberg - for you!
Posted: Nov 10, 2017 5:57 PM
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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 UTC-5, 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 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:

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