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Attempting to find the maximum curvature of the curve $e^x$ I have to differentiate kappa $$k(t)=\frac{e^t}{(1+e^t)^{3/2}}$$

of course, I use quotient rule and chain rule to find the derivative;

$$k^1(t)=\frac{e^t*(1-e^{2t})^{3/2}-e^t*3/2(1-e^{2t})*2e^{2t}}{((1+e^{2t})^{3/2})^2}$$

and simplify etc.

However, multiple sources including an exam key, and even a highly upvoted post on Stack Exchange Math has the derivative of kappa as

$$k^1(t)=\frac{e^t*(1-e^{2t})^{3/2}-e^t*3/2(1-e^{2t})*2t}{((1+e^{2t})^{3/2})^2}$$

Notice the blatant abuse of chain rule to get the last factor in the numerator.

Why are they doing this? am I missing something? are these people blatantly wrong? please help

If you don't believe me look at these pages:

question # 4 http://www.math.washington.edu/~conroy/m126-general/exams/mt2SolMath126Win2006.pdf

And look at chris' post: How to find the point where the curvature is maximum

fixed the typos, thanks to Lee and Mattos for pointing them out~

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helpmeh
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    You have so many errors in your formulas, it's impossible to tell what you are talking about. – Lee Mosher Dec 02 '16 at 01:27
  • The $\kappa$ in the linked question is different to your $\kappa$. Theirs has an $e^{2t}$ in the denominator, yours has an $e^{t}$. I'm assuming this is a typo on your part. Differentiating the $\kappa$ you provided at the top yields $$\kappa' = \frac{e^{t} - 1/2e^{2t}}{(1+e^{t})^{5/2}}$$ – Matthew Cassell Dec 02 '16 at 01:31
  • yes typo, thanks for pointing this out – helpmeh Dec 02 '16 at 01:31
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    In this post, Chris appears to have written $2x$ instead of $e^{2x}$ in the first quotient for $\kappa'(x)$. However, the next quotient in the equation ignores this and appears correct, indicating a mistype in the preceding expression. Your post contains many errors, as your $\kappa(t)$, $\kappa^1 (t)$, and quote of Chris' equation are not consistent or accurate as of right now. – Corellian Dec 02 '16 at 01:35
  • using quotient rule and chain rule $$(\frac{u}{v})'=\frac{u'v-uv'}{v^2}$$ take $u=e^t$ and $v=(1+e^{2t})^{3/2}$

    notice chain rule must be used when $v$ is differentiated.

     – helpmeh Dec 02 '16 at 01:39
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    Interestingly, the exam key you linked seems to exactly copy that post by Chris, and it contains the same error (if I'm correct about there being a mistype). – Corellian Dec 02 '16 at 01:43
  • @brody, I think i fixed all the errors, you see another mistake? why would multiple sources have this same error? these are not the only two sources displaying this error? is this a blatant case of plagiarism where the error was inadvertently copied? – helpmeh Dec 02 '16 at 01:44
  • oddly, another student in my class also must have copied this mistake down............. what a circle of plagiarism. – helpmeh Dec 02 '16 at 01:46
  • So are we in agreement that chris, and the author of that test solution made a mistake in derivation? – helpmeh Dec 02 '16 at 01:48
  • In my personal opinion, it is at least in poor taste to copy verbatim (without credit or citation) someone's original work when it is identified with that person, as with Chris' post. We already see one of the potential risks, i.e. inheriting the original's flaws. I'm not willing (or able) to say whether this counts as plagiarism or not. – Corellian Dec 02 '16 at 01:53
  • you're right i really should not be toying with that notion considering it is heavy. I think I will delete mentions of the p word. – helpmeh Dec 02 '16 at 01:54
  • Let us continue this discussion in chat. – Corellian Dec 02 '16 at 01:55
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    @Brody: The exam key is from a 2006 midterm, and Chris's post is from 2013, so it's clearly Chris who is plagiarizing. And that factor $2x$ should be $2 e^{2x}$ and nothing else. – Hans Lundmark Dec 02 '16 at 08:58

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