How Do I Reach g'(\epsilon) in Variational Calculus?

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The discussion centers on the challenge of deriving g'(\epsilon) in variational calculus, specifically referencing a post on the NRICH forum. The user expresses confusion regarding the left-hand side (LHS) of g' and seeks clarification on the derivation process. Recommendations include utilizing WolframAlpha's "show steps" feature to aid in understanding the differentiation of functions involving derivatives, such as d/dx (u'(x)/(1+(u'(x))^2)^0.5).

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I asked this question in here:
https://nrich.maths.org/discus/messages/7601/151442.html?1310911861
(post 4), and still haven't got any reply, does someone know how to get to [tex]g'(\epsilon)[/tex] in the text cause I get something different with another terms plus the terms in the text.

I guess i'll need to procceed without understanding how did he get the LHS fo g'.

)-:
 
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I have no idea which is "post 4" that you're asking about.
Perhaps a date and time of your post? Even better, edit the relevant image with MS Paint to add line numbers?

Have you tried WolframAlpha on your step?
Particularly, the "show steps" feature.

I processed through half of
d/dx ( u'(x)/(1+(u'(x))^2)^.5 )
when I discovered you were asking about an entirely different line.
I did find it helpful to ask Wolfram,
d/dx ( w'(x)/(1+(w'(x))^2)^.5 )
because the show steps, uses d/dx (uv) = uv' + vu' and that uses u for too many different ideas.

I'm running out of concentration ability.
 

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