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amistre64Best ResponseYou've already chosen the best response.0
\[[ln(u)]'=\frac{u'}{u}\]
 2 years ago

calyneBest ResponseYou've already chosen the best response.1
what are those brackets or 
 2 years ago

calyneBest ResponseYou've already chosen the best response.1
because it's not lnu it's lnu
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
those are just for grouping; it means that the whole of it is derived and not just the (u) part
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
i believe it goes piece wise in the end
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
or you can use the sqrt(u)^2 trick
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
\[\frac{d}{dx}ln(\sqrt{u\ }^2)=\frac{(\sqrt{u\ }^2)'}{ \sqrt{u\ }^2}\to\ \frac{2\sqrt{u\ }*\sqrt{u}'}{ \sqrt{u\ }^2}\]
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
if i follow that derivatives i think it would look like this ... but best to check with the wolf
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=derivative+ln+%7C2x5x%5E2%7C the wolf doesnt seem to go thru the hoopla tho
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
\[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\sqrt{u}'\] \[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\frac{u'}{2\sqrt{u}}=\frac{u'}{ \sqrt{u\ }^2}\to\ \frac{u'}{ u}\]
 2 years ago

calyneBest ResponseYou've already chosen the best response.1
whatevr it's 1/(2x5x^2) * (10x+1) = 10x+1 / 5x^2+x2 woohoo
 2 years ago
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