The discussion revolves around calculating the derivative g`(π/3) for the function g(u) = (e^cosu - 1/2) - 6sin(2u). Participants are attempting to derive the correct expression for the derivative and evaluate it at π/3.
The discussion is ongoing, with participants expressing uncertainty about their results and questioning where errors may have occurred in their differentiation process. Some guidance has been offered regarding the potential misuse of the chain rule.
There are indications of ambiguity in the function's notation, which may affect the interpretation of the problem. Participants are encouraged to clarify these aspects before proceeding further.
SwedishFred said:Homework Statement
Deside g`(π/3) when g(u)= (e^cosu-1/2)-6sin2u
Homework Equations
g`(π/3) when g(u)= (e^cosu-1/2)-6sin2u
3. The attempended solutions
ended upp with g`(u) = -12cosu-(e^cosu) sinu
g´(π/3)= -12cos(π/3)-(e^cos(π/3)) sin(π/3)
= -7.43 that dosen´t seem right..
It's not right.SwedishFred said:Oki
g(u)=ecos(u)-1/2-6sin(2u)
g´(π/3)= -12cos(π/3)-ecos(π/3)sin(π/3)
= -7.43 that dosen´t seem right..
SwedishFred said:So i figure, where did i mess up , all is it all wrong ??