Calculating g`(π/3): An Attempted Solution

[SwedishFred]

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..

 

[SammyS]

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..

You will need to use enough parentheses to remove ambiguity regarding your function.

Is the 1/2 an exponent?

Is that sin(2u) or is it sin²(u) ?

etc.
...

.

 

[SwedishFred]

Oki
g(u)=e^cos(u)-1/2-6sin(2u)

g´(π/3)= -12cos(π/3)-e^cos(π/3)sin(π/3)
= -7.43 that dosen´t seem right..

 

[SammyS]

Oki
g(u)=e^cos(u)-1/2-6sin(2u)

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]

So i figure, where did i mess up , all is it all wrong ??

 

[scurty]

So i figure, where did i mess up , all is it all wrong ??

You made errors when using the chain rule (twice). I would go back and review that and revisit this problem after.