i hate this paper, its got so many questions that im like "omg, help!" ...but im really stuck at number 6 ..."If f(x)=cos(2x)-3sin(4x), find the exact value of f'(pie/6).


2003 paper 2 Q6
Started by Azie, May 14 2007 02:25 PM
8 replies to this topic
#1
Posted 14 May 2007 - 02:25 PM
#2
Posted 14 May 2007 - 02:51 PM
you would find out what f'(x) is which is differentiating the sinx and cosx, then sub in pie/6 and you should know what your values are by changing the pie/6 into degrees (which is 30 degrees) and then find exact values from the exact values table or triangle.
hope that makes a bit of sense to you !
hope that makes a bit of sense to you !
#3
Posted 14 May 2007 - 02:55 PM
actually just ignore me, i think im talking a lot of rubbish there! sorry!
#4
Posted 14 May 2007 - 02:58 PM
i think u do differenciate but i jus dnt no how or where 2 start ...i dunno if u use the double angle formulas or the standard derivatives :s
#5
Posted 14 May 2007 - 03:01 PM
You differentiate the function and then sub pie/6 into the equation and then simplify
#6
Posted 14 May 2007 - 03:04 PM
To do the differentiation, just use the standard derivatives from the front of the exam paper, e.g.
, then sub in
to the result.


#7
Posted 14 May 2007 - 04:06 PM
nope, i still dnt get it
...any chance u's cud show me an example using the actual numbers with dumbed down solutions coz i can b really slow an get confused easily when it cums 2 questions like these :s

#8
Posted 14 May 2007 - 04:13 PM
okay
f(x) = cos2x - 3sin4x
f'(x) = -2sin2x - 12cosx
f'(pi/6) = -2sin(2pi/6) - 12cos(4pi/6)
= -2sin60 - 12cos120
=-2(root3/2) - 12(-0.5)
= -2root3/2 + 6
= -root3 + 6
or
6 - root3
f(x) = cos2x - 3sin4x
f'(x) = -2sin2x - 12cosx
f'(pi/6) = -2sin(2pi/6) - 12cos(4pi/6)
= -2sin60 - 12cos120
=-2(root3/2) - 12(-0.5)
= -2root3/2 + 6
= -root3 + 6
or
6 - root3
#9
Posted 14 May 2007 - 04:28 PM
aw, i think i get it now, thanks a lot

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