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2003 paper 2 Q6

Azie

Posted 14 May 2007 - 02:25 PM

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).

x_Jo_x

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 !

x_Jo_x

Posted 14 May 2007 - 02:55 PM

actually just ignore me, i think im talking a lot of rubbish there! sorry!

Azie

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

Chas

Posted 14 May 2007 - 03:01 PM

You differentiate the function and then sub pie/6 into the equation and then simplify

George

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. 

\begin{align*}\frac{d}{dx}\bigl( \cos(2x) \bigr) = -2\sin(2x)\end{align*}

, then sub in \frac\pi6 to the result.

Azie

Posted 14 May 2007 - 04:06 PM

nope, i still dnt get it sad.gif ...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

x_Jo_x

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



Azie

Posted 14 May 2007 - 04:28 PM

aw, i think i get it now, thanks a lot smile.gif

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