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# C3 Differentiating trig.. watch

1. y = (2x – 1) tan 2x, 0< x <pi/4 .

The curve has a minimum at the point P. The x-coordinate of P is k.

(a)Show that k satisfies the equation
4k + sin 4k – 2 = 0

So I got
u= 2x-1
du=2

v=tan2x
dv=sec^2 2x

V x du = 2tan2x
U x dv = (2x-1)(sec^2 2x)

Giving 2tan2x + 2x-1 sec^2 2x

This is bit correct?

2(sin2x/cos2x) + (2x-1)(1/cos^2 2x)

And I'm lost from there....(it said my latex formulas were dangerous?!)
2. yes your diff is correct....procede from here
3. thats on a solomon paper i think, and you need to insert K as the X co ordinate and proceed.
4. Its in the June 2006 paper and the mark scheme put a 2 infront of the (2x-1) when it's differentiated and I don't know why...
5. (Original post by jaz_jaz)
Its in the June 2006 paper and the mark scheme put a 2 infront of the (2x-1) when it's differentiated and I don't know why...
you get the factor of 2 when you differentiate the tan 2x
6. Can someone please tell me how to do it, I can't even do figure it out looking at the mark scheme!
7. tan 2x

diffs to 2sec^2 2x, cus the 2 comes it being 2x. just like sin 2x would diff to 2cos 2x.
8. The derrivative of tan 2x is 2sec^2(2x)
9. ok thanks but i still have no idea how to get to want they want
10. y = (2x – 1) tan 2x

dy/dx = 2(2x - 1)sec^2 + 2tan 2x

x=k

2(2k - 1)sec^2 2k + 2tan 2k = 0

(2k - 1)sec 2k + sin 2k = 0

sin2k = -(2k - 1)sec 2k

sin2kcos2k = -2k + 1

2sin2kcos2k = -4k + 2

[formula sheet states 2sin2kcos2k is equal to sin4k]

sin4k = -4k + 2

4k + sin4k - 2 = 0
11. Basically when you differentiate, you will get:

2tan2x + 2(2x-1)sec2x

Then you change everything into terms of sine and cosine:

(2sin2x)/(2cos2x) + (4x-2)/cos2x

to find the turning point, dy/dx = 0

therefore:

(2sin2x)/(2cos2x) + (4x-2)/cos2x=0

now, you can multiply through by cosx to get rid of the fractions:

2sin2xcos2x+4x-2=0

2sin2xcos2x is the identity sin2A

that should help.

edit: ojpowermake got there before me!
12. (Original post by ojpowermake)
y = (2x – 1) tan 2x

dy/dx = 2(2x - 1)sec^2 + 2tan 2x

x=k

2(2k - 1)sec^2 2k + 2tan 2k = 0

(2k - 1)sec 2k + sin 2k = 0

sin2k = -(2k - 1)sec 2k

sin2kcos2k = -2k + 1

2sin2kcos2k = -4k + 2

[formula sheet states 2sin2kcos2k is equal to sin4k]

sin4k = -4k + 2

4k + sin4k - 2 = 0
Thanks
13. (Original post by unamed)
Basically when you differentiate, you will get:

2tan2x + 2(2x-1)sec2x

Then you change everything into terms of sine and cosine:

(2sin2x)/(2cos2x) + (4x-2)/cos2x

to find the turning point, dy/dx = 0

therefore:

(2sin2x)/(2cos2x) + (4x-2)/cos2x=0

now, you can multiply through by cosx to get rid of the fractions:

2sin2xcos2x+4x-2=0

2sin2xcos2x is the identity sin2A

that should help.

edit: ojpowermake got there before me!
thanks i understand yours much better
14. for c4 paper people, i feel this shall definitely help..........
generally... to differentiate.. sinx ---> cosx ---> - sinx
to integrate .. -sinx ---> cosx ---> sinx ( so to integrate sinx you get -cos x ) + constant obvs...

hope this helps you further on..

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