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# ANother C3 Question!! Urgent watch

1. Given that x= 4sin(2y+6), find dy/dx in terms of x. please work it out for me simple and clear(step by step explaining how you got each equation).
2. dx/dy=8cos(2y+6)

dy/dx=1/(dx/dy)
3. (Original post by gilbo)
dx/dy=8cos(2y+6)

dy/dx=1/(dx/dy)
i got dah aswell
4. (Original post by gilbo)
dx/dy=8cos(2y+6)

dy/dx=1/(dx/dy)

can you do the full working please? because simply 1/dy/dx is not enough
it wants it interms of x.
5. 1/8cos(2y+6) i think..

dy/dx sinx = cosx

dy/dx 4sin(2x+6) would give you 8sin(2x+6)

therefore

dx/dy 4sin(2y+6) = 8cos(2y+6)

dy/dx= 1/dx/dy = 1/ 8cos (2y+6)
6. (Original post by sahil112)
i got dah aswell
is incomplete, the question wants in terms of x and i am stuck on here.
7. oh **** yeh mu bad i think you need to do a substitution. u = _______ blady blah
8. (Original post by zeb191)
1/8cos(2y+6) i think..

dy/dx sinx = cosx

dy/dx 4sin(2x+6) would give you 8sin(2x+6)

therefore

dx/dy 4sin(2y+6) = 8cos(2y+6)

dy/dx= 1/dx/dy = 1/ 8cos (2y+6)
but the question is askin in terms of x(your working is correct but incomplete and im stuck on here)
9. (Original post by gilbo)
oh **** yeh mu bad i think you need to do a substitution. u = _______ blady blah
care to do it please i need to know how do get the final answer in clear step by step
10. He needs it in terns of x and ive bloody forgotten how to do it, im getting the poster dw.
11. (Original post by gilbo)
He needs it in terns of x and ive bloody forgotten how to do it, im getting the poster dw.
Urgently! please I'm going to bed soon and i want to get this out of my head lol

12. By making y the subject.
13. x=4sin(2y+6)
so x/4=sin w if w=2y+6

so dy/d= 1/8cos w
we know cos^2w+sin^2w=1 so cos w=1- sqrt sin2w

so cs w = 1- sqrt x2/4^2

so dy/dx=1/8 sqrt 1- x^2/16

so dy/dx = 1/2 sqrt 16-x^2

hope that kinda made sense... sqrt = square root, and so answer = 1/ 2 times square root of 16- x^2
14. x=4sin(2y+6)
I'm going to use the chain rule.
u=2y+6
du/dy=2

x=4sinu
dx/du=4cosu

Dx/dy=du/dy*dx/du=2x4cosU=8cosU
Dy/Dx is the inverse of this so 1/(8cosU)

To find in terms of x, simply find U in terms of x. (there may be a prettier method but I think this works)
x=4sinU
x/4=sinu
u=arcsin(x/4)
so dy/dx in terms of x is
1/(8cos(arcsin(x/4)))

15. (Original post by tgodkin)

By making y the subject.
how did you make y the subject???

how did you get y out of the bracket???

edit: i know how you got x/4 but then how did you get y on its own...

x/4=sin(2y+6)....
16. (Original post by Remarqable M)
how did you make y the subject???

how did you get y out of the bracket???

17. (Original post by tgodkin)

THANK YOU SO MUCH! It makes sense now! REP+++++

This thread is no longer needed lol
18. One more thing say we have ln instead of sin example,

x=ln(2y+6) if i want y is it y=e^x-6/2 ??
19. yes.
PS. according to my source, his differentiation is wrong. at any rate, you can't differentiate inverse sine yet.

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Updated: January 20, 2010
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