## cwrw238 2 years ago Differentiate sinsinsin2x

1. cwrw238

i'm confused with this one i guess you use the chain rule?

2. Noliec

You do, it's horrifying! :(

3. Noliec

d(sin(sin(sin(2x))))/dx=dsinu/du*du/dx u=sin(sin(2x)) and dsinu/du=cosu

4. Noliec

And then you apply it once more but that gets too messy for me to type - bound to make a mistake. Hehe, hope my answer helped to some extent atleast; good luck!

5. Noliec

*twice more

6. Noliec

Using the chain rule three times.

7. cwrw238

ugh - yes! ill try that

8. .Sam.

you should be getting $2 \cos (2 x) \cos (\sin (2 x)) \cos (\sin (\sin (2 x)))$

9. cwrw238

i got 2cos(sinsin2x)*cos(sin2x)*2cos2x

10. cwrw238

not quite i've got one extra '2'

11. Noliec

Oh right, missed that one, sorry

12. cwrw238

i think sam is right though

13. cwrw238

yea the first 2 is wrong

14. cwrw238

thnx guys

15. cwrw238

i did it in steps from left to right

16. Chlorophyll

Sam's correct! You should take derivative from RIGHT to left: ( sin2x )' = 2 cosx sin ( sin2x) ' = 2 cosx * cos ( sin2x ) sin [ sin ( sin2x)] ' = 2 cosx * cos ( sin2x ) * cos [ sin ( sin2x)] Just go step by step, you won't miss it :)

17. Aron_West