## En one year ago PLEASE HELP! find the derivative of w=sin^4ycos^4y. my final answer is 4sin^3cos^3y * cos2y. but the answer written in my book is 1/2sin^3 (2y)cos(2y)

1. freckles

I recommend using the double angle identity first to rewrite w then differentiate

2. freckles

$\sin(2y)=2 \sin(y) \cos(y) \\ \frac{1}{2} \sin(2y)=\sin(y) \cos(y) \\ w=( \frac{1}{2} \sin(2y))^4$

3. En

w = sin^4y cos^4y dw/dy = 4 sin^3y* cos^5y - 4 cos^3y* sin^5y = 4 sin^3y cos^3y[ cos^2y - sin^2y] = 4 sin^3y cos^3y * cos2y = 1/2*(2siny cosy)^3 * cos2y = 1/2 sin^3(2y) cos2y someone answere it this way... i just dont get where the 1/2 came from.. please explain

4. freckles

$2\sin(y)\cos(y)=\sin(2y)$

5. freckles

oh you used that

6. anonymous

@En dont copy and past from yahoo answers.. its best to solve them yourself for you can learn more.

7. anonymous
8. freckles

$4=\frac{1}{2}(2)^3 \text{ they just rewrote 4 }$

9. En

@freckles tnx!

10. freckles

np i was kind of confused at first i thought it was your solution

11. freckles

@En I feel like that first way I suggested might be a little easier

12. En

i solved the 4sin^3cos^3y * cos2y i just got confused in the simplifying process :)

13. freckles

$w=(\frac{1}{2} \sin(2y))^4 \\ w'=4(\frac{1}{2}\sin(2y))^{3}(\frac{1}{2} \sin(2y))' \\ w'=4(\frac{1}{2}\sin(2y))^3 \frac{1}{2} \cdot 2 \cos(2y) \\ w'=4 (\frac{1}{2})^3 \sin^3(2y) \cos(2y) \\ w'=\frac{4}{8} \sin^3(2y) \cos(2y) \\ w'=\frac{1}{2} \sin^3(2y) \cos(2y)$ just a whole bunch of chain rule

14. En

Where did you get the w=1/2sin(2y))^4? @freckles

15. freckles

the double angle identity 2sin(y)cos(y)=sin(2y) so sin(y)cos(y)=1/2*sin(2y)

16. freckles

so (sin(y) cos(y))^4=(1/2*sin(2y))^4

17. freckles

and (sin(y) cos(y))^4=sin^4(y)*cos^4(y)

18. En

where did the "1/2" came from?

19. freckles

2sin(y)cos(y) is equal to sin(2y) do you understand this part?

20. freckles

divide both sides by 2 sin(y)cos(y) is equal to 1/2*sin(2y)

21. En

yes