Question

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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

\[2\sin(y)\cos(y)=\sin(2y)\]

Answer 5

oh you used that

Answer 6

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

Answer 7

http://www.wolframalpha.com/input/?i=derivative+of+sin^4y+cos^4y

Answer 8

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

Answer 9

@freckles tnx!

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

where did the "1/2" came from?

Answer 19

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

Answer 20

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

Answer 21

yes