En
  • En
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)
Mathematics
jamiebookeater
  • jamiebookeater
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freckles
  • freckles
I recommend using the double angle identity first to rewrite w then differentiate
freckles
  • freckles
\[\sin(2y)=2 \sin(y) \cos(y) \\ \frac{1}{2} \sin(2y)=\sin(y) \cos(y) \\ w=( \frac{1}{2} \sin(2y))^4 \]
En
  • 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

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freckles
  • freckles
\[2\sin(y)\cos(y)=\sin(2y)\]
freckles
  • freckles
oh you used that
anonymous
  • anonymous
@En dont copy and past from yahoo answers.. its best to solve them yourself for you can learn more.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=derivative+of+sin^4y+cos^4y
freckles
  • freckles
\[4=\frac{1}{2}(2)^3 \text{ they just rewrote 4 }\]
En
  • En
freckles
  • freckles
np i was kind of confused at first i thought it was your solution
freckles
  • freckles
@En I feel like that first way I suggested might be a little easier
En
  • En
i solved the 4sin^3cos^3y * cos2y i just got confused in the simplifying process :)
freckles
  • 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
En
  • En
Where did you get the w=1/2sin(2y))^4? @freckles
freckles
  • freckles
the double angle identity 2sin(y)cos(y)=sin(2y) so sin(y)cos(y)=1/2*sin(2y)
freckles
  • freckles
so (sin(y) cos(y))^4=(1/2*sin(2y))^4
freckles
  • freckles
and (sin(y) cos(y))^4=sin^4(y)*cos^4(y)
En
  • En
where did the "1/2" came from?
freckles
  • freckles
2sin(y)cos(y) is equal to sin(2y) do you understand this part?
freckles
  • freckles
divide both sides by 2 sin(y)cos(y) is equal to 1/2*sin(2y)
En
  • En
yes

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