anonymous
  • anonymous
find dy/dt using the chain rule. y=1/6(1+cos^2(7t)^3
Mathematics
jamiebookeater
  • jamiebookeater
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jennychan12
  • jennychan12
6(1+cos^2(7t))^-3 6(-3(1+cos^2(7t))(-sin(7t))(7) Simplify.
anonymous
  • anonymous
how did you get the -3?
jennychan12
  • jennychan12
for example, 1/x is x^-1.

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anonymous
  • anonymous
\[y=1/6(1+\cos ^{2}(7t))^{3}\]
jennychan12
  • jennychan12
you can rewrite it as 6(1+cos^2(7t))^-3 6(-3(1+cos^2(7t))(-sin(7t))(7) Simplify and rewrite.
anonymous
  • anonymous
cAN YOU EXPLAIN HOW YOU GOT THAT?
anonymous
  • anonymous
Assuming you have: \[y=\frac{1}{6}(1+\cos^2(7t))^3\] Then you have quite a few chain rules. One from the cos(7t) one from cos^2(7t) and one from (1+cos^2(7t))^3. So lets move from the outside in: \[\dot{y}=\frac{1}{6}(3)(1+\cos^2(7t))^2 \frac{d}{dt}(1+\cos^2(7t))=\frac{1}{6}(3)(1+\cos^2(7t))^2 \frac{d}{dt}\cos^2(7t)\] \[\implies \frac{1}{6}(3)(1+\cos^2(7t))^2\left[(2)(7)\cos(7t)(-\sin(7t) \right]\] Then simplify.
anonymous
  • anonymous
You're still isn't right thought. When you take the inner derivative of cos^2(7t) you should get a 2(7)cos(7t)(-sin(7t))
anonymous
  • anonymous
okay,

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