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roselin
 3 years ago
find dy/dt
using the chain rule.
y=1/6(1+cos^2(7t)^3
roselin
 3 years ago
find dy/dt using the chain rule. y=1/6(1+cos^2(7t)^3

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jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.06(1+cos^2(7t))^3 6(3(1+cos^2(7t))(sin(7t))(7) Simplify.

roselin
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get the 3?

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0for example, 1/x is x^1.

roselin
 3 years ago
Best ResponseYou've already chosen the best response.0\[y=1/6(1+\cos ^{2}(7t))^{3}\]

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0you can rewrite it as 6(1+cos^2(7t))^3 6(3(1+cos^2(7t))(sin(7t))(7) Simplify and rewrite.

roselin
 3 years ago
Best ResponseYou've already chosen the best response.0cAN YOU EXPLAIN HOW YOU GOT THAT?

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.0Assuming 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.

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.0You'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))
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