## anonymous 4 years ago find dy/dt using the chain rule. y=1/6(1+cos^2(7t)^3

1. jennychan12

6(1+cos^2(7t))^-3 6(-3(1+cos^2(7t))(-sin(7t))(7) Simplify.

2. anonymous

how did you get the -3?

3. jennychan12

for example, 1/x is x^-1.

4. anonymous

$y=1/6(1+\cos ^{2}(7t))^{3}$

5. jennychan12

you can rewrite it as 6(1+cos^2(7t))^-3 6(-3(1+cos^2(7t))(-sin(7t))(7) Simplify and rewrite.

6. anonymous

cAN YOU EXPLAIN HOW YOU GOT THAT?

7. 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.

8. 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))

9. anonymous

okay,