Question

find dy/dt

using the chain rule.
y=1/6(1+cos^2(7t)^3

Answer 1

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

Answer 2

how did you get the -3?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

cAN YOU EXPLAIN HOW YOU GOT THAT?

Answer 7

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.

Answer 8

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))

Answer 9

okay,