Question

h'(t)=?
h(t)=2cot^2(πt +2)

Answer 1

Seems like it could be a chain-rule.

Answer 2

h'(t)=2cot(πt+2)^2 ?

Answer 3

What class is this for?

Answer 4

ok are you there?

Answer 5

Please be sure our problem is correct:\[h(t)=2cot^2(\pi t+2)\]\[Find\ h'(t).\]

Answer 6

h'(t)=4cot(πt+2)*-csc(πt+2)(π)

Answer 7

yes the problem is correct

Answer 8

Have you tried product rule? Separating the function in cot()cot()?

Answer 9

oh

Answer 10

Hmm... I am still toying around with this trying to find a good explanation. Are you still here @dbrown41294?

Answer 11

yes

Answer 12

Let's rewind and consider just \[d(cot^2x)/dx\]\[u=cot(x)\]\[d(u^2)/du=2u(u')\]\[u'=-csc^2x\]\[d(cot^2x)/dx=-2cot(x)csc^2x\]

Answer 13

So can you figure out the derivative now? I used the chain rule to find d(cot^2)

Answer 14

is it -4π cot(πt+2)csc^2(πt+2)?

Answer 15

@dbrown41294 That is correct, well done.

Answer 16

ok thanks