Question

im having trouble
Find dy/dt.
y = cos^6(piΔt - 9)

A) -6 cos^5(pi* t - 9) sin(pi*t - 9)
B) -6 picos^5(pi *t - 9) sin(pi*t - 9)
C) 6 cos^5(pi *t - 9)
D) -6 sin^5(pi *t - 9)

Answer 1

Use the chain rule.

Answer 2

take me through the steps?

Answer 3

none of these look right to me, but maybe i don't understand the question

Answer 4

uhh let me check

Answer 5

The 14) should be on a separate line.

Answer 6

what is the 14 doing out at the end?

Answer 7

yeh it shouldnt be there sorry

Answer 8

those triangle should be empty spaces

Answer 9

There should be a factor of pi multiplying the answer.

Answer 10

ok now what about the \(\pi t\) ? if it is there, there should be a coefficient including \(\pi\) out front somewhere

Answer 11

in the question there isnt one

Answer 12

Which is the problem we (answering it) are having.

Answer 13

cos^6(pit-9)

Answer 14

well what is dy/dt?

Answer 15

\[f(x)=\cos^6(\pi t-9)\] ?

Answer 16

\[y=\cos^6 (\pi t -9)\]
\[ =[\cos(\pi t -9)]^6\]

Answer 17

so what does that mean exactly?

Answer 18

Ah nvm what I said before. You just use the chain rule.

Answer 19

the answer that i have doesnt match any of hte available

Answer 20

6 π cos^5(9-π t) sin(9-π t)

Answer 21

In the parenthesis it should be pi t minus nine, not nine minus pi t.

Answer 22

Yep there should be a pi in front of the 6.

Answer 23

so im clueless

Answer 24

A) -6 cos^5(pi* t - 9) sin(pi*t - 9)
B) -6 cos^5(pi *t - 9) sin(pi*t - 9)
Both A and B are the same.

Answer 25

You must of done a typo youself or if you copied correctly, one of them should have a pi in front of the -6.

Answer 26

yeh i see one has a pi after the six and a doesnt

Answer 27

A) -6 cos^5(pi* t - 9) sin(pi*t - 9)
B) -6 picos^5(pi *t - 9) sin(pi*t - 9)

Answer 28

Now that's better.

Answer 29

Could you show me you working now?

Answer 30

So you use the chain rule and what did you get?

Answer 31

umm

Answer 32

Remember:
\[\frac{d}{dx}\cos x =-\sin x\]

Answer 33

That's for reference to when you're differentiating cosine.

Answer 34

d/dt(cos^6(pit-9))
=d/dt(cos^6(9-pit))
=6 cos^5(9-pit)[d/dt(cos(9-pit))]
=-d/dt(9-pit)sin(9-pit) 6cos^5(9-pit)
=-6 sin(9-pit) cos^5(9-pit)[d/dt(9-pit)]
=d/dt(9)-pid/dt(t) -6 cos^5(9-pit) sin (9-pit)
=-6 sin(9-pit) cos^5(9-pit) [0-pi[d/dt(t)]]
=6 pi sin(9-piy) cos^5(9-pit)[d/dt(t)]
=1 6 pi cos^5(9-pit) sin(9-pit)
=6 pi sin(9-pit)cos^5(9-pit)

Answer 35

Why'd you change the pit-9 to 9-pit? Seems like a habit for you.

Answer 36

You don't need ot change it around and do unnecessary switcheroos.

Answer 37

need to*

Answer 38

ill be honest that is what the computation calculator does

Answer 39

Computation calculator. Did you do it yourself using pen and paper? It's much easier for you to do this with pen and paper.

Answer 40

i used the calculator because i dont know how to do it

Answer 41

Okay, let's see:
\[y = \cos^6(\pi t - 9)\]
\[y=[\cos(\pi t -9)]^6\]
Have you learnt the chain rule yet? If you have, then have you paid attention in class about moving the power to the front and differentiating the inside, etc.?

Answer 42

no i did not pay attention

Answer 43

Is this familiar with you or not? Because I need to know whether you've done questions that involve differentiation with the chain rule.

Answer 44

Okay.

Answer 45

This is tricky, because if you fall behind in maths, you will struggle. Okay...
Can you differentiate this using the chain rule?
\[\frac{d}{dx}(x-1)^2=?\]

Answer 46

maybe the calculator can

Answer 47

No, please don't use the calculator...

Answer 48

Would you be allowed to have a calculator of this kind when you do a test/exam?

Answer 49

The answer would most likely be NO. So for your benefit, you need to do this with pen and paper.

Answer 50

this is one question out of like 28 for my final which is due tomorrow by midnight. i had 6 chapters to read this semester and only read 3. it is an online class so i submit tomorrow night

Answer 51

i never finished chain rule i dont think but it is too late now

Answer 52

Did procrastination lead to this? SHould of started when you got the final.

Answer 53

i have to try to pass this final with as much correct answers. I KNOW i got it on like last weekend but waited until like today or yesterday to start

Answer 54

Don't teachers teach you? I find many students on OpenStudy learn online.

Answer 55

We're not responsible for your wrongdoing. We're only here to help guide you with questions.

Answer 56

That's why people with that attitude won't do well. You have to ASK the TEACHER for help and assistance when unsure of anything. Even if it was simple knowledge. the teacher gets paid for all this.

Answer 57

The chain rule is simple if you can follow this:
\[\frac{d}{dx}(x-1)^2=2(x-1)^{2-1}\times ?\]
First I rewrote the expression. Then I moved the power to the front. Then I subtracted 1 from the power. Now you can see there is a question mark which indicates somethings missing there when I'm differentiating using chain rule. I will talk about this in the next step.
Okay:
\[\frac{d}{dx}(x-1)^2=2(x-1)^{2-1}\times \frac{d}{dx}(x-1)\]
As you can see, I filled in the question mark with the derivative of x-1. That is simply what's inside (x-1)^2. So basically that is all you need to know for the chain rule. Here's a summary of the steps in using the chain rule.
1. Rewrite expression
2. Bring the power to the front.
3. Subtract one from the original power.
4. Multiply that result with the derivative of what's inside the paranthesis/brackets.

Answer 58

Apply this to your question as well as using your knowledge of differentiating trig functions.