Question

help!!!

Answer 1

lol what are you nuts? gotta use the quotient rule and it will be a big mess

Answer 2

use wolfram

Answer 3

Let


\[\Large y = \frac{u}{v}\]


that would mean


\[\Large u = 5 + \csc(x)\]
\[\Large v = 9 - \csc(x)\]

Answer 4

u = 5 + csc(x)

u ' = -csc(x)*cot(x) ... derive both sides with respect to x

-----------------

v = 9 - csc(x)

v ' = csc(x)*cot(x)

Answer 5

ahhh im so confused:(

Answer 6

jum_thompson5910 thank you!

Answer 7

and satellite 73, if he can't get it ill go to wolfram lol it always gives me wrong answers so i avoid that :/

Answer 8

Now we use the quotient rule

\[\Large y = \frac{u}{v}\]

\[\Large y^{\prime} = \frac{u^{\prime}*v - v^{\prime}*u}{v^2}\]

\[\Large y^{\prime} = \frac{-\csc(x)*\cot(x)*(9 - \csc(x)) - \csc(x)*\cot(x)*(5 + \csc(x))}{(9 - \csc(x))^2}\]


So yeah, it's a mess. However, once you plug in x = pi/6, a lot of things will simplify.

Answer 9

thats the final answer!!!>>>>

Answer 10

no

Answer 11

its a big mess, i don't even want an explanation because i will never know how to do this

Answer 12

you would then plug in x = pi/6

Answer 13

last bit got cut off, let me repost

Answer 14

\[\large y = \frac{u}{v}\]

\[\large y^{\prime} = \frac{u^{\prime}*v - v^{\prime}*u}{v^2}\]

\[\large y^{\prime} = \frac{-\csc(x)*\cot(x)*(9 - \csc(x)) - \csc(x)*\cot(x)*(5 + \csc(x))}{(9 - \csc(x))^2}\]

Answer 15

it got cut off again:(

Answer 16

one sec

Answer 17

take your time, no worries:) you've helped me a lot today!

Answer 18

lol told you you must be nuts to do this
what did you do to epenner?

Answer 19

i don't know, i tried logging in and it wouldn't let me!!!

Answer 20

you both should help me with my exam on sept 28:) I've been looking at that date like its my death sentence!!! :(

Answer 21

ok let's hope this pdf works

Answer 22

I recommend zooming in a bit

Answer 23

works perfectly let me submit the answer thank you so much!!!

Answer 24

that's not the answer though

Answer 25

we're almost there

Answer 26

After you derive, you plug in x = pi/6

\[\large y^{\prime} = \frac{-14\csc(x)*\cot(x)}{(9 - \csc(x))^2}\]

\[\large y^{\prime} = \frac{-14\csc(\pi/6)*\cot(\pi/6)}{(9 - \csc(\pi/6))^2}\]

\[\large y^{\prime} = \frac{-14*2*\sqrt{3}}{(9 - 2)^2}\]

\[\large y^{\prime} = \frac{-28*\sqrt{3}}{7^2}\]

\[\large y^{\prime} = \frac{-28*\sqrt{3}}{49}\]

\[\large y^{\prime} = \frac{-4*\sqrt{3}}{7}\]

Answer 27

okay! i didn't submit it:)

Answer 28

So \[\large \frac{-4\sqrt{3}}{7}\] is the slope of the tangent line at x = pi/6

Answer 29

so that is it?

Answer 30

yes

Answer 31

you just used the identities to condensed the last part right?

Answer 32

it was right:) thank you

Answer 33

Well in the pdf, I combined like terms to simplify (that's after I used the quotient rule)

Answer 34

<3 you helped me a lot!!! thank you!!!

Answer 35

when I plugged in x = pi/6, I used the unit circle to evaluate

Answer 36

i kinda followed this one more than the last one lol

Answer 37

so it's slowly starting to make more sense

that's good

Answer 38

yes!!!!! you explain better than my stupid online class!!!!