Question

find the y'' if y=6cscx

What is the rule to complete this question? Also how do you complete it?
Thanks.

Answer 1

take the derivative of the function twice.

Answer 2

If y' = -cot(x)csc(x)
take the derivative of that. Use product rule.

Answer 3

@abb0t So would the answer be -6cot(x)csc(x) ?

Answer 4

That is y', you still need to find y''.
so just take the derivative again of y' function I gave you there.

Answer 5

\[-6(-\csc^2x)(-\csc(x)\cot(x)) = -6(\csc^3x)(-\csc^2cotx) \]

Answer 6

Is this correct? but how do I continue on @abb0t

Answer 7

Um, almost! You should have cot\(^2\)(x)csc(x) idk if that is what you meant.

Answer 8

I don't know where to go once I get to −6(csc3x)(−csc2cotx)

Answer 9

6csc(x)[cot\(^2\)(x)+csc\(^2\)(x)]

Answer 10

\[\csc^2x(-6cscx+cotx)\]

Answer 11

Is that correct?

Answer 12

I have to go now, but that is your answer.

Answer 13

\[\Large y=6\csc x \qquad\qquad y'=-6\csc x \cot x\]
\[\Large y''=-6\color{royalblue}{(\csc x)'}\cot x-6\csc x\color{royalblue}{(\cot x)'}\]\[\Large y''=-6\color{royalblue}{(-\csc x \cot x)} \cot x-6 \csc x \color{royalblue}{(-\csc^2x)}\]

Answer 14

\[\Large y''=6\csc x \cot ^2x+6\cot x\csc^3x\]
So you get something like this yes? :o

Answer 15

Oops maybe I made a boo boo somewhere :D lemme check...

Answer 16

Wouldn't it be 6cscxcot^2x+6csc^3x

Answer 17

Without the cotx

Answer 18

Ya that sounds right :) woops, sily typo.

Answer 19

Thanks so much!