Question

find y'' for cotx

Answer 1

So ummm.. do you have the first derivative memorized or no?
If not, you can rewrite your expression as cosx/sinx and apply the quotient rule.

Answer 2

the first derivative is -csc^2x but i don't know how to get the derivative of that

Answer 3

Do you remember this derivative? \(\Large\bf\sf (csc x)'\quad=\quad ?\)

Answer 4

is it -cscx cotx

Answer 5

yes good good.
So we're starting with,\[\Large\bf\sf \left[-(\csc x)^2\right]'\quad=\quad -2(\csc x)\cdot \color{royalblue}{(\csc x)'}\]Power rule first, make sense?
Then chain.

Answer 6

-2(cscx)(cscx)*(-cscxcotx)?

Answer 7

You have one too many cscx in there..
Not sure what happened.\[\large\bf\sf \left[-(\csc x)^2\right]'\quad=\quad -2(\csc x)\cdot \color{royalblue}{(\csc x)'}\quad=\quad -2(\csc x)\cdot \color{royalblue}{(-\csc x \cot x)}\]