How do you find the derivative of #(cot^(2)x)#?

2 Answers

#-2"cosec"^2 x. cotx#

Explanation:

#f(x) = cot^2x#

#d [f(x)]/dx = 2cot x.(-"cosec"^2x).1#

#= -2"cosec"^2 x. cotx#

first you get the derivative of the indice, then the trig function, lastly the x.
it's easy if you remember it that way ;)

#-(2cotx)/sin^2x#

Explanation:

I like the previous answer, but I think it's simpler to use the product rule. Since #cot^2x=cotx*cotx#, #d/dx(cot^2x)=f'(x)g(x)+g'(x)f(x)#, where #f(x)=g(x)=cotx#. This evaluates to #cotx xx (-1/sin^2x)+(-1/sin^2x)xxcotx =2(-1/sin^2x)cotx#, #-(2cotx)/sin^2x#. This is equivalent to the previous answer, since #1/sin^2x="cosec" ^2x#, but I think that that answer is cleaner, don't you?