What is the derivative of # (cot^(2)x)#?

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Answer 1 · 2015-12-18T02:41:03

#-2csc^2xcotx#

Treat #cot^2x# like #u^2#, where #u=cotx#.

According to the chain rule,

#d/dx[u^2]=2u*u'#

Therefore,

#d/dx[cot^2x]=2cotx*d/dx[cotx]#

Know that #d/dx[cotx]=-csc^2x#

Thus,

#d/dx[cot^2x]=-2csc^2xcotx#