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What is the derivative of #y=cot^{-1}(x)#?

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Sep 14, 2014

The answer is #y'=-1/(1+x^2)#

We start by using implicit differentiation:

#y=cot^(-1)x#
#cot y=x#
#-csc^2y (dy)/(dx)=1#
#(dy)/(dx)=-1/(csc^2y)#
#(dy)/(dx)=-1/(1+cot^2y)# using trig identity: #1+cot^2 theta=csc^2 theta#
#(dy)/(dx)=-1/(1+x^2)# using line 2: #cot y = x#

The trick for this derivative is to use an identity that allows you to substitute #x# back in for #y# because you don't want leave the derivative as an implicit function; substituting #x# back in will make the derivative an explicit function.

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