What is the derivative of #f(x)=cot^-1(x)# ?

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Answer 1 · 2014-09-24T04:40:43

#f'(x)=-1/{1+x^2}#

Let us look at some details.

By replacing #f(x)# by #y#,

#y=cot^{-1}x#

by rewriting in terms of cotangent,

#Rightarrow coty=x#

by implicitly differentiating with respect to x,

#Rightarrow -csc^2ycdot{dy}/{dx}=1#

by dividing by #-csc^2y#,

#Rightarrow {dy}/{dx}=-1/{csc^2y}#

by the trig identity #csc^2y=1+cot^2y=1+x^2#,

#Rightarrow {dy}/{dx}=-1/{1+x^2}#

Hence,

#f'(x)=-1/{1+x^2}#