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

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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,

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

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