Question

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

Answer 1

By Implicit Differentiation,

`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}`