cot inverse ( 1 - x / 1 + x )

\cot^{- 1} (\frac{1 - x}{1 + x}) let x = \tan y \cot^{- 1} (\frac{1 - \tan (\frac{π}{4}) \tan y}{\tan \frac{π}{4} + \tan y}) = \cot^{- 1} (\frac{1}{\tan (\frac{π}{4} + y)}) since \tan (x + y) = \frac{\tan x + tany}{1 - \tan x \tan y} = \cot^{- 1} (\cot (\frac{π}{4} + y)) = \frac{π}{4} + y = \frac{π}{4} + \tan^{- 1} x On differentiating wrt x, we get \frac{d (\frac{π}{4} + \tan^{- 1} x)}{dx} = \frac{1}{1 + x^{2}} ANS . . .

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