If y=Tan−1⁡2x1−x2+Tan−1⁡3x−x31−3x2-Tan−1⁡4x−4x31−6x2+x4 then S.T. dydx=11+x2

Given y=Tan−1⁡2x1−x2+Tan−1⁡3x−x31−3x2−Tan−1⁡4x−4x31−6x2+x4put x=Tan⁡θ⇒θ=Tan−1⁡(x)y=Tan−1⁡2Tan⁡θ1+Tan2⁡θ+Tan−1⁡3Tan⁡θ−tan3⁡θ1−3Tan2⁡θ−Tan−1⁡4Tan⁡θ−4Tan3⁡θ1−6Tan2⁡θ+Tan4⁡θ=Tan−1⁡(Tan⁡2θ)+Tan−1⁡(Tan⁡3θ)−Tan−1⁡(Tan⁡4θ)=2θ+3θ−4θ ∵tan−1⁡(tan⁡θ)=θ∴y=θ⇒y=Tan−1⁡(x)Diff ‘y’ w.r.t. ‘x’ on B.S⇒dydx=11+x2

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If $y = {Tan}^{- 1} (\frac{2 x}{1 - x^{2}}) + {Tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) - {Tan}^{- 1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}}) then S.T. \frac{d y}{d x} = \frac{1}{1 + x^{2}}$

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Detailed Solution

Given $y = {Tan}^{- 1} (\frac{2 x}{1 - x^{2}}) + {Tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) - {Tan}^{- 1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}})$

put $x = Tan θ \Rightarrow θ = {Tan}^{- 1} (x)$

$y = {Tan}^{- 1} (\frac{2 Tan θ}{1 + {Tan}^{2} θ}) + {Tan}^{- 1} (\frac{3 Tan θ - \tan^{3} θ}{1 - 3 {Tan}^{2} θ}) - {Tan}^{- 1} (\frac{4 Tan θ - 4 {Tan}^{3} θ}{1 - 6 {Tan}^{2} θ + {Tan}^{4} θ})$