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

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

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