If \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2}), \frac{-1}{\sqrt{3}} < x < \frac{-1}{\sqrt{3}}\)(a) 3(b) \(\frac{3}{1+x}\)(c) –\(\frac{3}{1+x^2}\)(d) \(\frac{3}{1+x^2}\)I had been asked this question in an internship interview.I would like to ask this question from Differentiability in division Continuity and Differentiability of Mathematics – Class 12

If \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2}), \frac{-1}{\sqrt{3}} < x < \f

1.	If \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2}), \frac{-1}{\sqrt{3}} < x < \frac{-1}{\sqrt{3}}\)(a) 3(b) \(\frac{3}{1+x}\)(c) –\(\frac{3}{1+x^2}\)(d) \(\frac{3}{1+x^2}\)I had been asked this question in an internship interview.I would like to ask this question from Differentiability in division Continuity and Differentiability of Mathematics – Class 12
Answer» The correct choice is (d) \(\frac{3}{1+x^2}\) For explanation I would say: Given function is \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2})\), Now taking RHS and substituting x = tang in it and then we get, y = \(tan^{-1}(\frac{3tang-tang^3}{1-3tang^2})\), Now it becomes the expansion of the function tan3g, Hence the given function becomes y= tan^-1(tan3g). Which is equal to 3g, now substituting the VALUE of G= tan^-1x, now after differentiating both sides we get the answer \(\frac{3}{1+x^2}\).

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