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TS EAMCET · Maths · Differentiation

If \(y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)\) then \(\frac{d y}{d x}\) is equal to

  1. A \(\frac{2}{1+x^2}\)
  2. B \(\frac{4}{1+x^2}\)
  3. C \(\frac{6}{1+x^2}\)
  4. D \(\frac{7}{1+x^2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{7}{1+x^2}\)

Step-by-step Solution

Detailed explanation

We have, \[ y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right) \] Put \(x=\tan \theta\), we get…