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MHT CET · Maths · Differentiation

If \(\mathrm{u}=\frac{\tan ^{-1} x}{\tan ^{-1} x+1}\) and \(\mathrm{v}=\tan ^{-1}\left(\tan ^{-1} x\right)\) then \(\frac{\mathrm{du}}{\mathrm{dv}}=\ldots \ldots\)

  1. A 1
  2. B \(\frac{1+\left(\tan ^{-1} x\right)^2}{\left(1+\tan ^{-1} x\right)^2}\)
  3. C \(\frac{\tan ^{-1} x}{\left(1+\tan ^{-1} x\right)^2}\)
  4. D \(\frac{1}{\left(1+\tan ^{-1} x\right)^2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1+\left(\tan ^{-1} x\right)^2}{\left(1+\tan ^{-1} x\right)^2}\)

Step-by-step Solution

Detailed explanation

Let \(y = \tan^{-1} x\). \(u = \frac{y}{y+1}\)