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 \(\frac{\mathrm{du}}{\mathrm{dv}}=\ldots \ldots\)
- A 1
- B \(\frac{1+\left(\tan ^{-1} x\right)^2}{\left(1+\tan ^{-1} x\right)^2}\)
- C \(\frac{\tan ^{-1} x}{\left(1+\tan ^{-1} x\right)^2}\)
- D \(\frac{1}{\left(1+\tan ^{-1} x\right)^2}\)
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 \(y = \tan^{-1} x\). \(u = \frac{y}{y+1}\)
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