CUET · MATHS · PYQ PAPER 2023
The derivative of \(\frac{\tan ^{-1} x}{1+\tan ^{-1} x}\) with respect to \(\tan ^{-1}\) is :
- A \(\frac{\tan ^{-1} x}{\left(1+\tan ^{-1} x\right)^2}\)
- B \(\frac{1}{\left(1+\tan ^{-1} x\right)^2}\)
- C \(\left(1+\tan ^{-1} x\right)^2\)
- D \(\frac{\tan ^{-1} x}{\left(1-\tan ^{-1} x\right)^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\left(1+\tan ^{-1} x\right)^2}\)
Step-by-step Solution
Detailed explanation
Let \(u = \tan^{-1} x\). The expression becomes \(\frac{u}{1+u}\). The derivative with respect to \(u\) is \(\frac{d}{du}\left(\frac{u}{1+u}\right) = \frac{1 \cdot (1+u) - u \cdot 1}{(1+u)^2} = \frac{1}{(1+u)^2}\). Substituting back \(u = \tan^{-1} x\), the derivative is…
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