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  • Differential coefficient of <img alt="\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \text { w.r.t

Differential coefficient of \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \text { w.r.t. } \sin ^{-1} \frac{2 x}{1+x^2} \text { is equal to : }

Option: 1

0


Option: 2

-1


Option: 3

1


Option: 4

None of these


Answers (1)

best_answer

\text { Put } y=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \text { and } z=\sin ^{-1} \frac{2 x}{1+x^2}

\text { Put } \mathrm{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \mathrm{x}

\begin{aligned} & \therefore \quad y=\tan ^{-1}(\tan 2 \theta)=2 \theta, z=\sin ^{-1}(\sin 2 \theta)=2 \theta \\ & \frac{d y}{d \theta}=2, \frac{d z}{d \theta}=2 \Rightarrow \frac{d y}{d z}=1 \end{aligned}

Hence (C) is correct answer.

Posted by

Sanket Gandhi

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