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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If y= ^ -1 [ x- 1-x^ 2 x+ 1-x^ 2 ], then ((d y)/(d x))=

If y=tan⁡−1\mathrm{y}=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right], then (dydx)=\left(\frac{d y}{d x}\right)=

Options

  1. A−11−x2\frac{-1}{\sqrt{1-x^{2}}}
  2. B−x1−x2\frac{-x}{\sqrt{1-x^{2}}}
  3. C11−x2\frac{1}{\sqrt{1-x^{2}}}
  4. Dx1−x2\frac{x}{\sqrt{1-x^{2}}}

Correct answer

C. 11−x2\frac{1}{\sqrt{1-x^{2}}}

Step-by-step solution

Given y=tan⁡−1y=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right] Put x=cos⁡θ⇒θ=cos⁡−1xy=tan⁡−1[cos⁡θ−sin⁡θcos⁡θ+sin⁡θ]=tan⁡−1[1−tan⁡θ1+tan⁡θ]=tan⁡−1[tan⁡(π4−θ)]=π4−θ∴dydx=π4−cos⁡−1x=\cos \theta \Rightarrow \theta=\cos ^{-1} xy=\tan ^{-1}\left[\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right]=\tan ^{-1}\left[\frac{1-\tan \theta}{1+\tan \theta}\right]\quad=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\frac{\pi}{4}-\theta\therefore \frac{d y}{d x}=\frac{\pi}{4}-\cos ^{-1}xy and =11−x2y and =11−x2\begin{aligned}y \text{ and } =\frac{1}{\sqrt{1-x^{2}}}\\ y \text{ and } =\frac{1}{\sqrt{1-x^{2}}} \end{aligned}
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