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JEE Main 2021 · Mathematics · Differentiation and Application of Derivatives · medium

Let f(x)=sin ^-1 ((2x)/(1+x^2). Then f'(2) equals

Single correct option

Let f(x)=sin⁡−1 ⁣(2x1+x2)f(x)=\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right). Then f′(2)f'(2) equals
  1. A. 25\dfrac25
  2. B. −25-\dfrac25
  3. C. 45\dfrac45
  4. D. −45-\dfrac45
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Answer: (B) −25-\dfrac25

Put x=tan⁡θx=\tan\theta, so 2x1+x2=sin⁡2θ\dfrac{2x}{1+x^2}=\sin2\theta. The principal value sin⁡−1(sin⁡2θ)\sin^{-1}(\sin2\theta) equals 2θ2\theta only when 2θ∈[−π2,π2]2\theta\in[-\tfrac\pi2,\tfrac\pi2], i.e. ∣x∣≤1|x|\le1. For x>1x>1 we have 2θ∈(π2,π)2\theta\in(\tfrac\pi2,\pi), so sin⁡−1(sin⁡2θ)=π−2θ\sin^{-1}(\sin2\theta)=\pi-2\theta. Thus for x>1x>1, f(x)=π−2tan⁡−1xf(x)=\pi-2\tan^{-1}x and f′(x)=−21+x2.f'(x)=-\frac{2}{1+x^2}. At x=2x=2 this is −25-\dfrac25. The positive value 25\dfrac25 comes from wrongly using 2tan⁡−1x2\tan^{-1}x. The answer is negative since ff decreases for x>1x>1.
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