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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

On differentiating ^ -1 [ 1 + x^ 2 - 1 x ] with respect to x , the...

On differentiating tan⁡−1[1+x2−1x]\tan^{-1} [ \dfrac{\sqrt{1 + x^{2}} - 1}{x} ] with respect to xx, the result would be

Options

  1. A

    & 12.11+x2\dfrac{1}{2} . \dfrac{1}{1 + x^{2}}

  2. B

    & 11+x2\dfrac{1}{1 + x^{2}}

  3. C

    & 21+x2\dfrac{2}{1 + x^{2}}

  4. D

    & 12.11+2x\dfrac{1}{2} . \dfrac{1}{1 + 2x}

Correct answer

A.

& 12.11+x2\dfrac{1}{2} . \dfrac{1}{1 + x^{2}}

Step-by-step solution

Let us assume that,

& y=tan⁡−1[1+x2−1x]y = \tan^{- 1} [ \dfrac{\sqrt{1 + x^{2}} - 1}{x} ]

Put : x=tanθx = tan\theta

∴1+x2−1x=1+tan⁡2θ−1tanθ=secθ−1tanθ=1−cosθsinθ=2sin⁡2θ22sinθ2cosθ2=tanθ2\begin{array}{ll}∴ \dfrac{\sqrt{1 + x^{2}} - 1}{x} = \dfrac{\sqrt{1 + \tan^{2}\theta} - 1}{tan\theta} = \dfrac{sec\theta - 1}{tan\theta} \\ = \dfrac{1 - cos\theta}{sin\theta} = \dfrac{2\sin^{2}\dfrac{θ}{2}}{2sin\dfrac{θ}{2}cos\dfrac{θ}{2}} \\ = tan\dfrac{θ}{2}\end{array}

On differentiating with respect to x:

& ∴y=tan⁡−1(tanθ2)=12θ=12tan⁡−1x⇒dydx=12.11+x2.\begin{array}{ll}∴ y = \tan^{- 1} ( tan\dfrac{θ}{2} ) = \dfrac{1}{2} \theta = \dfrac{1}{2} \tan^{- 1} x \\ ⇒ \dfrac{dy}{dx} = \dfrac{1}{2} . \dfrac{1}{1 + x^{2}} .\end{array}

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