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MISCELLANEOUS EXERCISE 1 (II) · Q244

Q.Differentiate tan⁡−11+x2−1x\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x} w.r.t. tan⁡−12x1−x21−2x2\tan^{-1}\dfrac{2x\sqrt{1-x^2}}{1-2x^2}

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Reduce uu: let x=tan⁡θx=\tan\theta. Then 1+x2=sec⁡θ\sqrt{1+x^2}=\sec\theta, and

1+x2−1x=sec⁡θ−1tan⁡θ=(1−cos⁡θ)/cos⁡θsin⁡θ/cos⁡θ=1−cos⁡θsin⁡θ=tan⁡θ2.\frac{\sqrt{1+x^2}-1}{x}=\frac{\sec\theta-1}{\tan\theta}=\frac{(1-\cos\theta)/\cos\theta}{\sin\theta/\cos\theta}=\frac{1-\cos\theta}{\sin\theta}=\tan\frac{\theta}{2}.

So u=tan⁡−1 ⁣(tan⁡θ2)=θ2=12tan⁡−1xu=\tan^{-1}\!\left(\tan\dfrac{\theta}{2}\right)=\dfrac{\theta}{2}=\dfrac12\tan^{-1}x, giving dudx=12(1+x2)\dfrac{du}{dx}=\dfrac{1}{2(1+x^2)}.

Reduce vv: let x=sin⁡ϕx=\sin\phi. Then 2x1−x2=2sin⁡ϕcos⁡ϕ=sin⁡2ϕ2x\sqrt{1-x^2}=2\sin\phi\cos\phi=\sin2\phi and 1−2x2=1−2sin⁡2ϕ=cos⁡2ϕ1-2x^2=1-2\sin^2\phi=\cos2\phi, so …

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