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  • Differentiate tan inverse root 1+x^2 -1/x w.r.t. tan-1x when x not = 0

Differentiate \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right) w.r.t. tan^{-1}x when x\neq 0

Answers (1)

We have  \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

Let us Assume,

\begin{aligned} &\mathrm{p}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right) \text { and } \theta=\tan ^{-1}{ \mathrm{x}}\\ &\text { And, put } x=\tan \theta\\ &p=\tan ^{-1} \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\\ &p=\tan ^{-1} \frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\\ &p=\tan ^{-1} \frac{\sec \theta-1}{\tan \theta} \end{aligned}

\\ \mathrm{p}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ \mathrm{p}=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}\right) \\ \mathrm{p}=\tan ^{-1}\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right) \\ \mathrm{p}=\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \\ \mathrm{p}=\frac{\theta}{2}

\frac{dp}{d\theta} = \frac{1}{2}

Hence Differentiation of \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)w.r.t. \tan^{-1}x is \frac{1}{2}.

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