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\(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+1}{x}\right) \) |
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Answer» \(y = tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)\) Let \(x = tan \theta \) ⇒ \(\theta = tan^{-1}x\) \(\therefore y= tan^{-1}\left(\frac{\sqrt{1+tan^2\theta}+1}{tan\theta}\right)\) \(= tan^{-1}\left(\frac{sec\theta + 1}{tan\theta}\right)\) \(= tan^{-1}\left(\frac{1+cos\theta}{sin\theta}\right)\) \(= tan^{-1} \left(\frac{2cos^2\frac{\theta}{2}}{2sin\frac{\theta}2cos\frac{\theta}{2}}\right)\) \(= tan^{-1}\left(cot\frac{\theta}{2}\right)\) \(= tan^{-1}\left(tan\frac{\pi}{2}- \frac{\theta}{2}\right)\) ⇒ \(y = \frac{\pi}{2}- \frac{\theta}{2}\) \(\therefore \frac{dy}{dx} = \frac{-1}{2} \frac{d\theta}{dx}\) \(= \frac{-1}{2} \frac{d}{dx} tan^{-1}x\) \((\because \theta = tan^{-1}x)\) \(= \frac{-1}{2}\frac{1 }{1+x^2}\) \(\therefore \frac{d}{dx} tan^{-1} (\frac{\sqrt{1 +x^2} + 1}{x})= \frac{-1}{2(1 + x^2)}\) |
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