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\(y = \frac{x \,cos^{-1} x}{\sqrt{1- x^2}} - log(\sqrt{1 - x^2})\)Find dy/dx? |
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Answer» \(y = \frac{x \,cos^{-1} x}{\sqrt{1- x^2}} - log(\sqrt{1 - x^2})\) Let x = cosθ then cos-1x = θ and \(=\sqrt{1- x^2} = \sqrt{1 - cos^2\theta} = \sqrt{sin^2\theta} = sin\theta\) \(\therefore y = \frac{\theta cos\theta}{sin\theta} - log \, sin \theta \) \(= \theta cot \theta - log\, sin\theta\) \(\therefore \frac{dy}{dx} = \frac{dy}{d\theta} . \frac{d\theta}{dx}\) \(= \frac d{d\theta} (\theta cot\theta - log \, sin\theta) \frac{d\theta}{dx}\) \(= -\theta cosec^2\theta + cot \theta - \frac{cos\theta}{sin\theta} \;\frac{d\theta}{dx}\) \(= \frac{-\theta}{sin^2\theta} + \frac{-1}{sin\theta}\) \(\left(\because \frac{dx}{d\theta } = - sin\theta⇒\frac{d\theta}{dx } = \frac{-1}{sin\theta} \right)\) \(= \frac{\theta}{sin^3\theta}\) \(= \frac{cos^{-1}x}{(1-x^2)^{3/2}}\) \((\because sin \theta = \sqrt{1 - cos^2\theta} = \sqrt{1 - x^2})\) Hence, \(\frac{dy}{dx} = \frac{cos^{-1}x}{(1 - x^2)^{3/2}}\). |
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