If ysqrt{x^2 + 1} = log {sqrt{x^2 + 1} - x},t | vedclass

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(A) Given equation: $y\sqrt{x^2 + 1} = \log \{\sqrt{x^2 + 1} - x\}$Differentiating both sides with respect to $x$:$\frac{dy}{dx} \cdot \sqrt{x^2 + 1}

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(A) Given equation: $y\sqrt{x^2 + 1} = \log \{\sqrt{x^2 + 1} - x\}$Differentiating both sides with respect to $x$:$\frac{dy}{dx} \cdot \sqrt{x^2 + 1} + y \cdot \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( \frac{2x}{2\sqrt{x^2 + 1}} - 1 \right)$Simplify the right side:$\frac{dy}{dx} \sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right)$$\frac{dy}{dx} \sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = \frac{-( \sqrt{x^2 + 1} - x )}{(\sqrt{x^2 + 1} - x) \sqrt{x^2 + 1}}$$\frac{dy}{dx} \sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = -\frac{1}{\sqrt{x^2 + 1}}$Multiply the entire equation by $\sqrt{x^2 + 1}$:$(x^2 + 1)\frac{dy}{dx} + xy = -1$Rearranging the terms:$(x^2 + 1)\frac{dy}{dx} + xy + 1 = 0$

If $y\sqrt{x^2 + 1} = \log \{\sqrt{x^2 + 1} - x\}$,then $(x^2 + 1)\frac{dy}{dx} + xy + 1 = $

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