If log(sqrt{1+x^2}-x) = y(sqrt{1+x^2}),then ( | vedclass

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(D) Given,$\log(\sqrt{1+x^2}-x) = y(\sqrt{1+x^2})$. Differentiating both sides with respect to $x$: $\frac{d}{dx} [\log(\sqrt{1+x^2}-x)] = \frac{d}{dx

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$-1$

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(D) Given,$\log(\sqrt{1+x^2}-x) = y(\sqrt{1+x^2})$. Differentiating both sides with respect to $x$: $\frac{d}{dx} [\log(\sqrt{1+x^2}-x)] = \frac{d}{dx} [y(\sqrt{1+x^2})]$ $\Rightarrow \frac{1}{\sqrt{1+x^2}-x} \cdot \left( \frac{x}{\sqrt{1+x^2}} - 1 \right) = \sqrt{1+x^2} \frac{dy}{dx} + y \cdot \frac{x}{\sqrt{1+x^2}}$ $\Rightarrow \frac{1}{\sqrt{1+x^2}-x} \cdot \left( \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}} \right) = \sqrt{1+x^2} \frac{dy}{dx} + \frac{xy}{\sqrt{1+x^2}}$ $\Rightarrow \frac{-(\sqrt{1+x^2}-x)}{(\sqrt{1+x^2}-x) \sqrt{1+x^2}} = \sqrt{1+x^2} \frac{dy}{dx} + \frac{xy}{\sqrt{1+x^2}}$ $\Rightarrow -\frac{1}{\sqrt{1+x^2}} = \sqrt{1+x^2} \frac{dy}{dx} + \frac{xy}{\sqrt{1+x^2}}$ Multiplying both sides by $\sqrt{1+x^2}$: $-1 = (1+x^2) \frac{dy}{dx} + xy$ Therefore,$(1+x^2) \frac{dy}{dx} + xy = -1$.

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

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