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

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(A) Given $y=\log \left(x-\sqrt{x^2-1}\right)$.Taking exponential on both sides,$e^y = x-\sqrt{x^2-1}$.Differentiating with respect to $x$:$e^y y^{\pr

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(A) Given $y=\log \left(x-\sqrt{x^2-1}\right)$.Taking exponential on both sides,$e^y = x-\sqrt{x^2-1}$.Differentiating with respect to $x$:$e^y y^{\prime} = 1 - \frac{1}{2\sqrt{x^2-1}} \cdot 2x = 1 - \frac{x}{\sqrt{x^2-1}} = \frac{\sqrt{x^2-1}-x}{\sqrt{x^2-1}} = -\frac{(x-\sqrt{x^2-1})}{\sqrt{x^2-1}} = -\frac{e^y}{\sqrt{x^2-1}}$.Thus,$y^{\prime} = -\frac{1}{\sqrt{x^2-1}} = -(x^2-1)^{-1/2}$.Differentiating again with respect to $x$:$y^{\prime \prime} = -(-\frac{1}{2})(x^2-1)^{-3/2} \cdot 2x = \frac{x}{(x^2-1)^{3/2}}$.Now,substitute into the expression $(x^2-1)y^{\prime \prime} + x y^{\prime}$:$(x^2-1) \left( \frac{x}{(x^2-1)^{3/2}} \right) + x \left( -\frac{1}{\sqrt{x^2-1}} \right) = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1}} = 0$.

If $y=\log \left(x-\sqrt{x^2-1}\right)$,then $\left(x^2-1\right) y^{\prime \prime}+x y^{\prime}$ is equal to:

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