If y = frac{sqrt{x^2 + 1} + sqrt{x^2 - 1}}{sq | vedclass

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(A) Given $y = \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$.Rationalizing the denominator,we multiply the numerator and de

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$2x + \frac{2x^3}{\sqrt{x^4 - 1}}$

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(A) Given $y = \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})$:$y = \frac{(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})^2}{(\sqrt{x^2 + 1})^2 - (\sqrt{x^2 - 1})^2}$$y = \frac{(x^2 + 1) + (x^2 - 1) + 2\sqrt{(x^2 + 1)(x^2 - 1)}}{(x^2 + 1) - (x^2 - 1)}$$y = \frac{2x^2 + 2\sqrt{x^4 - 1}}{2}$$y = x^2 + \sqrt{x^4 - 1}$Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}((x^4 - 1)^{1/2})$$\frac{dy}{dx} = 2x + \frac{1}{2}(x^4 - 1)^{-1/2} \cdot (4x^3)$$\frac{dy}{dx} = 2x + \frac{2x^3}{\sqrt{x^4 - 1}}$.

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

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