If y = frac{sqrt{a + x} - sqrt{a - x}}{sqrt{a | vedclass

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

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$\frac{ay}{x\sqrt{a^2 - x^2}}$

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(A) Given $y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{a + x} - \sqrt{a - x})$:$y = \frac{(\sqrt{a + x} - \sqrt{a - x})^2}{(a + x) - (a - x)} = \frac{(a + x) + (a - x) - 2\sqrt{a^2 - x^2}}{2x} = \frac{2a - 2\sqrt{a^2 - x^2}}{2x} = \frac{a - \sqrt{a^2 - x^2}}{x} \dots (i)$Now,differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(a - \sqrt{a^2 - x^2}) - (a - \sqrt{a^2 - x^2}) \cdot \frac{d}{dx}(x)}{x^2}$$\frac{dy}{dx} = \frac{x \cdot [0 - \frac{1}{2\sqrt{a^2 - x^2}} \cdot (-2x)] - (a - \sqrt{a^2 - x^2})}{x^2}$$\frac{dy}{dx} = \frac{\frac{x^2}{\sqrt{a^2 - x^2}} - a + \sqrt{a^2 - x^2}}{x^2} = \frac{x^2 - a\sqrt{a^2 - x^2} + a^2 - x^2}{x^2\sqrt{a^2 - x^2}} = \frac{a^2 - a\sqrt{a^2 - x^2}}{x^2\sqrt{a^2 - x^2}}$$\frac{dy}{dx} = \frac{a(a - \sqrt{a^2 - x^2})}{x^2\sqrt{a^2 - x^2}} = \frac{a}{x\sqrt{a^2 - x^2}} \cdot \left( \frac{a - \sqrt{a^2 - x^2}}{x} \right)$Substituting from $(i)$,we get $\frac{dy}{dx} = \frac{ay}{x\sqrt{a^2 - x^2}}$.

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

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