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

Hint : Rationalise and Diff.w.r to x

$y=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}$

$=\frac{(\sqrt{a+x}-\sqrt{a-x}{)}^2}{(a+x)-(a-x)}$

$=\frac{(a+x)+(a-x)-2\left(\sqrt{a^2-x^2}\right)}{2x}$

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

$\frac{dy}{dx}=\frac{x\left[-\frac{1}{2\sqrt{a^2-x^2}}(-2x)\right]-\left(a-\sqrt{a^2-x^2}\right)}{x^2}$

$=\frac{x^2-a\sqrt{a^2-x^2}+a^2-x^2}{x^2\sqrt{a^2-x^2}}=\frac{a\left(a-\sqrt{a^2-x^2}\right)}{x^2\sqrt{a^2-x^2}}$

$=\frac{a}{x\sqrt{a^2-x^2}}\left[\frac{a-\sqrt{a^2-x^2}}{x}\right]=\frac{ay}{x\sqrt{a^2-x^2}}$