If y = tan^{-1}left( frac{sqrt{a} - sqrt{x}}{ | vedclass

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(C) Given $y = 	an^{-1}\left( \frac{\sqrt{a} - \sqrt{x}}{1 + \sqrt{a}\sqrt{x}} ight)$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\le

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$-\frac{1}{2(1 + x)\sqrt{x}}$

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(C) Given $y = 	an^{-1}\left( \frac{\sqrt{a} - \sqrt{x}}{1 + \sqrt{a}\sqrt{x}} ight)$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left( \frac{A - B}{1 + AB} ight)$,we can write:$y = 	an^{-1}(\sqrt{a}) - 	an^{-1}(\sqrt{x})$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an^{-1}(\sqrt{a})) - \frac{d}{dx}(	an^{-1}(\sqrt{x}))$.Since $\sqrt{a}$ is a constant,its derivative is $0$.Using the chain rule for $	an^{-1}(\sqrt{x})$:$\frac{dy}{dx} = 0 - \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x})$.$\frac{dy}{dx} = - \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}$.$\frac{dy}{dx} = - \frac{1}{2(1 + x)\sqrt{x}}$.

If $y = \tan^{-1}\left( \frac{\sqrt{a} - \sqrt{x}}{1 + \sqrt{ax}} \right)$,then $\frac{dy}{dx} = $

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