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

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(D) Given $y = 	an^{-1}\left( \frac{\sqrt{x} - x}{1 + x \cdot \sqrt{x}} ight)$.Using the formula $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}\left( \fr

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$-\frac{1}{4}$

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(D) Given $y = 	an^{-1}\left( \frac{\sqrt{x} - x}{1 + x \cdot \sqrt{x}} ight)$.Using the formula $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}\left( \frac{a - b}{1 + ab} ight)$,we let $a = \sqrt{x}$ and $b = x$.Thus,$y = 	an^{-1}(\sqrt{x}) - 	an^{-1}(x)$.Differentiating with respect to $x$:$y' = \frac{d}{dx}(	an^{-1}(\sqrt{x})) - \frac{d}{dx}(	an^{-1}(x))$$y' = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x}) - \frac{1}{1 + x^2}$$y' = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1 + x^2}$.Now,evaluating at $x = 1$:$y'(1) = \frac{1}{1 + 1} \cdot \frac{1}{2\sqrt{1}} - \frac{1}{1 + 1^2}$$y'(1) = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.

If $y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)$,then $y'(1)$ is

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