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If $\mathrm{y}=\tan ^{-1}\left(\frac{5-2 \tan \sqrt{\mathrm{x}}}{2+5 \tan \sqrt{\mathrm{x}}}\right)$, then what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to?
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The correct answer is:
$-\frac{1}{2 \sqrt{x}}$
$\begin{aligned} & y=\tan ^{-1}\left(\frac{5-2 \tan \sqrt{x}}{2+5 \tan \sqrt{x}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{5}{2}-\tan \sqrt{x}}{1+\left(\frac{5}{2}\right) \tan \sqrt{x}}\right) \\ &=\tan ^{-1} \frac{5}{2}-\tan ^{-1} \tan \sqrt{x} \\ &=\tan ^{-1} \frac{5}{2}-\sqrt{x} \\ \therefore & \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \end{aligned}$
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