Question:

Given the expression: $$2y = \cot^{-1}\left(\frac{3\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)^2$$ To simplify the argument of the inverse cotangent function, divide the numerator and denominator by $\sin x$: $$ \frac{3\cos x + \sin x}{\cos x - \sqrt{3}\sin x} = \frac{3\cot x + 1}{\cot x - \sqrt{3}} $$ Using the identity $\cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$, and recognizing that $\cot\frac{\pi}{6} = \sqrt{3}$, the expression is simplified as: $$ \frac{3\cot x + 1}{\cot x - \sqrt{3}} = \cot\left(\frac{\pi}{6} - x\right) $$ Therefore, the given expression becomes: $$ 2y = \cot^{-1}\left(\cot\left(\frac{\pi}{6} - x\right)\right)^2 $$ Given $x \in \left(0, \frac{\pi}{2}\right)$. For the simplification $\cot^{-1}(\cot \theta) = \theta$ to hold, the solution implies a restriction such that $\frac{\pi}{6} - x \in (0, \pi)$. If $0 < x < \frac{\pi}{6}$, then $\frac{\pi}{6} - x \in \left(0, \frac{\pi}{6}\right)$, and thus $\cot^{-1}\left(\cot\left(\frac{\pi}{6} - x\right)\right) = \frac{\pi}{6} - x$. So, $$ 2y = \left(\frac{\pi}{6} - x\right)^2 $$ To find $\frac{dy}{dx}$, first express $y$: $$ y = \frac{1}{2}\left(\frac{\pi}{6} - x\right)^2 $$ Differentiate $y$ with respect to $x$ using the chain rule: $$ \frac{dy}{dx} = \frac{1}{2} \cdot 2\left(\frac{\pi}{6} - x\right) \cdot \frac{d}{dx}\left(\frac{\pi}{6} - x\right) $$ $$ \frac{dy}{dx} = \left(\frac{\pi}{6} - x\right) \cdot (-1) $$ $$ \frac{dy}{dx} = -\frac{\pi}{6} + x $$ $$ \frac{dy}{dx} = x - \frac{\pi}{6} $$