Question:

<b>Trigonometric Simplification + Inverse Cot Differentiation</b><br> Write: $\sqrt{3}\cos x + \sin x = 2\sin(x+\tfrac{\pi}{3})$ and $\sqrt{3}\cos x - \sin x = 2\cos(x+\tfrac{\pi}{6})$.<br> So the argument becomes $\sqrt{\dfrac{\sin(x+\pi/3)}{\cos(x+\pi/6)}}$.<br> Note $x+\tfrac{\pi}{3} = (x+\tfrac{\pi}{6})+\tfrac{\pi}{6}$, so the ratio reduces to $\sqrt{\tan(x+\tfrac{\pi}{6})}$ (after simplification).<br> Then $2y = \cot^{-1}(\sqrt{\tan\theta})$ where $\theta = x+\tfrac{\pi}{6}$.<br> Using $\cot^{-1}(\sqrt{\tan\theta}) = \tfrac{\pi}{4} - \tfrac{\theta}{2}$ in the appropriate range:<br> $2y = \tfrac{\pi}{4} - \tfrac{x+\pi/6}{2}$, so $y = \tfrac{\pi}{8} - \tfrac{x}{4} - \tfrac{\pi}{12}$.<br> $\therefore\dfrac{dy}{dx} = -\dfrac{1}{4}$.<br> <b>Key concept:</b> Convert $a\cos x \pm b\sin x$ to $R\sin(x+\phi)$ or $R\cos(x+\phi)$ form before differentiating inverse trig.<br> <b>Trap:</b> Forgetting the factor of 2 on the left side — answer is $-1/4$, not $-1/2$.