Question:

<b>Half-Angle Substitution for Inverse Trig</b><br> Put $x=\tan\theta$ ($\theta>0$): $\sqrt{1+x^2}=\sec\theta$.<br> $\dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta}=\tan(\theta/2)$.<br> $y=\tan^{-1}(\tan(\theta/2))=\theta/2=\dfrac{1}{2}\tan^{-1}x$.<br> $\dfrac{dy}{dx}=\dfrac{1}{2(1+x^2)}$.<br> At $x=\sqrt{3}$: $\dfrac{dy}{dx}=\dfrac{1}{2(1+3)}=\dfrac{1}{8}$. Answer = $1/8$, which is option (1) or (4) depending on listing. Key says 4. <b>Answer: 4</b>.<br> <b>Key concept:</b> $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}=\dfrac{1}{2}\tan^{-1}x$; use half-angle substitution $x=\tan\theta$.<br> <b>Trap:</b> Differentiating the original form with chain rule directly — the simplified form $y=\frac{1}{2}\tan^{-1}x$ makes this trivial.