Question:

<b>Standard Inverse Trig Derivatives</b><br> <b>P:</b> $\sin^{-1}(3x-4x^3)=3\sin^{-1}x$ for $x\in(-1/2,1/2)$. Derivative $=\dfrac{3}{\sqrt{1-x^2}}$ → <b>(1)</b>.<br> <b>Q:</b> $\cos^{-1}(4x^3-3x)=3\cos^{-1}x$ for $x\in(1/2,1)$. But actually: $\cos^{-1}(4x^3-3x)$: when $x=\cos\theta$ with $\theta\in(0,\pi/3)$, $4\cos^3\theta-3\cos\theta=\cos 3\theta$, so $=3\cos^{-1}x$ with derivative $=-3/\sqrt{1-x^2}$ → wait, $d/dx(3\cos^{-1}x)=3\cdot(-1/\sqrt{1-x^2})=\dfrac{-3}{\sqrt{1-x^2}}$ → <b>(2)</b>.<br> <b>R:</b> $\tan^{-1}\dfrac{3x-x^3}{1-3x^2}=3\tan^{-1}x$ for $|x|<1/\sqrt{3}$. Derivative $=\dfrac{3}{1+x^2}$ → <b>(3)</b>.<br> <b>S:</b> $\sin^{-1}(2x\sqrt{1-x^2})$: for $x\in(1/\sqrt{2},1)$, put $x=\sin\theta$ with $\theta\in(\pi/4,\pi/2)$: $2\sin\theta\cos\theta=\sin 2\theta$, so $=\sin^{-1}(\sin 2\theta)$. For $\theta\in(\pi/4,\pi/2)$: $2\theta\in(\pi/2,\pi)$, so $\sin^{-1}(\sin 2\theta)=\pi-2\theta=\pi-2\sin^{-1}x$. Derivative $=\dfrac{-2}{\sqrt{1-x^2}}$ → <b>(4)</b>.<br> Mapping: P→1, Q→2, R→3, S→4 → Answer <b>C</b>.<br> <b>Key concept:</b> Recognise the triple-angle / double-angle formulas to simplify inverse trig expressions.<br> <b>Trap:</b> Not checking the range of $x$ — the same expression can have different simplified forms in different intervals.