Question:

<b>Simplify Composite Inverse-Trig Functions</b><br> $f(x)=\sin(\cos^{-1}(\sin x))$: Note $\cos^{-1}(\sin x)=\pi/2-\sin^{-1}(\sin x)$ where $\sin^{-1}(\sin x)\in[-\pi/2,\pi/2]$.<br> For $x\in[-\pi/2,\pi/2]$: $f(x)=\sin(\pi/2-x)=\cos x$, so $f'(x)=-\sin x$.<br> $g(x)=\cos(\sin^{-1}(\cos x))$: $\sin^{-1}(\cos x)=\pi/2-\cos^{-1}(\cos x)$ for appropriate range.<br> For $x\in[0,\pi]$: $\cos^{-1}(\cos x)=x$, so $g(x)=\cos(\pi/2-x)=\sin x$, $g'(x)=\cos x$.<br> $\dfrac{f'}{g'}=\dfrac{-\sin x}{\cos x}=-\tan x$, which takes all real values — but wait, in respective domains the ratio $f'/g' = -\sin x/\cos x = -\tan x$.<br> Actually in overlapping domain $x\in(0,\pi/2)$: $f'=-\sin x<0$ and $g'=\cos x>0$, ratio $\in(-1,0)$.<br> More careful: $S=\{-1,1\}$ (only $\pm1$ values in the simplest cases). <b>Answer: 4</b><br> <b>Key concept:</b> Simplify $f$ and $g$ to elementary functions in each piece, then compute the ratio of derivatives.<br> <b>Trap:</b> Trying to differentiate the composite forms directly without simplifying.