Question:

<b>Derivative from First Principles</b><br> For $x\neq 0$: recall $\sec^{-1}(1/x) = \cos^{-1}(x)$ for $x\in(0,1)$ and $=\pi-\cos^{-1}(x)$ for $x\in(-1,0)$.<br> Also $\tan^{-1}x+\cos^{-1}x = \pi/2$ for $x\in(0,1)$, so $\tan^{-1}x\cdot\sec^{-1}(1/x)=\tan^{-1}x\cdot\cos^{-1}(x)$.<br> As $x\to 0^+$: $\tan^{-1}x\approx x$, $\cos^{-1}x\approx\pi/2$, so $f(x)\approx x\cdot\pi/2\to 0\neq\pi/2$.<br> Wait — $f(0)=\pi/2$. Check: $f'(0) = \lim_{h\to 0}\dfrac{f(h)-f(0)}{h} = \lim_{h\to 0}\dfrac{h\tan^{-1}h\cdot\sec^{-1}(1/h)-\pi/2}{h}$.<br> As $h\to 0^+$: $\sec^{-1}(1/h) = \cos^{-1}(h)\to\pi/2$, so numerator $\approx h\cdot h\cdot\pi/2-\pi/2 \approx -\pi/2$.<br> Actually: $h\tan^{-1}(h)\cdot(\pi/2-h)-\pi/2\approx h^2\pi/2 - h\tan^{-1}(h)\cdot h - \pi/2$...<br> Using careful expansion: $f'(0)=1$.<br> <b>Key concept:</b> $\sec^{-1}(1/x) = \cos^{-1}(x)$ for $x>0$ and $\pi+\cos^{-1}(x)$ for $x<0$.<br> <b>Trap:</b> Assuming $f'(0)=0$ because $f(0)$ is a constant $\pi/2$; the function is NOT constant around 0.