Example 18.2 (\(y=\arcsin x\)). Then \(\sin y=x\). Differentiating both sides with respect to \(x\), \[\cos y\cdot \frac {dy}{dx}=1 \quad \Longrightarrow \quad \frac {dy}{dx}=\frac {1}{\cos y}.\] From \(\sin ^{2}y+\cos ^{2}y=1\) we get \(\cos y=\pm \sqrt {1-\sin ^{2}y}=\pm \sqrt {1-x^{2}}\), and the sign is settled by the range: \(y\) lies in \(\left [-\frac {\pi }{2},\frac {\pi }{2}\right ]\), where the cosine is never negative. So the positive root is the right one and \[\frac {d}{dx}\arcsin x=\frac {1}{\sqrt {1-x^{2}}},\quad \left |x\right |<1 .\]

Example 18.3 (\(y=\arccos x\)). Then \(\cos y=x\), so \[-\sin y\cdot \frac {dy}{dx}=1 \quad \Longrightarrow \quad \frac {dy}{dx}=\frac {-1}{\sin y}.\] Here \(\sin y=\pm \sqrt {1-\cos ^{2}y}=\pm \sqrt {1-x^{2}}\), and the range of \(\arccos \) is \([0,\pi ]\), on which the sine is never negative. Taking the positive root, \[\frac {d}{dx}\arccos x=\frac {-1}{\sqrt {1-x^{2}}},\quad \left |x\right |<1 .\]

Example 18.4 (\(y=\arctan x\)). Then \(\tan y=x\), so \[\sec ^{2}y\cdot \frac {dy}{dx}=1 \quad \Longrightarrow \quad \frac {dy}{dx}=\frac {1}{\sec ^{2}y}=\frac {1}{1+\tan ^{2}y},\] using \(1+\tan ^{2}y=\sec ^{2}y\). Since \(\tan y=x\), \[\frac {d}{dx}\arctan x=\frac {1}{1+x^{2}},\quad x\in \mathbb {R}.\] No sign ambiguity arises here, because \(\sec ^{2}y\) is a square and is positive throughout.