To find \(y=\arcsin \relax (x) \), always write as \(x=\sin \relax (y) \). Then \(\frac {dx}{dy}=\cos \relax (y) =\sqrt {1-\sin ^{2}y}=\sqrt {1-x^{2}}\). Then \(\frac {dy}{dx}=\frac {1}{\sqrt {1-x^{2}}}\), Hence\[ \frac {d}{dx}\arcsin \relax (x) =\frac {1}{\sqrt {1-x^{2}}}\] To find \(y=\arccos \relax (x) \), write as \(x=\cos \left ( y\right ) \). Then \(\frac {dx}{dy}=-\sin \relax (y) =-\sqrt {1-\cos ^{2}y}=-\sqrt {1-x^{2}}\). Then \(\frac {dy}{dx}=\frac {-1}{\sqrt {1-x^{2}}}\), Hence\[ \frac {d}{dx}\arccos \relax (x) =\frac {-1}{\sqrt {1-x^{2}}}\] To find \(y=\arctan \relax (x) \), write as \(x=\tan \left ( y\right ) \). Then \(\frac {dx}{dy}=\frac {1}{\cos ^{2}y}\), now need to use trick that \(\cos ^{2}y+\sin ^{2}y=1\) and divide both sides by \(\cos ^{2}y\), hence \(1+\tan ^{2}y=\frac {1}{\cos ^{2}y}\). Then \(\frac {dx}{dy}=1+\tan ^{2}y\). Hence \(\frac {dy}{dx}=\frac {1}{1+\tan ^{2}y}=\frac {1}{1+x^{2}}\). Therefore\[ \frac {d}{dx}\arctan \relax (x) =\frac {1}{1+x^{2}}\]