## Calculus (3rd Edition)

$$y'= -\frac{\sin^{-1}x +\cos^{-1}x }{(\sin^{-1}x)^2 \sqrt{1-x^2}}.$$
Since $y=\frac{\cos^{-1}x}{\sin^{-1}x}$, then by the qoutient rule, the derivative $y'$ is given by $$y'= \frac{\sin^{-1}x\left(\frac{-1}{\sqrt{1-x^2}}\right)-\cos^{-1}x\left(\frac{1}{\sqrt{1-x^2}}\right)}{(\sin^{-1}x)^2}\\=-\frac{\sin^{-1}x +\cos^{-1}x }{(\sin^{-1}x)^2 \sqrt{1-x^2}}.$$