Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 45

$$ y' =\frac{1}{t^2\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}.$$

Since $ y=\cos^{-1}t^{-1}-\sec^{-1}t $, then the derivative $ y'$ is given by $$ y'= \frac{-(t^{-1})'}{\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}\\ = \frac{-(-t^{-2})}{\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}\\ =\frac{1}{t^2\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}.$$