Thomas' Calculus 13th Edition

$0$ for $x \gt 1$
$y = tan^{-1}\sqrt{x^2-1}+csc^{-1}x$ $= tan^{-1}(x^2-1)^{1/2}+csc^{-1}x$ Derive using the chain rule: $\frac{dy}{dx}=\frac{\frac{(1)}{2}(x^2-1)^{-1/2}(2x)}{1+[(x^2-1)^{1/2}]^2}-\frac{1}{|x|\sqrt{x^2-1}}$ =$\frac{1}{x\sqrt{x^2-1}}-\frac{1}{|x|\sqrt{x^2-1}}=0$ for $x \gt 1$