This is loaded question, but let me take a stab at it:
[tex]\frac{d}{dx} \cot^{-1} \sqrt{x-1} = \frac{d\cot u}{du}\cdot \frac{du}{dx}[/tex]
with [tex]u=\sqrt{x-1}[/tex]
Start with the first part of the chain rule term:
[tex]y = \cot^{-1} u\\\frac{dy}{du} = \frac{d \cot^{-1}u}{du}[/tex]
because cot^-1 is inverse to cot we also know that
[tex]\cot y = u[/tex]
and use it in the earlier equation:
[tex]\frac{dy}{du} = \frac{d \cot^{-1}u}{du}\\\frac{1}{\frac{d\cot y}{dy}} = \frac{d \cot^{-1}u}{du}\\-\sin^2y=\frac{d \cot^{-1}u}{du}\\-\frac{1}{\cot^2 y +1} = \frac{d \cot^{-1}u}{du}\\-\frac{1}{u^2 +1} = \frac{d \cot^{-1}u}{du}[/tex]
Now, the second term of the chain rule:
[tex]\frac{du}{dx}=\frac{d\sqrt{x-1}}{dx} = \frac{1}{2\sqrt{x-1}}[/tex]
Putting it together:
[tex]\frac{d \cot^{-1}u}{du}\cdot\frac{du}{dx} = -\frac{1}{x-1 +1}\cdot\frac{1}{2\sqrt{x-1}} = - \frac{1}{2x\sqrt{x-1}}[/tex]
and that is the final derivative of your expression.
