Question

Find the derivative

Answer 1

This is loaded question, but let me take a stab at it:

$\frac{d}{dx} \cot^{-1} \sqrt{x-1} = \frac{d\cot u}{du}\cdot \frac{du}{dx}$

with $u=\sqrt{x-1}$

Start with the first part of the chain rule term:

$y = \cot^{-1} u\\\frac{dy}{du} = \frac{d \cot^{-1}u}{du}$

because cot^-1 is inverse to cot we also know that

$\cot y = u$

and use it in the earlier equation:

$\frac{dy}{du} = \frac{d \cot^{-1}u}{du}\\\frac{1}{\frac{d\cot y}{dy}} = \frac{d \cot^{-1}u}{du}\\-\sin^2y=\frac{d \cot^{-1}u}{du}\\-\frac{1}{\cot^2 y +1} = \frac{d \cot^{-1}u}{du}\\-\frac{1}{u^2 +1} = \frac{d \cot^{-1}u}{du}$

Now, the second term of the chain rule:

$\frac{du}{dx}=\frac{d\sqrt{x-1}}{dx} = \frac{1}{2\sqrt{x-1}}$

Putting it together:

$\frac{d \cot^{-1}u}{du}\cdot\frac{du}{dx} = -\frac{1}{x-1 +1}\cdot\frac{1}{2\sqrt{x-1}} = - \frac{1}{2x\sqrt{x-1}}$

and that is the final derivative of your expression.