$\displaystyle f(x) = arccosh\frac{1}{\sqrt{x^2-7x+x}} $

#1. Find the derivative of the function.

#2. State the interval where the function is differentiable and calculate the derivative.

here is my attempted solution so far:

For $\displaystyle y = cosh^{-1}x$, the domain is $\displaystyle x \geq 1. $ So,

$\displaystyle \frac{1}{\sqrt{x^2-7x+x}} \geq 1 $

$\displaystyle 1 \geq \sqrt{x^2-7x+x} $

$\displaystyle 1 \geq x^2 - 6x - 1 $

$\displaystyle \frac{6-\sqrt{40}}{2} \leq x \leq \frac{6+\sqrt{40}}{2} $

also $\displaystyle \sqrt{x^2-7x+x} \geq 0 $

$\displaystyle x^2 - 6x \geq 0 $

$\displaystyle x \geq 0, x \geq 6 $

I am also stumped at how to solve the second part of this question. Thanks for any help!