1. ## Hyperbolic Functions #2

$f(x) = arccosh\frac{1}{\sqrt{x^2-7x+x}}$
#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 $y = cosh^{-1}x$, the domain is $x \geq 1.$ So,
$\frac{1}{\sqrt{x^2-7x+x}} \geq 1$

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

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

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

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

$x^2 - 6x \geq 0$

$x \geq 0, x \geq 6$

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

2. For starters, you have a serious typo. Is the polynomial under the radical $x^2-7x-x$, $x^2-6x-1$, or $x^2-6x$? You interchange among all three in your question.

Secondly, rewrite $f(x)=a(b(c(x)))$ where $a(x)=\cosh^{-1}(x), b(x)=x^{-1/2}, c(x)=x^2-7x-x$, and you can derive using the chain rule: $f'(x)=a'(b(c(x)))b'(c(x))c'(x)$