# Thread: Hyperbolic Functions #2

1. ## Hyperbolic Functions #2

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

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

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