Originally Posted by

**xxlvh** So in this question I defined $\displaystyle g(x) = \frac{1}{\sqrt{x^2-6x}} $ and $\displaystyle h(x) = arccoshx$.

The function I am then evaluating is $\displaystyle f(x) = h(g(x))$. It is also given that $\displaystyle \color{red}\frac{dy}{dx}arccoshx = ln(x+\sqrt{x^2-1}), x \geq 1.$ The question asks to find the interval on which $\displaystyle f(x)$ is differentiable.

Solving for f'(x), I have:

$\displaystyle f'(x) = h'(g(x))*g'(x) $

$\displaystyle f'(x) = ln(\frac{1}{\sqrt{x^2-6x}} + \sqrt{\frac{1}{x^2-6x}-1})(\frac{x-3}{\sqrt{x^2-6x}}) $

Hopefully my work up to this point is correct, but to find out where the function is differentiable, do I just need to set $\displaystyle \frac{1}{\sqrt{x^2-6x}} \geq 1? $