inverse hyperbolic derivative

**xxlvh** · October 26th 2009, 07:19 PM

So in this question I defined $g(x) = \frac{1}{\sqrt{x^2-6x}}$ and $h(x) = arccoshx$ .
The function I am then evaluating is $f(x) = h(g(x))$ . It is also given that $\frac{dy}{dx}arccoshx = ln(x+\sqrt{x^2-1}), x \geq 1.$ The question asks to find the interval on which $f(x)$ is differentiable.

Solving for f'(x), I have:

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

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

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

**Opalg** · October 27th 2009, 01:24 AM

Originally Posted by xxlvh

So in this question I defined $g(x) = \frac{1}{\sqrt{x^2-6x}}$ and $h(x) = arccoshx$ .
The function I am then evaluating is $f(x) = h(g(x))$ . It is also given that $\color{red}\frac{dy}{dx}arccoshx = ln(x+\sqrt{x^2-1}), x \geq 1.$ The question asks to find the interval on which $f(x)$ is differentiable.

Solving for f'(x), I have:

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

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

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

There's a mistake here. It's the function arccosh(x) itself, not its derivative, that is equal to $\ln(x+\sqrt{x^2-1})$ . The derivative of arccosh(x) is $1/\sqrt{x^2-1}$ . For details, see here. Notice that arccosh(x) is only defined for $x\geqslant 1$ . So for x to be in the domain of h(g(x)) it is necessary that $g(x))\geqslant1$ . As you have concluded, this means finding out where $1/\sqrt{x^2-6x} \geqslant 1$ .

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