Derivative of $f(x)= {\sqrt{x^2-1}\over x}$

Question

I have the function following:

$$f(x)= {\sqrt{x^2-1}\over x}$$

And here is what I did:

$$f(x)= {\sqrt{x^2-1}\over x}$$

$$ = {{(x^2-1)^{1 \over 2}} \over x}$$ $$f'(x) = { x \cdot {1 \over 2} (x^2-1)^{-{1 \over 2}} (2x) - (x^2 - 1)^{1 \over 2}\over x^2}$$ $$=\frac {x^2 (x^2 -1)^{-{\frac {1}{2}}} - (x^2 -1)^{\frac {1}{2}}}{x^2}$$ And I'm pretty sure that this is wrong, and the answer book says it isn't either.

I think I messed up somewhere, or didn't do it properly.

I tried using the quotient rule. Do I need to make it into an exponent, and solve it as a chainrule?

Please help me find the steps and answer to this question. Thank you.

Or is there any other steps so it can match this?:

$$\frac {1}{x^2 \sqrt{x^2 -1}}$$

Answer 1

$$f(x) = \frac{\sqrt{x^2-1}}{x}$$ $$f(x) = \frac{u}{v}$$ Note that $u = \sqrt{x^2-1}$, $v = x$, and $v' = 1$We then note that $u' = \frac{x}{\sqrt{x^2-1}}$ $$f'(x) = \frac{u'v - uv'}{v^2}$$ $$ = \frac{\frac{x^2}{\sqrt{x^2-1}} - \sqrt{x^2-1}}{x^2}$$ $$ = \frac{\frac{x^2}{\sqrt{x^2-1}} + \frac{-(x^2-1)}{\sqrt{x^2-1}}} {x^2}$$ $$ = \frac{\frac{x^2 - x^2 - 1}{\sqrt{x^21}}}{x^2}$$ $$ = \frac{1}{x^2\sqrt{x^2-1}}$$

Answer 2

All your steps are fine, but your final answer can be simplified further to match the answer in your book. \begin{align*} \frac{x^2(x^2-1)^{-1/2}-(x^2-1)^2}{x^2} &= \frac{\frac{x^2}{\sqrt{x^2-1}}-\sqrt{x^2-1}}{x^2}\\ &=\frac{\frac{x^2}{\sqrt{x^2-1}}-\frac{x^2-1}{\sqrt{x^2-1}}}{x^2}\\ &=\frac{\frac{1}{\sqrt{x^2-1}}}{x^2}\\ &= \frac{1}{x^2{\sqrt{x^2-1}}} \end{align*}