Lovellss

2022-06-23

Derivative of $f\left(x\right)=\frac{\sqrt{{x}^{2}-1}}{x}$
I have the function following:
$f\left(x\right)=\frac{\sqrt{{x}^{2}-1}}{x}$
And here is what I did:
$f\left(x\right)=\frac{\sqrt{{x}^{2}-1}}{x}$
$=\frac{\left({x}^{2}-1{\right)}^{\frac{1}{2}}}{x}$
${f}^{\prime }\left(x\right)=\frac{x\cdot \frac{1}{2}\left({x}^{2}-1{\right)}^{-\frac{1}{2}}\left(2x\right)-\left({x}^{2}-1{\right)}^{\frac{1}{2}}}{{x}^{2}}$
$=\frac{{x}^{2}\left({x}^{2}-1{\right)}^{-\frac{1}{2}}-\left({x}^{2}-1{\right)}^{\frac{1}{2}}}{{x}^{2}}$
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?
Or is there any other steps so it can match this?:
$\frac{1}{{x}^{2}\sqrt{{x}^{2}-1}}$

Mateo Barajas

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

pokoljitef2

$\begin{array}{rl}\frac{{x}^{2}\left({x}^{2}-1{\right)}^{-1/2}-\left({x}^{2}-1{\right)}^{2}}{{x}^{2}}& =\frac{\frac{{x}^{2}}{\sqrt{{x}^{2}-1}}-\sqrt{{x}^{2}-1}}{{x}^{2}}\\ & =\frac{\frac{{x}^{2}}{\sqrt{{x}^{2}-1}}-\frac{{x}^{2}-1}{\sqrt{{x}^{2}-1}}}{{x}^{2}}\\ & =\frac{\frac{1}{\sqrt{{x}^{2}-1}}}{{x}^{2}}\\ & =\frac{1}{{x}^{2}\sqrt{{x}^{2}-1}}\end{array}$