Lovellss

2022-06-23

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

I have the function following:

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

And here is what I did:

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

$=\frac{({x}^{2}-1{)}^{\frac{1}{2}}}{x}$

${f}^{\prime}(x)=\frac{x\cdot \frac{1}{2}({x}^{2}-1{)}^{-\frac{1}{2}}(2x)-({x}^{2}-1{)}^{\frac{1}{2}}}{{x}^{2}}$

$=\frac{{x}^{2}({x}^{2}-1{)}^{-\frac{1}{2}}-({x}^{2}-1{)}^{\frac{1}{2}}}{{x}^{2}}$

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?:

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

I have the function following:

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

And here is what I did:

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

$=\frac{({x}^{2}-1{)}^{\frac{1}{2}}}{x}$

${f}^{\prime}(x)=\frac{x\cdot \frac{1}{2}({x}^{2}-1{)}^{-\frac{1}{2}}(2x)-({x}^{2}-1{)}^{\frac{1}{2}}}{{x}^{2}}$

$=\frac{{x}^{2}({x}^{2}-1{)}^{-\frac{1}{2}}-({x}^{2}-1{)}^{\frac{1}{2}}}{{x}^{2}}$

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?:

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

Mateo Barajas

Beginner2022-06-24Added 13 answers

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

$f(x)=\frac{u}{v}$

Note that $u=\sqrt{{x}^{2}-1}$, $v=x$ and ${v}^{\prime}=1$ We then note that ${u}^{\prime}=\frac{x}{\sqrt{{x}^{2}-1}}$

${f}^{\prime}(x)=\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$

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

$=\frac{\frac{{x}^{2}}{\sqrt{{x}^{2}-1}}+\frac{-({x}^{2}-1)}{\sqrt{{x}^{2}-1}}}{{x}^{2}}$

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

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

$f(x)=\frac{u}{v}$

Note that $u=\sqrt{{x}^{2}-1}$, $v=x$ and ${v}^{\prime}=1$ We then note that ${u}^{\prime}=\frac{x}{\sqrt{{x}^{2}-1}}$

${f}^{\prime}(x)=\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$

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

$=\frac{\frac{{x}^{2}}{\sqrt{{x}^{2}-1}}+\frac{-({x}^{2}-1)}{\sqrt{{x}^{2}-1}}}{{x}^{2}}$

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

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

pokoljitef2

Beginner2022-06-25Added 9 answers

All your steps are fine, but your final answer can be simplified further to match the answer in your book.

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

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