StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 54

Expert-verifiedFound in: Page 184

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use the definition of the derivative to find $f$ for each function $f$ in Exercises 39-54.

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

The derivative is ${f}^{\text{'}}\left(x\right)=\frac{2x\left({x}^{2}-x-2\right)-(2x-1)\left({x}^{2}-1\right)}{{\left({x}^{2}-x-2\right)}^{2}}$

Given function $f\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}-x-2}$

Calculating, we get

$f\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}-x-2}\phantom{\rule{0ex}{0ex}}f\text{'}\left(x\right)=\frac{({x}^{2}-x-2)\left(2x\right)-({x}^{2}-1)(2x-1)}{{({x}^{2}-x-2)}^{2}}\phantom{\rule{0ex}{0ex}}f\text{'}\left(x\right)=\frac{2x({x}^{2}-x-2)-(2x-1)({x}^{2}-1)}{{({x}^{2}-x-2)}^{2}}\phantom{\rule{0ex}{0ex}}$

94% of StudySmarter users get better grades.

Sign up for free