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Q. 54

Expert-verified
Found in: Page 184

### Calculus

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

See the step by step solution

## Step 1: Given information

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

## Step 2: Use the chain rule and calculate

Calculating, we get

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