## Calculus: Early Transcendentals (2nd Edition)

$f'(x)=1$ $f''(x)=0$ $f'''(x)=0$
Simplify the expression: $f(x)=\dfrac{x^2-7x-8}{x+1}=\dfrac{(x+1)(x-8)}{x+1}=x-8$ Apply power rule to the function to find the first derivate: $f'(x)=1$ Apply power rule to the first derivate to find the second derivate: $f''(x)=0$ Apply power rule to the second derivate to find the third derivate: $f'''(x)=0$