Chapter 3: Derivatives - Section 3.2 - The Derivative as a Function - Exercises 3.2 - Page 115: 13

$f'(x) =1-9x^{-2}$ and $f'(-3)=0$

Consider $f(x) = x +\dfrac{9}{x} \implies f(-3) = -3 -\dfrac{9}{3} = -6$ Now, $f'(x) = 1+9(-1)x^{-1-1}=1-9x^{-2}$ $f'(-3) =1-9(-3)^{-2}=1-1=0$ Hence, we have $f'(x) =1-9x^{-2}$ and $f'(-3)=0$