Answer
$f’(3) = -\frac{1}{169} $
Work Step by Step
$f(x) = \frac{1}{x+10} = (x+10) ^{-1}$
Power Rule:
$f’(x) = -1(x+10)^{-2} $
$f’(x) = -\frac{1}{(x+10)^2} $
Evaluate f’(x) at x=3:
$f’(3) = -\frac{1}{(3+10)^2} = -\frac{1}{(13)^2} $
$f’(3) = -\frac{1}{169} $
