## Calculus: Early Transcendentals (2nd Edition)

$f'(s)=\dfrac{1}{8\sqrt{s}}$
Rewrite the function: $f(s)=\dfrac{1}{4}s^{1/2}$ Apply the power rule: $f'(s) =\dfrac{1}{4}(\dfrac{1}{2})s^{1/2-1}$ $f'(s)=\dfrac{1}{8}s^{-1/2}$ $f'(s)=\dfrac{1}{8\sqrt{s}}$