Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 16

$g'(t) =\frac{3}{\sqrt t}$

$g(t) = 6\sqrt t$ $g'(t) = 8\times (\sqrt t)'$ $(\sqrt t)'=(t^{0.5})'= 0.5t^{-0.5}=\frac{1}{2t^{0.5}}=\frac{1}{2\sqrt t}$ Therefore $g'(t) = 6 \times (\sqrt t)'= 6\times \frac{1}{2\sqrt t}= \frac{3}{\sqrt t}$