Answer
$g'(t)=\frac{1}{4\sqrt[4]{t^3}}-6\csc(t)\cot(t)$
Work Step by Step
$g(t)=f(t)+h(t)\rightarrow f(t)=\sqrt[4]{t}$ ; $ h(t)=6\csc(t)$
Using the Power Rule: $f'(t)=\frac{1}{4}t^{\frac{1}{4}-1}=\frac{1}{4\sqrt[4]{t^3}}$
By Theorem 2.9 and Constant Multiple Rule:
$h'(t)=6(\dfrac{d}{dt}\csc(t))$
$=6(-\csc(t)\cot(t))$
$=-6\csc(t)\cot(t)$
Using the Sum Rule: $g'(t)=f'(t)+h'(t)=\frac{1}{4\sqrt[4]{t^3}}-6\csc(t)\cot(t)$
