Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 45

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: $g'(t)=f'(t)+h'(t)=\frac{1}{4\sqrt[4]{t^3}}-6\csc(t)\cot(t)$
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