Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 22


$$ h'(t)=-\frac{2t(\cot t+1)\csc^2t}{t^2}.$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(\csc x)'=-\csc x \cot x$. Since $ h(t)=\frac{\csc^2t}{t}$, then using the quotient rule, we have $$ h'(t)=\frac{t(t(\csc^2t)'-(t^2)'\csc^2t}{t^2}\\ =\frac{t(-2\csc t \csc t\cot t)-2t\csc^2t}{t^2} \\=-\frac{2t(\cot t+1)\csc^2t}{t^2}.$$
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