## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 10

#### Answer

$$h'(t)=-9\csc t \cot t+ \cot t -t \csc^2 t.$$

#### Work Step by Step

Since $h(t)=9\csc t+t \cot t$, then, using the product rule: $(uv)'=u'v+uv’$ and also, using the facts that $(\csc x)'=-\csc x \cot x$ and $(\cot x)'=-\csc^2 x$, we have $$h'(t)=(9\csc t)' +(t \cot t)'\\ =-9\csc t \cot t+ \cot t -t \csc^2 t.$$

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