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: 59

Answer

$y'=-4\sqrt{3}$.

Work Step by Step

Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=1+\csc(x); u'(x)=-\csc(x)\cot(x)$ $v(x)=1-\csc(x); v'(x)=\csc(x)\cot(x)$ $y'=\frac{-\csc(x)\cot(x)-\csc(x)\cot(x)}{(1-\csc(x))^2}$ $=-\frac{2\csc(x)\cot(x)}{(1-\csc(x))^2}$. To evaluate, substitute in $x=\frac{\pi}{6}$: $y'=-\dfrac{2\csc(\frac{\pi}{6})\cot(\frac{\pi}{6})}{(1-\csc(\frac{\pi}{6}))^2}=-4\sqrt{3}$
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