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: 47

Answer

$y'=\dfrac{3\sin(x)-3}{2\cos^2(x)}$

Work Step by Step

Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(1-\sin(x)); u'(x)=-\cos(x)$ $v(x)=\cos(x); v'(x)=-\sin(x)$ $y'=\frac{3}{2}(\frac{(-\cos(x))(cos(x))-(-\sin(x))(1-\sin(x))}{\cos^2(x)})$ $=\frac{3}{2}(\frac{-\cos^2(x)+\sin(x)-\sin^2(x)}{cos^2(x)})$ $=\frac{3}{2}(\frac{-(\cos^2(x)+\sin^2(x))+\sin(x)}{\cos^2(x)})$ $=\frac{3}{2}(\frac{\sin(x)-1}{\cos^2(x)})$ $=\frac{3\sin(x)-3}{2\cos^2(x)}$
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