Calculus 10th Edition

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

Chapter 2 - Differentiation - Review Exercises: 35

Answer

$y'=\dfrac{x^3(4\cos{x}+x\sin{x})}{\cos^2{x}}.$

Work Step by Step

Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^4; u'(x)=4x^3$ $v(x)=\cos{x}; v'(x)=-\sin{x}$ $y'=\dfrac{(4x^3)(\cos{x})-(-\sin{x})(x^4)}{\cos^2{x}}$ $=\dfrac{x^3(4\cos{x}+x\sin{x})}{\cos^2{x}}.$
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