Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises: 28

Answer

$y'=\dfrac{4-x^4}{\sqrt{(x^4+4)^3}}.$

Work Step by Step

Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x; u'(x)=1$ $v(x)=\sqrt{x^4+4}.$ $v'(x)$ is found using the Chain Rule: $u=x^4+4$; $\dfrac{du}{dx}=4x^3$ $\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$ $\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=(4x^3)(\dfrac{1}{2\sqrt{x^4+4}})=\dfrac{2x^3}{\sqrt{x^4+4}}.$ $y'=\dfrac{(1)(\sqrt{x^4+4})-(x)(\dfrac{2x^3}{\sqrt{x^4+4}})}{(\sqrt{x^4+4})^2}$ $=\dfrac{(\dfrac{1}{\sqrt{x^4+4}})(x^4+4-2x^4)}{x^4+4}$ $=\dfrac{4-x^4}{\sqrt{(x^4+4)^3}}.$
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