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: 34

Answer

$g'(x)=24x(2+(x^2+1)^4)^2(x^2+1)^3.$

Work Step by Step

Using the Chain Rule: $u=2+(x^2+1)^4=f(x)+h(x)\rightarrow f(x)=2$; $h(x)=(x^2+1)^4$ $f'(x)=0$. $h'(x)=(4)(2x)(x^2+1)^{4-1}=8x(x^2+1)^3.$ $\dfrac{du}{dx}=8x(x^2+1)^3+0=8x(x^2+1)^3.$ $g(u)=u^3;\dfrac{d}{du}g(u)=3u^2$. $\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}$. $=3(2+(x^2+1)^4)^2(8x(x^2+1)^3)$ $=24x(2+(x^2+1)^4)^2(x^2+1)^3$.
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