Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 18

Answer

$$g'(x)=6x(x^2+1)^2(x^2+2)^5(3x^2+4)$$

Work Step by Step

$$g'(x)=((x^2+1)^3(x^2+2)^6)'=((x^2+1)^3)'(x^2+2)^6+(x^2+1)^3((x^2+2)^6)'= 3(x^2+1)^2(x^2+1)'(x^2+2)^6+(x^2+1)^36(x^2+2)^5(x^2+2)'= 3(x^2+1)^2\cdot2x(x^2+2)^6+6(x^2+1)^3(x^2+2)^5\cdot2x= 6x(x^2+1)^2(x^2+2)^5(x^2+2+2(x^2+1))= 6x(x^2+1)^2(x^2+2)^5(3x^2+4)$$
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