University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 75



Work Step by Step

$$y=x(2x+1)^4$$ - Find $y'$: $$y'=(x)'(2x+1)^4+x\Big((2x+1)^4\Big)'$$ $$y'=1\times(2x+1)^4+x\Big(4(2x+1)^3(2x+1)'\Big)$$ $$y'=(2x+1)^4+4x(2x+1)^3\times2$$ $$y'=(2x+1)^4+8x(2x+1)^3$$ - Find $y''$: $$y''=\Big((2x+1)^4\Big)'+8\Big(x(2x+1)^3\Big)'$$ $$y''=4(2x+1)^3(2x+1)'+8\Bigg[(x)'(2x+1)^3+x\Big((2x+1)^3\Big)'\Bigg]$$ $$y''=8(2x+1)^3+8\Bigg[(2x+1)^3+3x(2x+1)^2(2x+1)'\Bigg]$$ $$y''=8(2x+1)^3+8\Bigg[(2x+1)^3+6x(2x+1)^2\Bigg]$$ $$y''=8(2x+1)^3+8(2x+1)^2(2x+1+6x)$$ $$y''=8(2x+1)^3+8(2x+1)^2(8x+1)$$ $$y''=8(2x+1)^2(2x+1+8x+1)=8(2x+1)^2(10x+2)$$ $$y''=16(2x+1)^2(5x+1)$$
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