## Calculus: Early Transcendentals 8th Edition

$g'(x)=6x(x^{2}+1)^{2}(x^{2}+2)^{5}(3x^{2}+4)$
$g(x)=(x^{2}+1)^{3}(x^{2}+2)^{6}$ Differentiate using the product rule: $g'(x)=[(x^{2}+1)^{3}][(x^{2}+2)^{6}]'+[(x^{2}+2)^{6}][(x^{2}+1)^{3}]'=...$ Use the chain rule to find $[(x^{2}+2)^{6}]'$ and $[(x^{2}+1)^{3}]'$: $...=[(x^{2}+1)^{3}][6(x^{2}+2)^{5}(x^{2}+2)']+[(x^{2}+2)^{6}][3(x^{2}+1)^{2}(x^{2}+1)']=...$ $...=[(x^{2}+1)^{3}][6(x^{2}+2)^{5}(2x)]+[(x^{2}+2)^{6}][3(x^{2}+1)^{2}(2x)]=...$ $...=12x(x^{2}+1)^{3}(x^{2}+2)^{5}+6x(x^{2}+2)^{6}(x^{2}+1)^{2}$ Take out common factors $6x$, $(x^{2}+1)^{2}$ and $(x^{2}+2)^{5}$ to present a better looking answer (Optional) $g'(x)=6x(x^{2}+1)^{2}(x^{2}+2)^{5}[2(x^{2}+1)+(x^{2}+2)]=...$ $...=6x(x^{2}+1)^{2}(x^{2}+2)^{5}(3x^{2}+4)$