Answer
$f'(x)=(2x-3)^{3}(x^{2}+x+1)^{4}[5(2x-3)(2x+1)+8(x^{2}+x+1)]$
Work Step by Step
$f(x)=(2x-3)^{4}(x^{2}+x+1)^{5}$
Differentiate using the product rule:
$f'(x)=[(2x-3)^{4}][(x^{2}+x+1)^{5}]'+[(x^{2}+x+1)^{5}][(2x-3)^{4}]'=...$
Use the chain rule to find $[(x^{2}+x+1)^{5}]'$ and $[(2x-3)^{4}]'$:
$...=[(2x-3)^{4}][5(x^{2}+x+1)^{4}(x^{2}+x+1)']+[(x^{2}+x+1)^{5}][4(2x-3)^{3}(2x-3)']=...$
$...=5(2x-3)^{4}(x^{2}+x+1)^{4}(2x+1)+4(x^{2}+x+1)^{5}(2x-3)^{3}(2)=...$
$...=5(2x-3)^{4}(x^{2}+x+1)^{4}(2x+1)+8(x^{2}+x+1)^{5}(2x-3)^{3}$
Take out common factor $(2x-3)^{3}$ and $(x^{2}+x+1)^{4}$: (Optional)
$f'(x)=(2x-3)^{3}(x^{2}+x+1)^{4}[5(2x-3)(2x+1)+8(x^{2}+x+1)]$
