Answer
$y '= 8(x^2+2x+7)^7\times (x+2)$
Work Step by Step
$y =(x^2+2x+7)^{8}$
$y ' =((x^2+2x+7)^{8})' $
The inner function sis $g(x) = x^2+2x+7$
and the outer function is $f(u) = u^{8}$
$f'(u) = 8u^7$
$((x^2+2x+7)^{8})' = 8(g(x))^7\times g'(x)$
$= 8(x^2+2x+7)^7\times (x+2)$
