Answer
$f^{'}(x)=(3x^2+15x+7)[96x^2 ]+(32x^3+49)[6x+15]$
Work Step by Step
$f(x)=(3x^2+15x+7)(32x^3+49)$
Taking derivative ,using product rule
$f{'}(x)=(3x^2+15x+7)\frac{d(32x^3+49)}{dx}+(32x^3+49)\frac{d(3x^2+15x+7)}{dx}$
$f^{'}(x)=(3x^2+15x+7)[\frac{d(32x^3)}{dx}+\frac{d(49)}{dx} ]+(32x^3+49)[\frac{d(3x^2)}{dx}+\frac{d(15x)}{dx}+\frac{d(7)}{dx}]$
$f^{'}(x)=(3x^2+15x+7)[32\frac{d(x^3)}{dx}+0 ]+(32x^3+49)[3\frac{d(x^2)}{dx}+15\frac{d(x)}{dx}+0]$
$f^{'}(x)=(3x^2+15x+7)[32(3x^2) ]+(32x^3+49)[3(2x)+15]$
$f^{'}(x)=(3x^2+15x+7)[32(3x^2) ]+(32x^3+49)[3(2x)+15]$
$f^{'}(x)=(3x^2+15x+7)[96x^2 ]+(32x^3+49)[6x+15]$
