Answer
(a)$f(x)=5e^{2x}(20x^3-30)$
(b)$f^{'}(x)=[10e^{2x}](20x^3-30)+5e^{2x} [60x^2] $
Work Step by Step
$g(x)=5e^{2x};h(x)=20x^3-30$
(a)
Let
$f(x)=g(x) h(x)$
$f(x)=5e^{2x}(20x^3-30)$
(b)
Taking derivative of f(x) with respect to x, using product rule
$f^{'}(x)=g^{'}(x)h(x)+g(x)h^{'}(x)$
$f^{'}(x)=[5(2)e^{2x}](20x^3-30)+5e^{2x} [20(3)x^2-0] $
$f^{'}(x)=[10e^{2x}](20x^3-30)+5e^{2x} [60x^2] $
