Answer
Sum/difference rule,
$4x^{3}-12x^{2}+2x-480$
Work Step by Step
With f(x)$=x^{4},\ g(x)=(x^{2}+120)(4x-1),$
the last operation we would perform on calculator is
$f(x)-g(x)$,
so we use the SUM/DIFFERENCE rule.
$[f(x)-g(x)]^{\prime}=f^{\prime}(x)-g^{\prime}(x)$
$ f^{\prime}(x)=4x^{3}\quad$ (power rule)
Calculating $g(x)$, the last operation would be a product of
$u(x)=x^{2}+120$ and $v(x)=4x-1$.
For $g^{\prime}(x)$ we use the product rule.
$g^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$=(2x)(4x-1)+(x^{2}+120)(4)$
$=8x^{2}-2x+4x^{2}+480$
$=12x^{2}-2x+480$
Finally,
$[f(x)-g(x)]^{\prime}=4x^{3}-(12x^{2}-2x+480)$
$=4x^{3}-12x^{2}+2x-480$
