Answer
a. and b:
$f^{\prime}(x)=4x$ (in both cases)
Work Step by Step
a.
$c=2, \ \ u(x)=x^{2}, \ \ f(x)=c\cdot u(x)$
Constant Multiple Rule$:$
$f^{\prime}(x)=2\cdot u^{\prime}(x)=$ ... Power Rule
$=2(2x)=4x$
b.
$u(x)=2 ,\ \ \ v(x)=x^{2}$
$f(x)=2x^{2}=u(x)v(x)$
$u^{\prime}(x)=0,\ \ \ v^{\prime}(x)=2x$
Product Rule:
$f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)=0(x^{2})+2(2x)=4x$
(the results are equal)
