Answer
In both cases (a. and b.):
$h^{\prime}(x)=2x+3$
Work Step by Step
a.
Since $ x\cdot(x+3)=x^{2}+3x$,
Using the Product , Constant Multiple and Sum rules,
$h(x)=x^{2}+3x$,
$h^{\prime}(x)=2x+3$
b.
$u(x)=x ,\ \ \ v(x)=x+3$
$h(x)=x\cdot x=u(x)v(x)$
$u^{\prime}(x)=1,\ \ \ v^{\prime}(x)=1$
Product Rule:
$h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$= 1(x+3)+x(1)=x+3+x=2x+3$
