Answer
In both cases (a. and b.):
$h^{\prime}(x)=1+4x$
Work Step by Step
a.
Since $ x\cdot(1+2x)=x+ 2x^{2}$,
Using the Power, Constant Multiple, and Sum rules,
$h(x)= x+2x^{2}$,
$h^{\prime}(x)=1+2(2x)=1+4x$
b.
$u(x)=x ,\ \ \ v(x)=1+2x$
$h(x)=u(x)v(x)$
$u^{\prime}(x)=1,\ \ \ v^{\prime}(x)=2$
Product Rule:
$h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$= 1(1+2x)+2(x)=1+2x+2x=1+4x$
