Answer
In both cases (a. and b.):
$f^{\prime}(x)=2x$
Work Step by Step
a.
Since $ x\cdot x=x^{2}$, we use the Power Rule:
$g(x)=x^{2}$
$g^{\prime}(x)=2\cdot x$
b.
$u(x)=x ,\ \ \ v(x)=x$
$f(x)=x\cdot x=u(x)v(x)$
$u^{\prime}(x)=1,\ \ \ v^{\prime}(x)=1$
Product Rule:
$f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$= 1(x)+x(1)=x+x=2x$
