Answer
In both cases (a. and b.):
$u^{\prime}(x)=\displaystyle \frac{2x}{3}$
Work Step by Step
a.
$ u(x)=\displaystyle \frac{x^{2}}{3}=\frac{1}{3}\cdot x^{2}\qquad$
(constant multiple,\ power rules)
$u^{\prime}(x)=\displaystyle \frac{1}{3}(2x)=\frac{2x}{3}$
b.
$f(x)=x^{2} ,\displaystyle \ \ \ g(x)=3,\ \ \ u(x)=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=2x,\ \ \ g^{\prime}(x)=0$
Quotient Rule:
$u^{\prime}(x)=\displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}=$
$= \displaystyle \frac{2x\cdot 3-0(x^{2})}{3^{2}}=\frac{6x}{9}=\frac{2x}{3}$
