Answer
In both cases (a. and b.):
$s^{\prime}(x)=-\displaystyle \frac{2}{x^{2}}$
Work Step by Step
a.
Using the Power and Constant Multiple rules,
$c=2,\displaystyle \quad u(x)=\frac{1}{x}=x^{-1}$
$s(s)=c\cdot u(x)$
$s^{\prime}(x)=c\cdot u^{\prime}(x)$
$=2\displaystyle \cdot(-1x^{-1-1})=-2x^{-2}=-\frac{2}{x^{2}}$
b.
$u(x)=2 ,\displaystyle \ \ \ v(x)=x,\ \ \ s(x)=\frac{u(x)}{v(x)}$
$u^{\prime}(x)=0,\ \ \ v^{\prime}(x)=1$
Quotient Rule:
$s^{\prime}(x)=\displaystyle \frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{[v(x)]^{2}}=$
$= \displaystyle \frac{0\cdot x-2(1)}{x^{2}}=-\frac{2}{x^{2}}$
