Answer
In both cases (a. and b.):
$s^{\prime}(x)=-\displaystyle \frac{6}{x^{3}}$
Work Step by Step
a.
$ s(x)=\displaystyle \frac{3}{x^{2}}=3\cdot x^{-2}\qquad$
(constant multiple, power rules)
$s^{\prime}(x)=3(-2x^{-2-1})=-6x^{-3}=-\displaystyle \frac{6}{x^{3}}$
b.
$f(x)=3 ,\displaystyle \ \ \ g(x)=x^{2},\ \ \ s(x)=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=0,\ \ \ g^{\prime}(x)=2x$
Quotient Rule:
$s^{\prime}(x)=\displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}=$
$= \displaystyle \frac{0(x^{2})-3(2x)}{[x^{2}]^{2}}=\frac{-6x}{x^{4}}=-\frac{6}{x^{3}}$
