Answer
(a) $-16\;,\; $(b) $\frac{-20}{9}\;\;,$ (c) 20
Work Step by Step
(a) $\; (fg)(x)=f(x)g(x)$
Differentiating with respect to $x$ using product rule
$(fg)'(x)=f'(x)g(x)+f(x)g'(x)$
$(fg)'(5)=f'(5)g(5)+f(5)g'(5)$
$(fg)'(5)=(6)(-3)+(1)(2)$
$(fg)'(5)=-18+2=-16$
(b)$\; \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$
Differentiating with respect to $x$ using quotient rule
$\left(\frac{f}{g}\right)'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$
$\Rightarrow \left(\frac{f}{g}\right)'(5)=\frac{f'(5)g(5)-f(5)g'(5)}{(g(5))^2}$
$\Rightarrow \left(\frac{f}{g}\right)'(5)=\frac{(6)(-3)-(1)(2)}{9}$
$\Rightarrow \left(\frac{f}{g}\right)'(5)=\frac{-18-2}{9}=\frac{-20}{9}$
(c)$\; \left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}$
Differentiating with respect to $x$ using quotient rule
$\left(\frac{g}{f}\right)'(x)=\frac{g'(x)f(x)-g(x)f'(x)}{(f(x))^2}$
$\Rightarrow \left(\frac{g}{f}\right)'(5)=\frac{g'(5)f(5)-g(5)f'(5)}{(f(5))^2}$
$\Rightarrow \left(\frac{g}{f}\right)'(5)=\frac{(2)(1)-(-3)(6)}{(1)^2}$
$\Rightarrow \left(\frac{g}{f}\right)'(5)=2+18=20$
