Answer
(a) $F'(2) =-\frac{3}5; y=-\frac{3}5+\frac{16}5$
(b) (see graph)
Work Step by Step
(a) $F'(x)=\frac{5(1+x^2)-5x(2x)}{(1+x^2)^2}=\frac{5(1-x^2)}{(1+x^2)^2}$
$F'(2)=\frac{5[1-(2)^2]}{[1+(2)^2]^2}=-\frac{15}5=-\frac{3}5$
$F(2)=\frac{5(2)}{1+(2)^2}=\frac{10}5=2$
$F(x)=F'(x)x+b$
$2=-\frac{3}5(2)+b$
$b=\frac{16}5$
$y=-\frac{3}5x+\frac{16}5$