## Calculus 8th Edition

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