Answer
$(\displaystyle \frac{1}{3.2}-\frac{3.2}{x^{2}})\cdot(x^{2}+1)+2x( \frac{x}{3.2}+\frac{3.2}{x})$
Work Step by Step
$f(x)= \displaystyle \frac{1}{3.2}x+3.2x^{-1}, \ \ \quad g(x)=x^{2}+1,\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=\displaystyle \frac{1}{3.2}+3.2(-x^{-2})=\frac{1}{3.2}-3.2x^{-2}$
$g^{\prime}(x)=2x$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=(\displaystyle \frac{1}{3.2}-3.2x^{-2})\cdot(x^{2}+1)+( \frac{1}{3.2}x+3.2x^{-1})\cdot 2x$
$=(\displaystyle \frac{1}{3.2}-\frac{3.2}{x^{2}})\cdot(x^{2}+1)+2x( \frac{x}{3.2}+\frac{3.2}{x})$
... we do not need to expand the answer...