#### Answer

$F'(y)=5+\frac{14}{y^2}+\frac{9}{y^4}$

#### Work Step by Step

$F(y)=(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)$
Rewrite into exponent form.
$F(y)=(y^{-2}-3y^{-4})(y+5y^3)$
Use the product rule.
$F'(y)=(y^{-2}-3y^{-4})(1+15y^2)+(y+5y^3)(-2y^{-3}+12y^{-5})$
Distribute.
$F'(y)=y^{-2}+15-3y^{-4}-45y^{-2}-2y^{-2}+12y^{-4}-10+60y^{-2}$
Combine like terms.
$F'(y)=5+14y^{-2}+9y^{-4}$
Rewrite.
$F'(y)=5+\frac{14}{y^2}+\frac{9}{y^4}$