Answer
$F'(y)=\dfrac{14}{y^{2}}+\dfrac{9}{y^{4}}+5$
Work Step by Step
$F(y)=(\dfrac{1}{y^{2}}-\dfrac{3}{y^{4}})(y+5y^{3})$
Write the function like this for an easier differentiation process:
$F(y)=(y^{-2}-3y^{-4})(y+5y^{3})$
Differentiate using the product rule
$F'(y)=(y^{-2}-3y^{-4})(y+5y^{3})'+(y+5y^{3})(y^{-2}-3y^{-4})'=...$
$...=(y^{-2}-3y^{-4})(1+15y^{2})+(y+5y^{3})(-2y^{-3}+12y^{-5})=...$
Evaluate the products and simplify
$y^{-2}+15-3y^{-4}-45y^{-2}-2y^{-2}+12y^{-4}-10+60y^{-2}=...$
$..=14y^{-2}+9y^{-4}+5=\dfrac{14}{y^{2}}+\dfrac{9}{y^{4}}+5$
