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v.sBest ResponseYou've already chosen the best response.0
F(y) = (1/y^2 −3/y^4)(y + 5y^3)
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
is this the correct equation?
 one year ago

v.sBest ResponseYou've already chosen the best response.0
F(y) = (1/y^2 −3/y^4)(y + 5y^3)
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
\[F(y)=\frac{ 1 }{ y ^{2} \frac{ 3 }{ y ^{4} }}(y+5y ^{3})\]
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
how does this one look?
 one year ago

v.sBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ y ^{2} }\frac{ 3 }{ y ^{4} }(y+5y ^{3})\]
 one year ago

v.sBest ResponseYou've already chosen the best response.0
both fractions r in a bracket together
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
\[F(y)=y ^{2}3y ^{4}(y+5y ^{3})\] I would first rewrite it like this and then use the product/chainrule.
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
\[F'(y)=2y ^{3}3y ^{4}(1+15y ^{2})+(y+5y ^{3})(12y ^{5})\]
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
you can combine like terms in order to get a simplified answer
 one year ago

VeritasVosLiberabitBest ResponseYou've already chosen the best response.0
you're welcome
 one year ago
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