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

v.s Group TitleBest ResponseYou've already chosen the best response.0
use product rule?
 one year ago

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

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

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

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

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

VeritasVosLiberabit Group TitleBest ResponseYou've already chosen the best response.0
ok
 one year ago

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

VeritasVosLiberabit Group TitleBest 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

v.s Group TitleBest ResponseYou've already chosen the best response.0
(y^−2−3y^−4)(y+5y3)
 one year ago

VeritasVosLiberabit Group TitleBest 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

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

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