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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)
 2 years ago

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

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

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

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

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

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

VeritasVosLiberabit Group TitleBest ResponseYou've already chosen the best response.0
ok
 2 years ago

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

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(y^−2−3y^−4)(y+5y3)
 2 years ago

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

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

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