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v.s
 3 years ago
Differentiate.
F(y) =
1/(y^2 −3/y^4)(y + 5y^3)
v.s
 3 years ago
Differentiate. F(y) = 1/(y^2 −3/y^4)(y + 5y^3)

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

VeritasVosLiberabit
 3 years ago
Best ResponseYou've already chosen the best response.0is this the correct equation?

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

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

VeritasVosLiberabit
 3 years ago
Best ResponseYou've already chosen the best response.0how does this one look?

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

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

VeritasVosLiberabit
 3 years ago
Best 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.

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

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

VeritasVosLiberabit
 3 years ago
Best ResponseYou've already chosen the best response.0you're welcome
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