Question

Differentiate.
F(y) =

1/(y^2 −3/y^4)(y + 5y^3)

Answer 1

F(y) = (1/y^2 −3/y^4)(y + 5y^3)

Answer 2

use product rule?

Answer 3

is this the correct equation?

Answer 4

no

Answer 5

F(y) = (1/y^2 −3/y^4)(y + 5y^3)

Answer 6

\[F(y)=\frac{ 1 }{ y ^{2} -\frac{ 3 }{ y ^{4} }}(y+5y ^{3})\]

Answer 7

how does this one look?

Answer 8

\[\frac{ 1 }{ y ^{2} }-\frac{ 3 }{ y ^{4} }(y+5y ^{3})\]

Answer 9

ok

Answer 10

both fractions r in a bracket together

Answer 11

\[F(y)=y ^{-2}-3y ^{-4}(y+5y ^{3})\]
I would first rewrite it like this and then use the product/chain-rule.

Answer 12

okkay

Answer 13

(y^−2−3y^−4)(y+5y3)

Answer 14

\[F'(y)=-2y ^{-3}-3y ^{-4}(1+15y ^{2})+(y+5y ^{3})(12y ^{-5})\]

Answer 15

you can combine like terms in order to get a simplified answer

Answer 16

thanks man

Answer 17

you're welcome