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v.s Group Title

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

  • one year ago
  • one year ago

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  1. v.s Group Title
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    F(y) = (1/y^2 −3/y^4)(y + 5y^3)

    • one year ago
  2. v.s Group Title
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    use product rule?

    • one year ago
  3. VeritasVosLiberabit Group Title
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    is this the correct equation?

    • one year ago
  4. v.s Group Title
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    no

    • one year ago
  5. v.s Group Title
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    F(y) = (1/y^2 −3/y^4)(y + 5y^3)

    • one year ago
  6. VeritasVosLiberabit Group Title
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    \[F(y)=\frac{ 1 }{ y ^{2} -\frac{ 3 }{ y ^{4} }}(y+5y ^{3})\]

    • one year ago
  7. VeritasVosLiberabit Group Title
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    how does this one look?

    • one year ago
  8. v.s Group Title
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    \[\frac{ 1 }{ y ^{2} }-\frac{ 3 }{ y ^{4} }(y+5y ^{3})\]

    • one year ago
  9. VeritasVosLiberabit Group Title
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    ok

    • one year ago
  10. v.s Group Title
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    both fractions r in a bracket together

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

    • one year ago
  12. v.s Group Title
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    okkay

    • one year ago
  13. v.s Group Title
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    (y^−2−3y^−4)(y+5y3)

    • one year ago
  14. VeritasVosLiberabit Group Title
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    \[F'(y)=-2y ^{-3}-3y ^{-4}(1+15y ^{2})+(y+5y ^{3})(12y ^{-5})\]

    • one year ago
  15. VeritasVosLiberabit Group Title
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    you can combine like terms in order to get a simplified answer

    • one year ago
  16. v.s Group Title
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    thanks man

    • one year ago
  17. VeritasVosLiberabit Group Title
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    you're welcome

    • one year ago
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