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v.s

  • 2 years ago

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

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

  2. v.s
    • 2 years ago
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    use product rule?

  3. VeritasVosLiberabit
    • 2 years ago
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    is this the correct equation?

  4. v.s
    • 2 years ago
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    no

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

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

  7. VeritasVosLiberabit
    • 2 years ago
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    how does this one look?

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

  9. VeritasVosLiberabit
    • 2 years ago
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    ok

  10. v.s
    • 2 years ago
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    both fractions r in a bracket together

  11. VeritasVosLiberabit
    • 2 years ago
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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.

  12. v.s
    • 2 years ago
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    okkay

  13. v.s
    • 2 years ago
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    (y^−2−3y^−4)(y+5y3)

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

  15. VeritasVosLiberabit
    • 2 years ago
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    you can combine like terms in order to get a simplified answer

  16. v.s
    • 2 years ago
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    thanks man

  17. VeritasVosLiberabit
    • 2 years ago
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    you're welcome

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