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anonymous

  • 5 years ago

what is the d'(y)= (y^2)^2+(y-3)^2?? any help?

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  1. anonymous
    • 5 years ago
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    Are you asking for the first derivative with respect to y? If so, then you can do this many ways, the easiest is to simplify first. y^2^2=y^4, (y-3)^2 = y^2-6y+9. Now just take the derivative of y^4+y^2-6y+9 with respect to y. This becomes 4y^3+2y-6

  2. anonymous
    • 5 years ago
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    how would i find what y is worth from that equation .. because thats what i had before and i dont know where im going wrong?>

  3. anonymous
    • 5 years ago
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    What do you mean find out what y is worth? Y is a variable, it does not have a set value.

  4. anonymous
    • 5 years ago
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    critical points i meant of the derivative

  5. anonymous
    • 5 years ago
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    If you want critical points of the original function, then set the derivative equal to zero.

  6. anonymous
    • 5 years ago
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    ya i know this ... but i cant find the right answer to it .. how would you do that when you have a third root of somthing ...

  7. anonymous
    • 5 years ago
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    Take factors of the first term and then take factors of the second term. After that, you have to do some sythetic division.

  8. anonymous
    • 5 years ago
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    sorry not second term i meant last term

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