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

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

Mathematics
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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
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?>
What do you mean find out what y is worth? Y is a variable, it does not have a set value.

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critical points i meant of the derivative
If you want critical points of the original function, then set the derivative equal to zero.
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 ...
Take factors of the first term and then take factors of the second term. After that, you have to do some sythetic division.
sorry not second term i meant last term

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