Find the derivative of the function.y= 1/(x^4+3)^3  ............y=1 divided by x to the power of 4 plus 3 all in the paratheses and all of the bottom half is to the 3rd power

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`y= 1/(x^4+3)^3 `

`==gt y= (x^4 + 3)^-3 `

`==gt y'= -3(x^3+3)' (x^4 +3)^-4 `

`==gt y'= -3(3x^2)(x^4+3)^-4 `

`==gt y'= (-9x^2)/(x^4+3)^4`


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