let x^3+y^3=65 at the point (4,1). y double prime(4)=?

Expert Answers

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Differentiating once we get

`3x^2 + 3y^2 y' = 0`

y'=-(3x^2)/(3y^2)=-x^2/y^2

use the quotient property

`y'' = -(y^2(2x)-x^2(2y)y')/(y^4)=-(2xy(y-x y'))/y^4=-2x(y-xy')/y^3`

y' at (4,1) `= -4^2/1^2 = -16`

y'' at (4,1) `= -2(4)(1-4(-16))/(1^3)=-8(1+64)/1=-8(65)=-520`

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