Let x^3 + y^3 = 65. Find y′′(x) at the point (4,1). y′′ (4) = ?

Answer 1

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Answer 2

y^3 = 65-x^3

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

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

y"(x)= -x^2(-2y^-3)y'(x) + (y^-2)(-2x)

y"(x)= -x^2(-2y^-3)x^2/y^2 + (y^-2)(-2x)

= -2x^4(y^-5) - 2x(y^-2)

at (4,1)

y"(4) = -2(4^4)/(1^5) -2(4)/(1^-2)

=-520

(4,1) is on the curve as

4^3 + 1^3 = 65