What is the second derivative of xy+y^3=1 using implicit differentiation?

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Quote:

$\text{Find }y''\text{ for: }\:xy+y^3\:=\:1$

We have: . $1\!\cdot\!y + x\!\cdot\!y' + 3y^2\!\cdot\!y' \:=\:0$

. . . . . . . . . . . . $x\!\cdot\!y' + 3y^2\!\cdot\!y' \:=\:-y$

. . . . . . . . . . . . . $(x + 3y^2)y' \:=\:-y$

. . . . . . . . . . . . . . . . . . . $y' \:=\:\frac{\text{-}y}{x+3y^2}$ .[1]

Then: . $y" \;=\;\frac{(x+3y^2)(\text{-}y') - (\text{-}y)(1 + 6yy')}{(x+3y^2)^2}$

. . . . . $y'' \;=\;\frac{-xy' - 3y^2y' + y + 6y^2y'}{(x+3y^2)^2}$

. . . . . $y'' \;=\;\frac{y - xy + 3y^2y'}{(x+3y^2)^2}$

Substitute [1]:
. . $y'' \;=\;\frac{y - xy + 3y^2\left(\frac{\text{-}y}{x+3y^2}\right)}{(x+3y^2)^2}$

. . $y'' \;=\;\frac{y(x+3y^2) - xy(x+3y^2) - 3y^3}{(x+3y^2)^3}$

. . $y'' \;=\;\frac{xy + 3y^3 - x^2y - 3xy^3 - 3y^3}{(x+3y^2)^3}$

. . $y'' \;=\;\frac{xy - x^2y - 3xy^3}{(x+3y^2)^3}$