I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = – {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second derivative.
How would I do it to find the second derivative? apparently, it is supposed to be$$-{2x \over y^5}$$
Best Answer
$1=x^3+y^3\implies 0=3 x^2+3 y^2 y',$ so $$0=x^2+y^2 y'.$$ Differentiate this to get $$0=2 x +2 y y'^2+y^2 y''.$$ Therefore for $y\ne 0$ we have $$y''=-y^{-2}(2 x+2 y y'^2).$$ From the first differentiation we have $$y'=-x^2/y^2.$$ Therefore for $y\ne 0$ (equivalently, for $x\ne 1$),$$y''=-y^{-2}(2 x +2 y y'^2)=- y^{-2}(2 x +2 y (x^4/y^4))=-2 y^{-5} x (y^3+x^3)=-2y^{-5}x$$ because $y^3+x^3=1.$ This can also be written as $y''=-2 y x y^{-6}=-2 y x (1-x^3)^{-2}.$