Second derivative of $x^3+y^3=1$ using implicit differentiation 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$$-{2x \over y^5}$$
 A: Throughout this solution, I will use the Leibniz notation $\frac{dy}{dx}$ to represent the derivative of $y$ with respect to $x.$ Notice that the second derivative is notated $\frac{d^{2}y}{dx^{2}}.$ We begin as you did - find the first derivative.
Given: $x^{3} + y^{3} = 1$
$3x^{2} + 3y^{2} \cdot \frac{dy}{dx} = 0$ (*)
$\frac{dy}{dx} = -\frac{x^{2}}{y^{2}}$ (**)
We differentiate the (*) equation with respect to $x.$ This yields the following:
$6x + \frac{dy}{dx} \cdot (6y \cdot \frac{dy}{dx}) + (3y^{2}) \cdot (\frac{d^{2}y}{dx^{2}}) = 0$
$6x + 6y \cdot (\frac{dy}{dx})^{2} + 3y^2 \cdot (\frac{d^{2}y}{dx^{2}}) = 0$
$3y^2 \cdot (\frac{d^{2}y}{dx^{2}}) = -6x - 6y \cdot (\frac{dy}{dx})^{2}$
Substituting the (**) equation for $\frac{dy}{dx}$ here, we find our answer:
$\frac{d^{2}y}{dx^{2}} = \frac{-2x - 2y \cdot (-\frac{x^{2}}{y^{2}})^{2}}{y^{2}}$
$\frac{d^{2}y}{dx^{2}} = \boxed{-\frac{2x}{y^{2}}(1 + \frac{x^{3}}{y^{3}})}.$
A: $1=x^3+y^3\implies 0=3 x^2+3 y^2 y',$ 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 $y''=-2 y x y^{-6}=-2 y x (1-x^3)^{-2}.$
A: If $Dx$ is the first derivative and $Dx^2$ is the second derivative, than your first derivative is correct. For the second derivative we have:
$$
\frac{d}{dx}y'=\frac{d}{dx}\left(\frac{-x^2}{y^2}\right)$$
that, using fraction rule and chain rule for $y$, becomes:
$$\frac{-2xy^2+2x^2y(y')}{y^4}$$
substituting $y'=-x^2/y^2$ and wi a bit of algebra:
$$\frac{-2xy^3-2x^4}{y^5}=\frac{-2x(y^3+x^3)}{y^5}$$
finally, using $x^3+y^3=1$:
$$\frac{d}{dx}y'=y''=\frac{-2x}{y^5}
$$
A: If 
$x^n + y^n =1
$
then
$nx^{n-1} +ny^{n-1}y' = 0$
or
$y'
=-\frac{x^{n-1}}{y^{n-1}}
$.
Differentiating again,
$\begin{array}\\
0
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y^{n-2}(y')^2\\
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y(\frac{x^{n-1}}{y^{n-1}})^2\\
&=y^{n-1}y''+(n-1)x^{n-2}(1+\frac{x^n}{y^{2n-3}})\\
&=y^{n-1}y''+(n-1)x^{n-2}\frac{x^n+y^{2n-3}}{y^{2n-3}}\\
\end{array}
$
so
$y''
=-\frac{(n-1)x^{n-2}\frac{x^n+y^{2n-3}}{y^{2n-3}}}{y^{n-1}}
=-\frac{(n-1)x^{n-2}(x^n+y^{2n-3})}{y^{3n-4}}
$
If $n=3$,
$y'
=-\frac{x^2}{y^2}
$
and
$y''
=-\frac{2x(x^3+y^3)}{y^5}
=-\frac{2x}{y^5}
$
since
$x^3+y^3 = 1$.
