### Home > CALC > Chapter 6 > Lesson 6.3.2 > Problem6-92

6-92.

Consider the relation $y^3-3y = x^3 + 1$.

1. Find $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ in terms of $x$ and $y$.

Find $\frac{dy}{dx}$:
$\frac{d}{dx}(y^{3}-3y)=\frac{d}{dx}(x^{3}+1)$
$3y^{2}\frac{dy}{dx}-3\frac{dy}{dx}=3x^{2}\frac{dx}{dx}$
$\frac{dy}{dx}(3y^{2}-3)=3x^{2}$
$\frac{dy}{dx}=\frac{3x^{2}}{3y^{2}-3}=\frac{x^{2}}{y^{2}-1}$

Find $\frac{d^{2}y}{dx^{2}}$:
$\frac{dy}{dx}=\frac{x^{2}}{y^{2}-1}$
$\frac{d^{2}y}{dx^{2}}=\frac{2x\frac{dx}{dx}(y^{2}-1)-2y\frac{dy}{dx}(x^{2})}{(y^{2}-1)^{2}}=\frac{2x(y^{2}-1)-2y\frac{x^{2}}{y^{2}-1}(x^{2})}{(y^{2}-1)^{2}}$
You could simplify now, if you choose.

2. Find $\frac { d ^ { 2 } x } { d x ^ { 2 } } | _ { x = 0 }$

Notice that  $\frac{d^{2}y}{dx^{2}}$, which you found in part (a), has both $x$ and $y$ variables. So, before you evaluate, use the original equation
$(y^3 − 3y = x^3 + 1)$ to find the $y$-value that corresponds with $x = 0$. It is possible that there will be more than one.

$y^3 − 3y = (0)^3 + 1$
$y^3 − 3y − 1 = 0$
Use a calculator to find the THREE $y$-values that correspond with $x = 0$.

Evaluate $\frac{d^{2}y}{dx^{2}}$ at those THREE coordinate points.

3. Find all points where the derivative is undefined.

The derivative,  $\frac{dy}{dx} = \frac{x^2}{y^2-1}$, will be undefined where the denominator equals $0$.