### Home > APCALC > Chapter 6 > Lesson 6.5.1 > Problem6-175

6-175.

Consider the equation $y^3 − 3y = x^3 + 1$.

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

$\text{Find } \frac{dy}{dx}:$

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

$3y^{2}\frac{dy}{dx}-3\frac{dy}{dx}=3x^{2}\frac{dx}{dx}$

$\frac{dy}{dx}(3y^{2}-3)=3x^{2}$

$\frac{dy}{dx}=\frac{3x^{2}}{3y^{2}-3}=\frac{x^{2}}{y^{2}-1}$

$\text{Find } \frac{d^{2}y}{dx^{2}}:$

$\frac{dy}{dx}=\frac{x^{2}}{y^{2}-1}$

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

You could simplify now, if you choose.

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

$\text{Notice that } \frac{d^{2}y}{dx^{2}},\text{ which you found in part (a),}$

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 $x = 0$.

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

3. Determine all points where the derivative is undefined.

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