Lipossig

2021-03-31

Consider the equation ${y}^{3}-\frac{3}{2}{x}^{2}=1$.

jlo2niT

Step 1
Consider the equation
${y}^{3}-\frac{3}{2}{x}^{2}=1$
Let us find the second derivative implicitly.
Step 2
Differentiate each term with respect to x
$\frac{d}{dx}\left({y}^{3}-\frac{3}{2}{x}^{2}\right)=\frac{d}{dx}$...(1)
$\frac{d}{dx}\left({y}^{3}\right)-\frac{3}{2}\frac{d}{dx}\left({x}^{2}\right)=0$
$3{y}^{2}\frac{dy}{dx}-3x=0$
$\frac{dy}{dx}=\frac{x}{{y}^{2}}\to \left(1\right)$
Step 3
Differentiate second time with respect to x
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{d}{dx}\left(\frac{x}{{y}^{2}}\right)$
$=\frac{\frac{d}{dx}\left(x\right){y}^{2}-\frac{d}{dx}\left({y}^{2}x\right)}{{\left({y}^{2}\right)}^{2}}$
$=\frac{1\cdot {y}^{2}-2y\frac{dy}{dx}x}{{y}^{4}}$
$=\frac{y-2x\frac{dy}{dx}}{{y}^{3}}$
Step 4
From equation (1) we have $\frac{dy}{dx}=\frac{x}{{y}^{2}}$. Substituting this we get
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{y-2x\left(\frac{x}{{y}^{2}}\right)}{{y}^{3}}$
$=\frac{{y}^{3}-2{x}^{2}}{{y}^{5}}$

Jeffrey Jordon