Question

If $y^3=x^2+y^2$, find $\frac{d^{2}y}{d{x}^2}$ at y = 1

Answer 1

$y^3=x^2+y^{2}y=1,1^2=x^2+1^2\Rightarrow x=0\therefore wheny=1,x=0y^3=x^2+y^2$
Differentiate w.r.t x,
$3y^2\frac{dy}{dx}=2x+2y.\frac{dy}{dx}\to \left(1\right)$
Differentiate w.r.t x,
$3y.{\left(\frac{dy}{dx}\right)}^2+3y^2.\frac{d^{2}y}{d{x}^2}=2+2{\left(\frac{dy}{dx}\right)}^2+2y.\frac{d^{2}y}{d{x}^2}\therefore \frac{d^{2}y}{d{x}^2}=\frac{2+2{\left(\frac{dy}{dx}\right)}^2-3y{\left(\frac{dy}{dx}\right)}^2}{3y^2+2y}\frac{d^{2}y}{d{x}^2}{|}_{y=1}=\frac{2+{\left(\frac{dy}{dx}\right)}^2{|}_{y=1}-3\left(1\right){\left(\frac{dy}{dx}\right)}^2{|}_{y=1}}{3(1{)}^2+2(1)}\to \left(2\right)$
From (1), $\frac{dy}{dx}=\frac{2x}{3y^2-2y}$
$y=1andx=0,\frac{dy}{dx}=0\to \left(3\right)$
Substitute (3) in (2),
$\frac{d^y}{d{x}^2}{|}_{y=1}=\frac{2+2(0{)}^2-3(1\left)\right(0)}{5}\therefore \frac{d^{2}y}{d{x}^2}{|}_{y=1}=\frac{2}{5}$