use implicit differentiation to find $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$.\n$2x^{2}+y^{2}=1$\n$\frac{d…

use implicit differentiation to find $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$.\n$2x^{2}+y^{2}=1$\n$\frac{dy}{dx}=-\frac{2x}{y}$\n$\frac{d^{2}y}{dx^{2}}=square$

use implicit differentiation to find $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$.\n$2x^{2}+y^{2}=1$\n$\frac{dy}{dx}=-\frac{2x}{y}$\n$\frac{d^{2}y}{dx^{2}}=square$

Answer

Explanation:

Step1: Differentiate the first - derivative result

We know that $\frac{dy}{dx}=-\frac{2x}{y}$. Use the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = 2x$, $u'=2$, $v = y$, and $v'=\frac{dy}{dx}$.

Step2: Substitute $\frac{dy}{dx}$

Substitute $\frac{dy}{dx}=-\frac{2x}{y}$ into the quotient - rule formula. $\frac{d^{2}y}{dx^{2}}=\frac{2y-2x\frac{dy}{dx}}{y^{2}}=\frac{2y - 2x\left(-\frac{2x}{y}\right)}{y^{2}}$.

Step3: Simplify the expression

$\frac{d^{2}y}{dx^{2}}=\frac{2y^{2}+4x^{2}}{y^{3}}$. Since $2x^{2}+y^{2}=1$, then $4x^{2}=2 - 2y^{2}$. Substitute $4x^{2}=2 - 2y^{2}$ into the above formula: $\frac{d^{2}y}{dx^{2}}=\frac{2y^{2}+2 - 2y^{2}}{y^{3}}=\frac{2}{y^{3}}$.

Answer:

$\frac{2}{y^{3}}$