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 $\frac{dy}{dx}$ with respect to $x$

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

Step2: Substitute $\frac{dy}{dx}=-\frac{2x}{y}$

Substitute $\frac{dy}{dx}$ into the above formula: $\frac{-2y + 2x\left(-\frac{2x}{y}\right)}{y^{2}}=\frac{-2y^{2}-4x^{2}}{y^{3}}$. Since $2x^{2}+y^{2}=1$, then $2x^{2}=1 - y^{2}$, and $-2y^{2}-4x^{2}=-2y^{2}-2(2x^{2})=-2y^{2}-2(1 - y^{2})=-2$.

Answer:

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