find $\frac{d^{2}y}{dx^{2}}$.\n$14x^{2}+y^{2}=2$\n$\frac{d^{2}y}{dx^{2}}=square$
find $\frac{d^{2}y}{dx^{2}}$.\n$14x^{2}+y^{2}=2$\n$\frac{d^{2}y}{dx^{2}}=square$
Answer
Explanation:
Step1: Differentiate implicitly once
Differentiate $14x^{2}+y^{2}=2$ with respect to $x$. Using the power - rule and chain - rule, we have $28x + 2y\frac{dy}{dx}=0$. Then solve for $\frac{dy}{dx}$: [2y\frac{dy}{dx}=-28x] [\frac{dy}{dx}=-\frac{14x}{y}]
Step2: Differentiate $\frac{dy}{dx}$ with respect to $x$
Use the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = - 14x$ and $v = y$. $u'=-14$ and $v'=\frac{dy}{dx}$. [\frac{d^{2}y}{dx^{2}}=\frac{-14y-(-14x)\frac{dy}{dx}}{y^{2}}] Substitute $\frac{dy}{dx}=-\frac{14x}{y}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-14y + 14x\cdot\frac{14x}{y}}{y^{2}}\ &=\frac{-14y^{2}+196x^{2}}{y^{3}} \end{align*} ] Since $14x^{2}+y^{2}=2$, then $14x^{2}=2 - y^{2}$, and $196x^{2}=14(2 - y^{2})=28 - 14y^{2}$. [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-14y^{2}+28 - 14y^{2}}{y^{3}}\ &=\frac{28 - 28y^{2}}{y^{3}}\ &=\frac{28(1 - y^{2})}{y^{3}} \end{align*} ]
Answer:
$\frac{28(1 - y^{2})}{y^{3}}$