find $\frac{d^{2}y}{dx^{2}}$.\n$6x^{2}+y^{2}=6$\n$\frac{d^{2}y}{dx^{2}}=square$
find $\frac{d^{2}y}{dx^{2}}$.\n$6x^{2}+y^{2}=6$\n$\frac{d^{2}y}{dx^{2}}=square$
Answer
Explanation:
Step1: Differentiate implicitly once
Differentiate $6x^{2}+y^{2}=6$ with respect to $x$. Using the power - rule and chain - rule, we have $12x + 2y\frac{dy}{dx}=0$. Then solve for $\frac{dy}{dx}$: $2y\frac{dy}{dx}=-12x$, so $\frac{dy}{dx}=-\frac{6x}{y}$.
Step2: Differentiate $\frac{dy}{dx}$ implicitly
Differentiate $\frac{dy}{dx}=-\frac{6x}{y}$ with respect to $x$ using the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = - 6x$ and $v = y$. $\frac{d^{2}y}{dx^{2}}=\frac{(-6)y-(-6x)\frac{dy}{dx}}{y^{2}}$.
Step3: Substitute $\frac{dy}{dx}$
Substitute $\frac{dy}{dx}=-\frac{6x}{y}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-6y + 6x\cdot\frac{6x}{y}}{y^{2}}\ &=\frac{-6y^{2}+ 36x^{2}}{y^{3}}\ &=\frac{-6(y^{2}-6x^{2})}{y^{3}} \end{align*} ] Since $6x^{2}+y^{2}=6$, then $y^{2}-6x^{2}=6 - 12x^{2}$. So $\frac{d^{2}y}{dx^{2}}=\frac{-6(6 - 12x^{2})}{y^{3}}=\frac{72x^{2}- 36}{y^{3}}$.
Answer:
$\frac{72x^{2}-36}{y^{3}}$