find $\frac{d^{2}y}{dx^{2}}$ implicitly in terms of $x$ and $y$.\n$x^{2}y - 5x = 2$

Explanation:

Step1: Differentiate both sides with product - rule

Differentiate $x^{2}y - 5x=2$ with respect to $x$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v = y$. The derivative of $x^{2}y$ is $2xy + x^{2}y^\prime$, and the derivative of $-5x$ is $-5$, and the derivative of the constant $2$ is $0$. So we have $2xy+x^{2}y^\prime - 5=0$. Solve for $y^\prime$: $x^{2}y^\prime=5 - 2xy$, then $y^\prime=\frac{5 - 2xy}{x^{2}}$.

Step2: Differentiate $y^\prime$ to get $y^{\prime\prime}$

Using the quotient - rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = 5 - 2xy$ and $v = x^{2}$. First, find $u^\prime$: $u^\prime=-2y-2xy^\prime$. And $v^\prime = 2x$. Then $y^{\prime\prime}=\frac{(-2y - 2xy^\prime)x^{2}-(5 - 2xy)\times2x}{x^{4}}$. Substitute $y^\prime=\frac{5 - 2xy}{x^{2}}$ into the above formula: [ \begin{align*} y^{\prime\prime}&=\frac{(-2y-2x\times\frac{5 - 2xy}{x^{2}})x^{2}-(5 - 2xy)\times2x}{x^{4}}\ &=\frac{(-2yx^{2}-2(5 - 2xy))x^{2}-2x(5 - 2xy)}{x^{4}}\ &=\frac{-2yx^{2}-10 + 4xy-10x + 4x^{2}y}{x^{3}}\ &=\frac{2x^{2}y-10x - 10 + 4xy}{x^{3}} \end{align*} ]

Answer:

$\frac{2x^{2}y-10x - 10 + 4xy}{x^{3}}$