find $\\frac{d^{2}y}{dx^{2}}$, where $\\sqrt{y}+2xy = 1$.\n$\\frac{d^{2}y}{dx^{2}}=\\square$

find $\\frac{d^{2}y}{dx^{2}}$, where $\\sqrt{y}+2xy = 1$.\n$\\frac{d^{2}y}{dx^{2}}=\\square$

find $\\frac{d^{2}y}{dx^{2}}$, where $\\sqrt{y}+2xy = 1$.\n$\\frac{d^{2}y}{dx^{2}}=\\square$

Answer

Explanation:

Step1: Differentiate the equation (\sqrt{y}+2xy = 1) with respect to (x)

Using the chain - rule ((\sqrt{y})^\prime=\frac{1}{2\sqrt{y}}y^\prime) and the product - rule ((2xy)^\prime=2y + 2xy^\prime). The derivative of the left - hand side is (\frac{y^\prime}{2\sqrt{y}}+2y + 2xy^\prime), and the derivative of the right - hand side is (0). So, (\frac{y^\prime}{2\sqrt{y}}+2y + 2xy^\prime=0). Multiply through by (2\sqrt{y}) to get (y^\prime+4y\sqrt{y}+4xy\sqrt{y}y^\prime = 0). Solve for (y^\prime): [ \begin{align*} y^\prime(1 + 4xy\sqrt{y})&=-4y\sqrt{y}\ y^\prime&=\frac{-4y\sqrt{y}}{1 + 4xy\sqrt{y}} \end{align*} ]

Step2: Differentiate (y^\prime=\frac{-4y\sqrt{y}}{1 + 4xy\sqrt{y}}) with respect to (x) using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}})

Let (u=-4y^{\frac{3}{2}}), then (u^\prime=-4\times\frac{3}{2}y^{\frac{1}{2}}y^\prime=-6\sqrt{y}y^\prime) Let (v = 1+4xy^{\frac{3}{2}}), then (v^\prime=4y^{\frac{3}{2}}+4x\times\frac{3}{2}y^{\frac{1}{2}}y^\prime=4y^{\frac{3}{2}}+6xy^{\frac{1}{2}}y^\prime)

[ \begin{align*} y^{\prime\prime}&=\frac{(-6\sqrt{y}y^\prime)(1 + 4xy^{\frac{3}{2}})-(-4y^{\frac{3}{2}})(4y^{\frac{3}{2}}+6xy^{\frac{1}{2}}y^\prime)}{(1 + 4xy^{\frac{3}{2}})^{2}}\ \end{align*} ] Substitute (y^\prime=\frac{-4y\sqrt{y}}{1 + 4xy\sqrt{y}}) into the above formula:

[ \begin{align*} y^{\prime\prime}&=\frac{-6\sqrt{y}\times\frac{-4y\sqrt{y}}{1 + 4xy\sqrt{y}}(1 + 4xy\sqrt{y})+4y^{\frac{3}{2}}(4y^{\frac{3}{2}}+6xy^{\frac{1}{2}}\times\frac{-4y\sqrt{y}}{1 + 4xy\sqrt{y}})}{(1 + 4xy\sqrt{y})^{2}}\ &=\frac{24y^{2}+4y^{\frac{3}{2}}\times\frac{4y^{\frac{3}{2}}(1 + 4xy\sqrt{y})-24xy^{2}}{1 + 4xy\sqrt{y}}}{(1 + 4xy\sqrt{y})^{2}}\ &=\frac{24y^{2}(1 + 4xy\sqrt{y})+16y^{3}-96xy^{2}}{(1 + 4xy\sqrt{y})^{3}}\ &=\frac{24y^{2}+96xy^{\frac{5}{2}}+16y^{3}-96xy^{2}}{(1 + 4xy\sqrt{y})^{3}}\ &=\frac{16y^{3}-72y^{2}+96xy^{\frac{5}{2}}}{(1 + 4xy\sqrt{y})^{3}} \end{align*} ]

Answer:

(\frac{16y^{3}-72y^{2}+96xy^{\frac{5}{2}}}{(1 + 4xy\sqrt{y})^{3}})