find \\( \\frac { d ^ { 2 } y } { d x ^ { 2 } } \\), where \\( \\sqrt { y } + 3 x y = 4 \\).\n\\( \\frac { d…

Explanation:

Step1: Differentiate the equation implicitly for the first - order derivative

Differentiate (\sqrt{y}+3xy = 4) with respect to (x). Using the chain rule ((\sqrt{y})^\prime=\frac{1}{2\sqrt{y}}\cdot y^\prime) and the product rule ((3xy)^\prime=3y + 3xy^\prime). We get (\frac{y^\prime}{2\sqrt{y}}+3y + 3xy^\prime=0). Solve for (y^\prime): [ \begin{align*} y^\prime\left(\frac{1}{2\sqrt{y}}+3x\right)&=- 3y\ y^\prime&=\frac{-3y}{\frac{1}{2\sqrt{y}}+3x}=\frac{-6y^{\frac{3}{2}}}{1 + 6x\sqrt{y}} \end{align*} ]

Step2: Differentiate (y^\prime) to find (y^{\prime\prime})

Use the quotient rule (y^{\prime\prime}=\frac{(u/v)^\prime=u^\prime v - uv^\prime}{v^{2}}), where (u=-6y^{\frac{3}{2}}), (u^\prime=-6\times\frac{3}{2}y^{\frac{1}{2}}y^\prime=-9y^{\frac{1}{2}}y^\prime), and (v = 1+6x\sqrt{y}), (v^\prime=6\sqrt{y}+6x\times\frac{1}{2\sqrt{y}}y^\prime=6\sqrt{y}+\frac{3xy^\prime}{\sqrt{y}}).

[ \begin{align*} y^{\prime\prime}&=\frac{-9y^{\frac{1}{2}}y^\prime(1 + 6x\sqrt{y})+6y^{\frac{3}{2}}\left(6\sqrt{y}+\frac{3xy^\prime}{\sqrt{y}}\right)}{(1 + 6x\sqrt{y})^{2}}\ \end{align*} ] Substitute (y^\prime=\frac{-6y^{\frac{3}{2}}}{1 + 6x\sqrt{y}}) into the above formula:

[ \begin{align*} y^{\prime\prime}&=\frac{-9y^{\frac{1}{2}}\times\frac{-6y^{\frac{3}{2}}}{1 + 6x\sqrt{y}}(1 + 6x\sqrt{y})+6y^{\frac{3}{2}}\left(6\sqrt{y}+\frac{3x\times\frac{-6y^{\frac{3}{2}}}{1 + 6x\sqrt{y}}}{\sqrt{y}}\right)}{(1 + 6x\sqrt{y})^{2}}\ &=\frac{54y^{2}+6y^{\frac{3}{2}}\left(6\sqrt{y}-\frac{18xy^{2}}{(1 + 6x\sqrt{y})\sqrt{y}}\right)}{(1 + 6x\sqrt{y})^{2}}\ &=\frac{54y^{2}+36y^{2}-\frac{108xy^{3}}{(1 + 6x\sqrt{y})}}{(1 + 6x\sqrt{y})^{2}}\ &=\frac{90y^{2}(1 + 6x\sqrt{y})-108xy^{3}}{(1 + 6x\sqrt{y})^{3}}\ &=\frac{90y^{2}+540xy^{\frac{5}{2}}-108xy^{3}}{(1 + 6x\sqrt{y})^{3}} \end{align*} ]

Answer:

(\frac{90y^{2}+540xy^{\frac{5}{2}}-108xy^{3}}{(1 + 6x\sqrt{y})^{3}})