find $\\frac{d^{2}y}{dx^{2}}$, where $\\sqrt{y}+3xy = 2$.\n$\\frac{d^{2}y}{dx^{2}}=$

find $\\frac{d^{2}y}{dx^{2}}$, where $\\sqrt{y}+3xy = 2$.\n$\\frac{d^{2}y}{dx^{2}}=$

find $\\frac{d^{2}y}{dx^{2}}$, where $\\sqrt{y}+3xy = 2$.\n$\\frac{d^{2}y}{dx^{2}}=$

Answer

Explanation:

Step1: Differentiate the equation (\sqrt{y}+3xy = 2) with respect to (x)

Differentiate term - by - term. Using the chain rule, (\frac{d}{dx}(\sqrt{y})=\frac{1}{2\sqrt{y}}\frac{dy}{dx}). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = 3x) and (v = y), (\frac{d}{dx}(3xy)=3y + 3x\frac{dy}{dx}). Differentiating the constant (2) gives (0). So, (\frac{1}{2\sqrt{y}}\frac{dy}{dx}+3y + 3x\frac{dy}{dx}=0). Solve for (\frac{dy}{dx}): [ \begin{align*} \frac{dy}{dx}(\frac{1}{2\sqrt{y}}+3x)&=- 3y\ \frac{dy}{dx}&=\frac{-3y}{\frac{1}{2\sqrt{y}}+3x}=\frac{-6y\sqrt{y}}{1 + 6x\sqrt{y}} \end{align*} ]

Step2: Differentiate (\frac{dy}{dx}) with respect to (x) to find (\frac{d^{2}y}{dx^{2}})

Let (u=-6y\sqrt{y}=-6y^{\frac{3}{2}}) and (v = 1+6x\sqrt{y}=1 + 6xy^{\frac{1}{2}}). Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). First, find (u^\prime): Using the chain rule, (u^\prime=-6\times\frac{3}{2}y^{\frac{1}{2}}\frac{dy}{dx}=-9\sqrt{y}\frac{dy}{dx}). Then, find (v^\prime): Using the sum rule and product rule, (v^\prime=6y^{\frac{1}{2}}+6x\times\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}=6\sqrt{y}+3\frac{x}{\sqrt{y}}\frac{dy}{dx}).

Substitute (u), (u^\prime), (v), and (v^\prime) into the quotient rule: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{(-9\sqrt{y}\frac{dy}{dx})(1 + 6x\sqrt{y})-(-6y\sqrt{y})(6\sqrt{y}+3\frac{x}{\sqrt{y}}\frac{dy}{dx})}{(1 + 6x\sqrt{y})^{2}}\ \end{align*} ] Substitute (\frac{dy}{dx}=\frac{-6y\sqrt{y}}{1 + 6x\sqrt{y}}) into the above expression: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-9\sqrt{y}\times\frac{-6y\sqrt{y}}{1 + 6x\sqrt{y}}(1 + 6x\sqrt{y})+6y\sqrt{y}(6\sqrt{y}+3\frac{x}{\sqrt{y}}\times\frac{-6y\sqrt{y}}{1 + 6x\sqrt{y}})}{(1 + 6x\sqrt{y})^{2}}\ &=\frac{54y^{2}+6y\sqrt{y}(6\sqrt{y}-\frac{18xy^{2}}{1 + 6x\sqrt{y}})}{(1 + 6x\sqrt{y})^{2}}\ &=\frac{54y^{2}(1 + 6x\sqrt{y})+6y\sqrt{y}(6\sqrt{y}(1 + 6x\sqrt{y})-18xy^{2})}{(1 + 6x\sqrt{y})^{3}}\ &=\frac{54y^{2}+324xy^{\frac{5}{2}}+36y^{2}+432xy^{\frac{5}{2}}-108xy^{\frac{5}{2}}}{(1 + 6x\sqrt{y})^{3}}\ &=\frac{90y^{2}+648xy^{\frac{5}{2}}}{(1 + 6x\sqrt{y})^{3}} \end{align*} ]

Answer:

(\frac{90y^{2}+648xy^{\frac{5}{2}}}{(1 + 6x\sqrt{y})^{3}})