Question

This one is really hard I have been stuck for about 3 hours. PLEASE HELP

Answer 1

Given,

The expression is,

$\sqrt{y}+6xy=5$

Required

The double differentiation of the given function.

Differentiating the expression with respect to x then,

$\begin{gathered} \frac{d}{dx}\sqrt{y}+6xy=\frac{d}{dx}5 \\ \frac{1}{2\sqrt{y}}\frac{dy}{dx}+6x\frac{dy}{dx}+6y=0 \\ \frac{dy}{dx}(\frac{1}{2\sqrt{y}}+6x)+6y=0 \end{gathered}$

Differentiating the function again with respect to x then,

$\begin{gathered} \frac{d}{dx}(\frac{dy}{dx}(\frac{1}{2\sqrt{y}}+6x)+\frac{d}{dx}6y=0 \\ \frac{d}{dx}(\frac{1}{2\sqrt{y}}\frac{dy}{dx}+6x\frac{dy}{dx})+\frac{d}{dx}6y=0 \\ \frac{1}{2\sqrt{y}}\frac{d^2y}{dx^2}-(\frac{dy}{dx})^2\frac{1}{4y\sqrt{y}}+6x\frac{d^2y}{dx^2}+6\frac{dy}{dx}+6\frac{dy}{dx}=0 \\ \frac{d^2y}{dx^2}(\frac{1}{2\sqrt{y}}+6x)=\frac{1}{4y\sqrt{y}}(\frac{dy}{dx})^2-12\frac{dy}{dx} \end{gathered}$

Substituting the value of dy/dx then,

$\begin{gathered} \frac{d}{dx}(\frac{dy}{dx}(\frac{1}{2\sqrt{y}}+6x)+\frac{d}{dx}6y=0 \\ \frac{d}{dx}(\frac{1}{2\sqrt{y}}\frac{dy}{dx}+6x\frac{dy}{dx})+\frac{d}{dx}6y=0 \\ \frac{1}{2\sqrt{y}}\frac{d^2y}{dx^2}-(\frac{dy}{dx})^2\frac{1}{4y\sqrt{y}}+6x\frac{d^2y}{dx^2}+6\frac{dy}{dx}+6\frac{dy}{dx}=0 \\ \frac{d^2y}{dx^2}(\frac{1}{2\sqrt{y}}+6x)=\frac{1}{4y\sqrt{y}}(\frac{dy}{dx})^2-12\frac{dy}{dx} \end{gathered}$