Need help finding the second derivative

[kluda06]

X²+3y²=18

answer is -2/y³

I got to the point where my dy/dx=-x/3y

I used the quotient rule which led me to have

3y-3(dy/dx)/3y²

After i plug in my dy/dx I do not know how to keep going from there! Can someone help me please?

 

[soroban]

Hello, kluda06!

\(\displaystyle \text{Find the second derivative: }\: x^2 + 3y^2 \:=\:18\)

\(\displaystyle \text{Answer: }\:\dfrac{d^2y}{dx^2} \:=\:-\dfrac{2}{y^3}\)

We have: .\(\displaystyle 2x + 6y\left(\frac{dy}{dx}\right) \:=\:0 \quad\Rightarrow\quad \dfrac{dy}{dx} \:=\:\dfrac{-2x}{6y}\)

. . . . . . . \(\displaystyle \dfrac{dy}{dx} \:=\:-\dfrac{1}{3}\left(\dfrac{x}{y}\right)\)

Then: .\(\displaystyle \dfrac{d^2y}{dx^2} \:=\:-\dfrac{1}{3}\left(\dfrac{y\cdot1 - x\left(\frac{dy}{dx}\right)}{y^2}\right) \;=\; -\dfrac{1}{3}\left(\dfrac{y - x(\text{-}\frac{1}{3}\frac{x}{y})}{y^2}\right)\)

. . . . \(\displaystyle \dfrac{d^2y}{dx^2}\;=\;-\dfrac{1}{3}\left(\dfrac{y+\frac{x^2}{3y}}{y^2} \right) \;=\;-\dfrac{1}{3}\cdot\dfrac{\overbrace{3y^2+x^2}^{ \text{This is 18}}}{3y^3}\)

. . . . \(\displaystyle \dfrac{d^2y}{dx^2} \;=\;-\dfrac{1}{3}\cdot\dfrac{18}{3y^3} \;=\;-\dfrac{2}{y^3}\)