Evelyn Buchanan

2023-03-01

How to find y'' by implicit differentiation for $4{x}^{2}+3{y}^{2}=6$?

Giovanna Shaw

Beginner2023-03-02Added 3 answers

$4{x}^{2}+3{y}^{2}=6$

$\frac{d}{dx}(4{x}^{2}+3{y}^{2})=\frac{d}{dx}\left(6\right)$

$8x+6y\frac{dy}{dx}=0$

So $\frac{dy}{dx}=\frac{-4x}{3y}$

The second derivative follows:

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{-4x}{3y}\right)$

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{(-4)\left(3y\right)-(-4x)\left(3\frac{dy}{dx}\right)}{{\left(3y\right)}^{2}}$

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12y+12x\frac{dy}{dx}}{9{y}^{2}}$

Replacing $\frac{dy}{dx}$ by the expression above gives us:

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12y+12x\left(\frac{-4x}{3y}\right)}{9{y}^{2}}=\frac{-12y+\frac{-16{x}^{2}}{y}}{9{y}^{2}}$

No multiply by $\frac{y}{y}$ to get

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12{y}^{2}-16{x}^{2}}{9{y}^{3}}=\frac{-4(4{x}^{2}+3{y}^{2})}{9{y}^{3}}$

Now, use the initial equation: $4{x}^{2}+3{y}^{2}=6$ to write the answer as:

$\frac{{d}^{2}y}{{dx}^{2}}==\frac{-24}{9{y}^{3}}=\frac{-8}{3{y}^{3}}$

$\frac{d}{dx}(4{x}^{2}+3{y}^{2})=\frac{d}{dx}\left(6\right)$

$8x+6y\frac{dy}{dx}=0$

So $\frac{dy}{dx}=\frac{-4x}{3y}$

The second derivative follows:

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{-4x}{3y}\right)$

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{(-4)\left(3y\right)-(-4x)\left(3\frac{dy}{dx}\right)}{{\left(3y\right)}^{2}}$

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12y+12x\frac{dy}{dx}}{9{y}^{2}}$

Replacing $\frac{dy}{dx}$ by the expression above gives us:

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12y+12x\left(\frac{-4x}{3y}\right)}{9{y}^{2}}=\frac{-12y+\frac{-16{x}^{2}}{y}}{9{y}^{2}}$

No multiply by $\frac{y}{y}$ to get

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12{y}^{2}-16{x}^{2}}{9{y}^{3}}=\frac{-4(4{x}^{2}+3{y}^{2})}{9{y}^{3}}$

Now, use the initial equation: $4{x}^{2}+3{y}^{2}=6$ to write the answer as:

$\frac{{d}^{2}y}{{dx}^{2}}==\frac{-24}{9{y}^{3}}=\frac{-8}{3{y}^{3}}$