Evelyn Buchanan

2023-03-01

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

Giovanna Shaw

$4{x}^{2}+3{y}^{2}=6$
$\frac{d}{dx}\left(4{x}^{2}+3{y}^{2}\right)=\frac{d}{dx}\left(6\right)$
$8x+6y\frac{dy}{dx}=0$
So $\frac{dy}{dx}=\frac{-4x}{3y}$
The second derivative follows:
$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{-4x}{3y}\right)$
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{\left(-4\right)\left(3y\right)-\left(-4x\right)\left(3\frac{dy}{dx}\right)}{{\left(3y\right)}^{2}}$
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12y+12x\frac{dy}{dx}}{9{y}^{2}}$
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
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{-12{y}^{2}-16{x}^{2}}{9{y}^{3}}=\frac{-4\left(4{x}^{2}+3{y}^{2}\right)}{9{y}^{3}}$
Now, use the initial equation: $4{x}^{2}+3{y}^{2}=6$ to write the answer as:
$\frac{{d}^{2}y}{{dx}^{2}}==\frac{-24}{9{y}^{3}}=\frac{-8}{3{y}^{3}}$

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