Ideovedobpusf

2023-03-25

How to use implicit differentiation to find $\frac{dy}{dx}$ given $3{x}^{2}+3{y}^{2}=2$?

Padria4mia

Beginner2023-03-26Added 6 answers

Circle differentiation that is implicit $3{x}^{2}+{3}^{y}^2=2$ gives

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

Solving for $\frac{dy}{dx}$ gives

$\frac{dy}{dx}=-\frac{x}{y}$

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

Solving for $\frac{dy}{dx}$ gives

$\frac{dy}{dx}=-\frac{x}{y}$

inpuctists8f5

Beginner2023-03-27Added 5 answers

$\text{differentiate implicitly with respect to x}$

$\text{noting that}$

$\frac{d}{dx}\left(y\right)=\frac{dy}{dx}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\frac{d}{dx}\left({y}^{2}\right)=2y\frac{dy}{dx}$

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

$\Rightarrow 6y\frac{dy}{dx}=-6x$

$\Rightarrow \frac{dy}{dx}=\frac{-6x}{6y}=-\frac{x}{y}$

$\text{noting that}$

$\frac{d}{dx}\left(y\right)=\frac{dy}{dx}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\frac{d}{dx}\left({y}^{2}\right)=2y\frac{dy}{dx}$

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

$\Rightarrow 6y\frac{dy}{dx}=-6x$

$\Rightarrow \frac{dy}{dx}=\frac{-6x}{6y}=-\frac{x}{y}$

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