Aiyana Jenkins

2023-03-06

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

sayko11czj

On both sides, differentiate each term $\text{implicitly with respect to x}$
$\text{Note differentiate}\phantom{\rule{1ex}{0ex}}x{y}^{3}\phantom{\rule{1ex}{0ex}}\text{using}\phantom{\rule{1ex}{0ex}}\text{product rule}$
$\left(x.3{y}^{2}.\frac{dy}{dx}+{y}^{3}.1\right)=\frac{dy}{dx}+1$
$⇒3x{y}^{2}\frac{dy}{dx}-\frac{dy}{dx}=1-{y}^{3}$
$⇒\frac{dy}{dx}\left(3x{y}^{2}-1\right)=1-{y}^{3}$
$⇒\frac{dy}{dx}=\frac{1-{y}^{3}}{3x{y}^{2}-1}$

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