Aiyana Jenkins

2023-03-06

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

sayko11czj

Beginner2023-03-07Added 2 answers

On both sides, differentiate each term $\text{implicitly with respect to x}$

$\text{Note differentiate}\phantom{\rule{1ex}{0ex}}x{y}^{3}\phantom{\rule{1ex}{0ex}}\text{using}\phantom{\rule{1ex}{0ex}}{\text{product rule}}$

$(x.3{y}^{2}.\frac{dy}{dx}+{y}^{3}.1)=\frac{dy}{dx}+1$

$\Rightarrow 3x{y}^{2}\frac{dy}{dx}-\frac{dy}{dx}=1-{y}^{3}$

$\Rightarrow \frac{dy}{dx}(3x{y}^{2}-1)=1-{y}^{3}$

$\Rightarrow \frac{dy}{dx}=\frac{1-{y}^{3}}{3x{y}^{2}-1}$

$\text{Note differentiate}\phantom{\rule{1ex}{0ex}}x{y}^{3}\phantom{\rule{1ex}{0ex}}\text{using}\phantom{\rule{1ex}{0ex}}{\text{product rule}}$

$(x.3{y}^{2}.\frac{dy}{dx}+{y}^{3}.1)=\frac{dy}{dx}+1$

$\Rightarrow 3x{y}^{2}\frac{dy}{dx}-\frac{dy}{dx}=1-{y}^{3}$

$\Rightarrow \frac{dy}{dx}(3x{y}^{2}-1)=1-{y}^{3}$

$\Rightarrow \frac{dy}{dx}=\frac{1-{y}^{3}}{3x{y}^{2}-1}$