Lexi Holmes

2023-03-21

How to differentiate ${x}^{\frac{2}{3}}+{y}^{\frac{2}{3}}=4$?

Attagswalmec6

Beginner2023-03-22Added 6 answers

Consider that in this instance of implicit differentiation $y$ is function of $x$ so you get:

$\frac{2}{3}{x}^{\frac{2}{3}-1}+\frac{2}{3}{y}^{\frac{2}{3}-1}\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{{x}^{-\frac{1}{3}}}{{y}^{-\frac{1}{3}}}=-\sqrt[3]{\frac{y}{x}}$

Where $y={(4-{x}^{\frac{2}{3}})}^{\frac{3}{2}}$ from your original expression.

$\frac{2}{3}{x}^{\frac{2}{3}-1}+\frac{2}{3}{y}^{\frac{2}{3}-1}\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{{x}^{-\frac{1}{3}}}{{y}^{-\frac{1}{3}}}=-\sqrt[3]{\frac{y}{x}}$

Where $y={(4-{x}^{\frac{2}{3}})}^{\frac{3}{2}}$ from your original expression.

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