Find $\frac{dy}{dx}$, if $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

$-\left(\frac{y}{x}\right)^{\frac{1}{3}}$

The correct answer is Option (3) → $-\left(\frac{y}{x}\right)^{\frac{1}{3}}$ ##

Let $x = a \cos^3 \theta, y = a \sin^3 \theta$. Then

$x^{\frac{2}{3}} + y^{\frac{2}{3}} = (a \cos^3 \theta)^{\frac{2}{3}} + (a \sin^3 \theta)^{\frac{2}{3}}$

$= a^{\frac{2}{3}} (\cos^2 \theta + \sin^2 \theta) = a^{\frac{2}{3}}$

Hence, $x = a \cos^3 \theta, y = a \sin^3 \theta$ is parametric equation of $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

Now $\frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta \quad \text{and} \quad \frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta$

Therefore $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta = -\sqrt[3]{\frac{y}{x}}$