Question

If $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ , then $\frac{dy}{dx}$ is equal to

• A. $-\sqrt[3]{3\frac{y}{x}}$
• B. $-\sqrt[5]{5\frac{y}{x}}$
• C. $\sqrt{\frac{y}{x}}$
• D. $\sqrt[3]{3\frac{x}{y}}$

Answer 1

The correct option is A $-\sqrt[3]{3\frac{y}{x}}$

As $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Let $x=a{\sin }^3\theta ,y=a{\cos }^3\theta$
Therefore
$\frac{dx}{d\theta }=3{\sin }^{2}x\cos x\cdot a$
$\frac{dy}{d\theta }=-3{\cos }^{2}x\sin x\cdot a$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-3{\cos }^{2}x\sin x\cdot a}{3{\sin }^{2}x\cos x\cdot a}$
$\frac{dy}{dx}=-\cot \theta =-\sqrt[3]{3\frac{y}{x}}$

Alternative Solution

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

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