If \(x^{2/3} + y^{2/3} = a^{2/3}\), where a > 0, x > 0 and y > 0, then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\left(\displaystyle\frac{y}{x}\right)^{1/3}\)
By implicit differentiation (a is a constant):
\( \displaystyle \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} \)
\( \displaystyle = -\frac{y^{1/3}}{x^{1/3}} = -\left(\frac{y}{x}\right)^{1/3} \)