If $y=\frac{1}{\sqrt[3]{1-x^3}}$, then $\frac{dy}{dx}$ is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $x^2(1-x^3)^{-4/3}$
Given:
$y=(1-x^{3})^{-\frac{1}{3}}$
$\frac{dy}{dx}=-\frac{1}{3}(1-x^{3})^{-\frac{4}{3}}\cdot(-3x^{2})$
$\frac{dy}{dx}=x^{2}(1-x^{3})^{-\frac{4}{3}}$
Writing in radical form, $\frac{dy}{dx}=\frac{x^{2}}{(1-x^{3})^{\frac{4}{3}}}$.
∴ the derivative is $\frac{dy}{dx}=x^{2}(1-x^{3})^{-\frac{4}{3}}$