If \(y=\displaystyle\frac{1}{\sqrt[3]{1-x^3}}\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x^2(1-x^3)^{-4/3}\)
Writing y as a power:
\(y = (1 - x^3)^{-1/3}\)
By the chain rule:
\( \displaystyle \frac{dy}{dx} = -\frac{1}{3}(1 - x^3)^{-4/3} \cdot \frac{d}{dx}(1 - x^3) \)
\( \displaystyle = -\frac{1}{3}(1 - x^3)^{-4/3} \cdot (-3x^2) \)
\( = x^2(1 - x^3)^{-4/3} \)
Why the other options are incorrect
Option (1) \(x(1-x^3)^{-4/3}\) — Incorrect. The derivative of \(1 - x^3\) is \(-3x^2\), so the factor is \(x^2\), not x.
Option (3) \((1-x^3)^{-4/3}\) — Incorrect. It leaves out the factor \(x^2\) that comes from the derivative of the inside function \(1 - x^3\).
Option (4) \(-x^2(1-x^3)^{-4/3}\) — Incorrect. The two minus signs, from the power \(-\frac{1}{3}\) and from \(-3x^2\), cancel, so the derivative is positive.