Determine the derivative of $y(x) = \sqrt[3]{x}$.
$y'(x) = \dfrac{1}{2\sqrt{x}}$.
This cannot be determined by the derivatives of elementary functions.
None of the other options is correct.
$y'(x) = \dfrac{1}{3\sqrt[3]{x^2}}$.
Correct: We first write $y(x)$ as $x^a$, hence $y(x) = \sqrt[3]{x} = x^{\tfrac{1}{3}}$. Then we can apply the rule for power functions:
$$
y'(x) = \dfrac{1}{3} x^{\tfrac{1}{3}-1} = \dfrac{1}{3} x^{-\tfrac{2}{3}} = \dfrac{1}{3}\cdot\dfrac{1}{x^{\tfrac{2}{3}}} = \dfrac{1}{3} \cdot \dfrac{1}{(x^2)^{\tfrac{1}{3}} = \dfrac{1}{3}\cdot\dfrac{1}{\sqrt[3]{x^2}}} = \dfrac{1}{3\sqrt[3]{x^2}}.
$$
Wrong: $y(x)$ is a cube root, not a 'regular' square root.
See Example 2 and Power functions: extra explanation.
Wrong: This is possible, but you have to rewrite $y(x)$ as $x^k$.
See Example 2 en Power functions: extra explanation.
Wrong: The correct answer is shown. Maybe you need to rewrite your answer in order to find the correct one amongst them.