QuestionMY NOTES TANAP
Find the derivative of the function. $$\begin{array}{l} g(t)=\sqrt[3]{9-2 t^{3}} \\ g^{\prime}(t)=\square \end{array}$$

Studdy Solution
Simplify the expression for $g'(t)$:
$$g'(t) = \frac{1}{3} \cdot (-6t^2) \cdot (9 - 2t^3)^{-2/3}$$ $$g'(t) = -2t^2 \cdot (9 - 2t^3)^{-2/3}$$
The derivative of the function is:
$$g'(t) = -2t^2 (9 - 2t^3)^{-2/3}$$