Find the derivative of the function. \[ g(u)=\left(\frac{u^{3}-3}{u^{3}+3}\right)^{7} \] \( g^{\prime}(u)=\square \) Need Help? Read it

To find the derivative of the function \( g(u)=\left(\frac{u^{3}-3}{u^{3}+3}\right)^{7} \), we can use the chain rule. Let \( h(u) = \frac{u^{3}-3}{u^{3}+3} \), then \( g(u) = h(u)^7 \). The derivative will be: \[ g^{\prime}(u) = 7h(u)^6 \cdot h^{\prime}(u) \] Next, we need to compute \( h^{\prime}(u) \) using the quotient rule: \[ h^{\prime}(u) = \frac{(3u^2)(u^3 + 3) - (u^3 - 3)(3u^2)}{(u^3 + 3)^2} \] Simplifying gives: \[ h^{\prime}(u) = \frac{(3u^2)(u^3 + 3) - (3u^2)(u^3 - 3)}{(u^3 + 3)^2} = \frac{6u^2 \cdot 3}{(u^3 + 3)^2} = \frac{18u^2}{(u^3 + 3)^2} \] Putting it all together, we get: \[ g^{\prime}(u) = 7 \left(\frac{u^3 - 3}{u^3 + 3}\right)^6 \cdot \frac{18u^2}{(u^3 + 3)^2} \] So the final derivative is: \[ g^{\prime}(u) = \frac{126u^2 \left( \frac{u^3-3}{u^3+3} \right)^6}{(u^3 + 3)^2} \]