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

Let \[ g(u)=\left(\frac{u^3-3}{u^3+3}\right)^7. \] Define \[ T(u)=\frac{u^3-3}{u^3+3}, \] so that \[ g(u) = \left(T(u)\right)^7. \] **Step 1. Differentiate using the chain rule** Using the chain rule, we have \[ g'(u)=7\left(T(u)\right)^6\cdot T'(u). \] **Step 2. Differentiate \(T(u)\) using the quotient rule** The quotient rule gives \[ T'(u)=\frac{(u^3+3)\cdot (u^3-3)' - (u^3-3)\cdot (u^3+3)'}{(u^3+3)^2}. \] First, compute the derivatives: \[ (u^3-3)'=3u^2,\quad (u^3+3)'=3u^2. \] Substitute these derivatives into the quotient rule: \[ T'(u)=\frac{(u^3+3)\cdot 3u^2 - (u^3-3)\cdot 3u^2}{(u^3+3)^2}. \] Factor out \(3u^2\): \[ T'(u)=\frac{3u^2\left[(u^3+3) - (u^3-3)\right]}{(u^3+3)^2}. \] Simplify the numerator: \[ (u^3+3)-(u^3-3)=6, \] so \[ T'(u)=\frac{3u^2\cdot 6}{(u^3+3)^2}=\frac{18u^2}{(u^3+3)^2}. \] **Step 3. Combine the results** Substitute \(T'(u)\) into the expression for \(g'(u)\): \[ g'(u)=7\left(\frac{u^3-3}{u^3+3}\right)^6\cdot \frac{18u^2}{(u^3+3)^2}. \] Multiply the constants: \[ 7\cdot 18=126, \] so \[ g'(u)=\frac{126u^2\left(u^3-3\right)^6}{\left(u^3+3\right)^8}. \] Thus, the derivative is \[ g'(u)=\frac{126u^2\left(u^3-3\right)^6}{\left(u^3+3\right)^8}. \]