find the derivative of the function.\ng(u) = (\\frac{u^{3}-3}{u^{3}+3})^{7}\ng(u) =

Explanation:

Step1: Apply chain - rule

Let $y = g(u)=(\frac{u^{3}-3}{u^{3}+3})^{7}$, and let $t=\frac{u^{3}-3}{u^{3}+3}$. Then $y = t^{7}$. By the chain - rule $\frac{dy}{du}=\frac{dy}{dt}\cdot\frac{dt}{du}$. First, find $\frac{dy}{dt}$: $\frac{dy}{dt}=7t^{6}=7(\frac{u^{3}-3}{u^{3}+3})^{6}$.

Step2: Apply quotient - rule to find $\frac{dt}{du}$

The quotient - rule states that if $t=\frac{f(u)}{h(u)}$ where $f(u)=u^{3}-3$ and $h(u)=u^{3}+3$, then $\frac{dt}{du}=\frac{f^{\prime}(u)h(u)-f(u)h^{\prime}(u)}{(h(u))^{2}}$. Since $f^{\prime}(u) = 3u^{2}$ and $h^{\prime}(u)=3u^{2}$, we have $\frac{dt}{du}=\frac{3u^{2}(u^{3}+3)-(u^{3}-3)\cdot3u^{2}}{(u^{3}+3)^{2}}=\frac{3u^{5}+9u^{2}-3u^{5}+9u^{2}}{(u^{3}+3)^{2}}=\frac{18u^{2}}{(u^{3}+3)^{2}}$.

Step3: Calculate $\frac{dy}{du}$

By the chain - rule $\frac{dy}{du}=\frac{dy}{dt}\cdot\frac{dt}{du}$. Substitute $\frac{dy}{dt}$ and $\frac{dt}{du}$: $\frac{dy}{du}=7(\frac{u^{3}-3}{u^{3}+3})^{6}\cdot\frac{18u^{2}}{(u^{3}+3)^{2}}=\frac{126u^{2}(u^{3}-3)^{6}}{(u^{3}+3)^{8}}$.

Answer:

$\frac{126u^{2}(u^{3}-3)^{6}}{(u^{3}+3)^{8}}$