4. given that ( g(u)=left(\frac{u^{3}-1}{u^{3}+1}\right)^{8} ) show that ( g^{prime}(u)=\frac{48…

Explanation:

Step1: Use the chain rule

Let (y = g(u)=(\frac{u^{3}-1}{u^{3}+1})^{8}), let (t=\frac{u^{3}-1}{u^{3}+1}), then (y = t^{8}). By the chain rule (\frac{dy}{du}=\frac{dy}{dt}\cdot\frac{dt}{du}).

First, find (\frac{dy}{dt}): (\frac{dy}{dt}=8t^{7})

Step2: Use the quotient rule to find (\frac{dt}{du})

The quotient rule states that if (t=\frac{f(u)}{h(u)}) where (f(u)=u^{3}-1), (f^{\prime}(u) = 3u^{2}) and (h(u)=u^{3}+1), (h^{\prime}(u)=3u^{2}), then (\frac{dt}{du}=\frac{f^{\prime}(u)h(u)-f(u)h^{\prime}(u)}{h^{2}(u)})

[ \begin{align*} \frac{dt}{du}&=\frac{3u^{2}(u^{3}+1)-3u^{2}(u^{3}-1)}{(u^{3}+1)^{2}}\ &=\frac{3u^{5}+ 3u^{2}-3u^{5}+3u^{2}}{(u^{3}+1)^{2}}\ &=\frac{6u^{2}}{(u^{3}+1)^{2}} \end{align*} ]

Step3: Substitute (t) and (\frac{dt}{du}) into the chain - rule formula

(\frac{dy}{du}=8t^{7}\cdot\frac{6u^{2}}{(u^{3}+1)^{2}}), since (t=\frac{u^{3}-1}{u^{3}+1})

[ \begin{align*} \frac{dy}{du}&=8(\frac{u^{3}-1}{u^{3}+1})^{7}\cdot\frac{6u^{2}}{(u^{3}+1)^{2}}\ &=\frac{48u^{2}(u^{3}-1)^{7}}{(u^{3}+1)^{9}} \end{align*} ]

Answer:

We have shown that (g^{\prime}(u)=\frac{48u^{2}(u^{3}-1)^{7}}{(u^{3}+1)^{9}}) by using the chain rule and the quotient rule.