4. given that ( g(u)=left(\frac{u^{3}-1}{u^{3}+1}
ight)^{8} ) show that ( g^{prime}(u)=\frac{48 u^{2}left(u^{3}-1
ight)^{7}}{left(u^{3}+1
ight)^{9}} )

Question

Accepted Answer

# 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.

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QUESTION IMAGE

Question

Explanation:

Step1: Use the chain rule

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

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

Answer: