question 3 - of 11 step 1 of 1 02:55:32 use the product rule or quotient rule to find the derivative…

Explanation:

Step1: Identify functions for product - rule

Let $f(u)=6 + \frac{1}{u}=6+u^{-1}$ and $h(u)=u^{3}-\frac{1}{7}$.

Step2: Find derivatives of $f(u)$ and $h(u)$

The derivative of $f(u)$: $f^\prime(u)=-u^{-2}=-\frac{1}{u^{2}}$ using the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$. The derivative of $h(u)$: $h^\prime(u)=3u^{2}$ using the power - rule.

Step3: Apply the product rule

The product rule states that $(f\cdot h)^\prime=f^\prime h+fh^\prime$. So $g^\prime(u)=f^\prime(u)h(u)+f(u)h^\prime(u)$. Substitute $f(u), f^\prime(u), h(u), h^\prime(u)$: [ \begin{align*} g^\prime(u)&=-\frac{1}{u^{2}}\left(u^{3}-\frac{1}{7}\right)+\left(6 + \frac{1}{u}\right)\cdot3u^{2}\ &=-\frac{u^{3}}{u^{2}}+\frac{1}{7u^{2}}+18u^{2}+3u\ &=-u+\frac{1}{7u^{2}}+18u^{2}+3u\ &=18u^{2}+2u+\frac{1}{7u^{2}} \end{align*} ]

Answer:

$18u^{2}+2u+\frac{1}{7u^{2}}$