Question

let $f(x)=left(\frac{x + 6}{x + 2}
ight)^6$. $f(x)=$

Explanation:

Step1: Apply chain - rule

Let $u=\frac{x + 6}{x + 2}$, then $y = u^{6}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=6u^{5}=6(\frac{x + 6}{x + 2})^{5}$.

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

The quotient - rule states that if $u=\frac{v}{w}$, then $u^\prime=\frac{v^\prime w - vw^\prime}{w^{2}}$. Here, $v=x + 6$, $v^\prime=1$, $w=x + 2$, $w^\prime=1$. So $\frac{du}{dx}=\frac{1\cdot(x + 2)-(x + 6)\cdot1}{(x + 2)^{2}}=\frac{x + 2-x - 6}{(x + 2)^{2}}=\frac{-4}{(x + 2)^{2}}$.

Step3: Calculate $f^\prime(x)$

By the chain - rule $f^\prime(x)=\frac{dy}{du}\cdot\frac{du}{dx}=6(\frac{x + 6}{x + 2})^{5}\cdot\frac{-4}{(x + 2)^{2}}=\frac{-24(x + 6)^{5}}{(x + 2)^{7}}$.

Answer:

$\frac{-24(x + 6)^{5}}{(x + 2)^{7}}$