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section 2.6: chain rule (homework)
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let ( f(x)=left(\frac{x + 6}{x + 2}
ight)^{6})
( f(x)=)
(\frac{6(x + 6)^{5}(-4)}{(x + 2)^{3}})

Explanation:

Step1: Identify outer - inner functions

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

Step2: Find $\frac{du}{dx}$ using quotient rule

The quotient rule states that if $u=\frac{v}{w}$, where $v=x + 6$ and $w=x + 2$, then $\frac{du}{dx}=\frac{v'w - vw'}{w^{2}}$. Here, $v'=1$ and $w'=1$. 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: Apply chain rule

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=6(\frac{x + 6}{x + 2})^{5}\cdot\frac{-4}{(x + 2)^{2}}=\frac{6(x + 6)^{5}(-4)}{(x + 2)^{3}}$.

Answer:

$\frac{6(x + 6)^{5}(-4)}{(x + 2)^{3}}$