Question

hw14 the chain rule (§3.6)
score: 7/11 answered: 7/11
question 8
let $f(x)=\left(\frac{x + 3}{x + 2}\
ight)^7$
$f(x)=$
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Explanation:

Step1: Let \( u=\frac{x + 3}{x+2} \), then \( f(x)=u^{7} \)

Use the chain rule \( f^{\prime}(x)=\frac{df}{du}\cdot\frac{du}{dx} \).
First, find \( \frac{df}{du} \):
By the power rule \( \frac{d}{du}(u^{n})=nu^{n - 1} \), for \( n = 7 \), \( \frac{df}{du}=7u^{6}=7(\frac{x + 3}{x+2})^{6} \)

Step2: Find \( \frac{du}{dx} \)

Use the quotient rule \( \frac{d}{dx}(\frac{v}{w})=\frac{v^{\prime}w - vw^{\prime}}{w^{2}} \), where \( v=x + 3\), \(v^{\prime}=1\), \(w=x + 2\), \(w^{\prime}=1\)
\( \frac{du}{dx}=\frac{1\cdot(x + 2)-(x + 3)\cdot1}{(x + 2)^{2}}=\frac{x+2-(x + 3)}{(x + 2)^{2}}=\frac{x+2-x - 3}{(x + 2)^{2}}=\frac{-1}{(x + 2)^{2}} \)

Step3: Calculate \( f^{\prime}(x) \)

By the chain rule \( f^{\prime}(x)=\frac{df}{du}\cdot\frac{du}{dx} \)
\( f^{\prime}(x)=7(\frac{x + 3}{x+2})^{6}\cdot\frac{-1}{(x + 2)^{2}}=-\frac{7(x + 3)^{6}}{(x + 2)^{8}} \)

Answer:

\( -\frac{7(x + 3)^{6}}{(x + 2)^{8}} \)