Question

hw14 the chain rule (target c4; 53.6)
score 7/11 answered 7/11
question #
let ( f(x)=left(\frac{x + 6}{x + 1}
ight)^{8})
( f^{prime}(x)=)
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Explanation:

Step1: Let $u=\frac{x + 6}{x + 1}$, so $y = u^{8}$.

$y = u^{8}$ and $u=\frac{x + 6}{x + 1}$

Step2: First find $\frac{dy}{du}$ using the power - rule.

If $y = u^{8}$, then $\frac{dy}{du}=8u^{7}$

Step3: Then find $\frac{du}{dx}$ using the quotient - rule.

The quotient - rule states that if $u=\frac{v}{w}$, then $\frac{du}{dx}=\frac{v'w - vw'}{w^{2}}$. Here, $v=x + 6$, $v' = 1$, $w=x + 1$, $w' = 1$. So $\frac{du}{dx}=\frac{1\cdot(x + 1)-(x + 6)\cdot1}{(x + 1)^{2}}=\frac{x + 1-x - 6}{(x + 1)^{2}}=-\frac{5}{(x + 1)^{2}}$

Step4: Apply the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

$\frac{dy}{dx}=8u^{7}\cdot(-\frac{5}{(x + 1)^{2}})$

Step5: Substitute $u=\frac{x + 6}{x + 1}$ back into the expression.

$\frac{dy}{dx}=8(\frac{x + 6}{x + 1})^{7}\cdot(-\frac{5}{(x + 1)^{2}})=-\frac{40(x + 6)^{7}}{(x + 1)^{9}}$

Answer:

$-\frac{40(x + 6)^{7}}{(x + 1)^{9}}$