Question

let $f(x)=left(\frac{x + 2}{x + 1}
ight)^{6}$
$f(x)=$
question help: video

Explanation:

Step1: Use chain - rule and quotient - rule

Let $u=\frac{x + 2}{x + 1}$, so $f(x)=u^{6}$. By the chain - rule, $f^{\prime}(x)=\frac{d}{du}(u^{6})\cdot\frac{du}{dx}$. First, $\frac{d}{du}(u^{6}) = 6u^{5}$. Then, use the quotient - rule to find $\frac{du}{dx}$. The quotient - rule states that if $u=\frac{g(x)}{h(x)}$ where $g(x)=x + 2$ and $h(x)=x + 1$, then $\frac{du}{dx}=\frac{g^{\prime}(x)h(x)-g(x)h^{\prime}(x)}{h^{2}(x)}$.

Step2: Calculate $g^{\prime}(x)$ and $h^{\prime}(x)$

We know that $g^{\prime}(x)=1$ and $h^{\prime}(x)=1$. Then $\frac{du}{dx}=\frac{1\cdot(x + 1)-(x + 2)\cdot1}{(x + 1)^{2}}=\frac{x + 1-x - 2}{(x + 1)^{2}}=-\frac{1}{(x + 1)^{2}}$.

Step3: Substitute $u$ and $\frac{du}{dx}$ into $f^{\prime}(x)$

Since $u=\frac{x + 2}{x + 1}$ and $\frac{du}{dx}=-\frac{1}{(x + 1)^{2}}$, and $f^{\prime}(x)=6u^{5}\cdot\frac{du}{dx}$, we have $f^{\prime}(x)=6(\frac{x + 2}{x + 1})^{5}\cdot(-\frac{1}{(x + 1)^{2}})=-\frac{6(x + 2)^{5}}{(x + 1)^{7}}$.

Answer:

$-\frac{6(x + 2)^{5}}{(x + 1)^{7}}$