find the derivative.\n$\frac{d}{dx}ln(\frac{3x + 1}{3x - 1})$\n$\frac{d}{dx}ln(\frac{3x + 1}{3x - 1})=square$

Explanation:

Step1: Apply chain - rule

Let $u=\frac{3x + 1}{3x-1}$, then $\frac{d}{dx}\left[\ln\left(\frac{3x + 1}{3x-1}\right)\right]=\frac{1}{u}\cdot\frac{du}{dx}$.

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

The quotient - rule states that if $u=\frac{f(x)}{g(x)}$ where $f(x)=3x + 1$ and $g(x)=3x-1$, then $\frac{du}{dx}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g(x)^2}$. Here, $f^{\prime}(x)=3$ and $g^{\prime}(x)=3$. So, $\frac{du}{dx}=\frac{3(3x - 1)-3(3x + 1)}{(3x-1)^2}=\frac{9x-3-(9x + 3)}{(3x-1)^2}=\frac{9x-3 - 9x-3}{(3x-1)^2}=\frac{-6}{(3x-1)^2}$.

Step3: Substitute $u$ and $\frac{du}{dx}$ back

Since $u = \frac{3x + 1}{3x-1}$, we have $\frac{1}{u}\cdot\frac{du}{dx}=\frac{3x-1}{3x + 1}\cdot\frac{-6}{(3x-1)^2}=\frac{-6}{(3x + 1)(3x-1)}=\frac{-6}{9x^{2}-1}$.

Answer:

$\frac{-6}{9x^{2}-1}$